5.1 AREA BETWEEN CURVES

Transcription

1 3 CHAPTER 5.. Applications of the Definite Integral 5- that each of these plas in connecting the new problem to integration. B doing so, ou will see that the common thread running through so man diverse applications is the integral. 5. AREA BETWEEN CURVES 3 We initiall developed the definite integral (in Chapter ) to compute the area under a curve. In particular, let f be a continuous function defined on [a, b], where f () on[a, b]. To find the area under the curve = f () onthe interval [a, b], we begin b dividing (partitioning) [a, b] into n subintervals of equal size, = b a. The points in the partition n are then = a, = +, = + and so on. That is, i = a + i, for i =,,,...,n. a 3 5 FIGURE 5. Approimation of area f() b g() FIGURE 5. Area between two curves On each subinterval [ i, i ], we construct a rectangle of height f (c i ), for some c i [ i, i ], as indicated in Figure 5. and take the sum of the areas of the n rectangles as an approimation of the area A under the curve: A n f (c i ). i= As we take more and more rectangles, this sum approaches the eact area, which is A = lim n n f (c i ) = i= b a f () d. We now etend this notion to find the area bounded between the two curves = f () and = g() onthe interval [a, b] (see Figure 5.), where f and g are continuous and f () g()on[a, b]. We first use rectangles to approimate the area. In this case, on each subinterval [ i, i ], construct a rectangle, stretching from the lower curve = g()tothe upper curve = f (), as shown in Figure 5.3a. Referring to Figure 5.3b, the ith rectangle has height h i = f (c i ) g(c i ), for some c i [ i, i ]. So, the area of the ith rectangle is Area = length width = h i = [ f (c i ) g(c i )]. (c i, f(c i )) f() h i f(c i ) g(c i ) f() a b a c i b g() FIGURE 5.3a Approimate area g() (c i, g(c i )) FIGURE 5.3b Area of ith rectangle

2 5-3 SECTION 5... Area between Curves 33 The total area is then approimatel equal to the sum of the areas of the n indicated rectangles, n A [ f (c i ) g(c i )]. i= Finall, observe that if the limit as n eists, we will get the eact area, which we recognize as a definite integral: AREA BETWEEN TWO CURVES n b A = lim [ f (c i ) g(c i )] = [ f () g()] d. (.) n i= a REMARK. Formula (.) is valid onl when f () g() onthe interval [a, b]. In general, the area between = f () and = g() for a b is given b b f () g() d. a Notice that to evaluate this integral, ou must evaluate d [ f () g()] d on c all subintervals where f () g(), then evaluate d [g() f ()] d on all c subintervals where g() f () and finall, add the integrals together. EXAMPLE. Finding the Area between Two Curves Find the area bounded b the graphs of = 3 and = 9 (see Figure 5.) FIGURE 5. = 3 and = 9 Solution The region in Figure 5. is determined b the intersection of the two curves. The limits of integration will then correspond to the -coordinates of the points of intersection. To find the limits, we set the two functions equal and solve for.wehave 3 = 9 or = + = ( 3)( + ). Thus, the curves intersect at = and = 3. Be careful to notice from the graph which curve forms the upper boundar of the region and which one forms the lower boundar. In this case, the upper boundar is formed b = 3. So, for each fied value of, the height of a rectangle (such as the one indicated in Figure 5.) is h() = (3 ) ( 9). From (.), the area between the curves is then A = = = 3 3 [(3 ) ( 9)] d ( + ) = [ 3 3 ] [ (3) [ ( )3 3 ] 3 + ( ) ] + ( ) = 33 6.

3 3 CHAPTER 5.. Applications of the Definite Integral 5- Sometimes, a given upper or lower boundar is not defined b a single function, as in the following case of intersecting graphs. EXAMPLE. Finding the Area between Two Curves That Cross Find the area bounded b the graphs of = and = for. Solution Notice from Figure 5.5 that since the two curves intersect in the middle of the interval, we will need to compute two integrals, one on the interval where and one on the interval where.tofind the point of intersection, we solve =,sothat = or = or =±. Since = isoutside the interval of interest, the onl intersection of note is at =. From (.), the area is FIGURE 5.5 = and = A = = = [( ) ] d + ( ) d + ( 3 ) ( ) + [ ( )] d ( ) d = [ 3 ] [ 3 ] ( ) ( ) = =. In eample., the intersection point was eas to find. In eample.3, the intersection points must be approimated numericall. FIGURE 5.6 = cos and = cos EXAMPLE.3 A Case Where the Intersection Points Are Known Onl Approimatel Find the area bounded b the graphs of = cos and =. Solution The graph of = cos and = in Figure 5.6 indicates intersections at about = and =, where cos =.However, this equation cannot be solved algebraicall. Instead, we use a rootfinding method to find the approimate solutions =±.83. [For instance, ou can use Newton s method to find values of for which f () = cos =.] From the graph, we can see that between these two -values, cos and so, the desired area is given b.83 A (cos ) d = [sin 3 ] = sin.83 3 (.83)3 [sin(.83) 3 ] (.83) Note that we have approimated both the limits of integration and the final calculations. Finding the area of some regions ma require breaking the region up into several pieces, each having different upper and/or lower boundaries. FIGURE 5.7a = and = EXAMPLE. The Area of a Region Determined b Three Curves Find the area bounded b the graphs of =, = and =. Solution Asketch of the three defining curves is shown in Figure 5.7a. Notice that the top boundar of the region is the curve = on the first portion of the interval and the

4 5-5 SECTION 5... Area between Curves 35 A A FIGURE 5.7b = and = line = on the second portion. To determine the point of intersection, we solve = or = + = ( + )( ). Since = istothe left of the -ais, the intersection we seek occurs at =. We then break the region into two pieces, as shown in Figure 5.7b and find the area of each separatel. The total area is then A = A + A = ( ) d + = [ ] = 5 6. [( ) ] d Although it was certainl not difficult to break up the region in eample. into several pieces, we want to suggest an alternative that will prove to be surprisingl useful. Notice that if ou turn the page sidewas, Figure 5.7a will look like a region with a single curve determining each of the upper and lower boundaries. Of course, b turning the page sidewas, ou are essentiall reversing the roles of and, which is the ke to computing the area of this tpe of region: treat the left and right boundaries of the region as functions of. More generall, for two continuous functions, f and g of, where f () g() for all on the interval c d, tofind the area bounded between the two curves = f () and = g(), we first partition the interval [c, d] into n equal subintervals, each of width = d c (see Figure 5.8a). We denote the points in the partition b = c, n = +, = + and so on. That is, i = c + i, for i =,,,...,n. g() d g() d w i f(c i ) g(c i ) (g(c i ), c i ) c (f(c i ), c i ) i c f() c f() FIGURE 5.8a Area between = g() and = f () FIGURE 5.8b Area of ith rectangle On each subinterval [ i, i ] (for i =,,...,n), we then construct a rectangle of width w i = [ f (c i ) g(c i )], for some c i [ i, i ], as shown in Figure 5.8b. The area of the ith rectangle is given b Area = length width = [ f (c i ) g(c i )]. The total area between the two curves is then given approimatel b n A [ f (c i ) g(c i )]. i=

5 36 CHAPTER 5.. Applications of the Definite Integral 5-6 We get the eact area b taking the limit as n and recognizing the limit as a definite integral. We have Area between two curves A = lim n n [ f (c i ) g(c i )] = i= d c [ f () g()] d. (.) FIGURE 5.9 = and = EXAMPLE.5 An Area Computed b Integrating with Respect to Repeat eample., but integrate with respect to instead. Solution The area bounded b the graphs of =, = and = isshown in Figure 5.9. Notice that the left-hand boundar of the region is formed b the graph of = and the right-hand boundar of the region is formed b the line = and so, a single integral with respect to will suffice. To write these left- and right-hand boundaries as functions of, solve the equation = for. Weget = (since onl the right half of the parabola forms the left boundar). Likewise, = is equivalent to =. Finall, these curves intersect where =. Squaring both sides gives us = ( ) = + or = 5 + = ( )( ). So, the curves intersect at = and =. From Figure 5.9, it is clear that = isthe solution we need. (What does the solution = correspond to?) From (.), the area is given b A = [( ) [ ] d = ] 3 3/ = 3 = 5 6. FIGURE 5. = and = EXAMPLE.6 The Area of a Region Bounded b Functions of Find the area bounded b the graphs of = and =. Solution Of course, the graphs are parabolas opening to the right and left, respectivel, as indicated in Figure 5.. Notice that it s easiest to compute this area b integrating with respect to, since integrating with respect to would require us to break the region into two pieces. We must first find the two intersections of the curves. These occur where =,or =, so that =±. On the interval [, ], notice that (since the curve = stas to the right of the curve = ). So, from (.), the area is given b A = = [( ) ] d = [ 3 3 ] = ( 3 ) ( ) d ( + ) = In collisions between a tennis racket and ball, the ball changes shape, first compressing and then epanding. Let represent how far the ball is compressed, where m, and let f () represent the force eerted on the ball b the racket. Then, the energ transferred is proportional to the area under the curve = f (). Suppose that f c ()isthe force during

6 5-7 SECTION 5... Area between Curves 37 compression of the ball and f e () isthe force during epansion of the ball. Energ is transferred to the ball during compression and transferred awa from the ball during epansion, so that the energ lost b the ball in the collision (due to friction) is proportional to m [ f c() f e ()] d. The percentage of energ lost in the collision is then given b m [ f c() f e ()] d m. f c () d EXAMPLE.7 Estimating the Energ Lost b a Tennis Ball Suppose that test measurements provide the following data on the collision of a tennis ball with a racket. Estimate the percentage of energ lost in the collision. (in.) f c () (lb) f e () (lb) f c () f e () Solution The data are plotted in Figure 5., connected b line segments. We need to estimate the area between the curves and the area under the top curve. Since we don t have a formula for either function, we must use a numerical method such as Simpson s Rule. For. f c () d,weget. f c () d. [ + (5) + (5) + (9) + 6] =. 3 To use Simpson s Rule to approimate. [ f c () f e ()] d,weneed a table of function values for f c () f e (). Subtraction gives us...3. FIGURE 5. Force eerted on a tennis ball from which Simpson s Rule gives us f c () f e () [ f c () f e ()] d. 3 [ + () + () + () + ] = The percentage of energ lost is then (6./3) = 8.9%. With over 9% of its energ retained in the collision, this is a livel tennis ball. BEYOND FORMULAS In eample.5, we viewed the given graphs as functions of and set up the area as an integral of. This idea indicates the direction that much of the rest of the course takes. The derivative and integral remain the two most important operations, but we diversif our options for working with them, often b changing variables. The fleible thinking that this promotes is ke in calculus, as well as in other areas of mathematics and science. We develop some general techniques and often the first task in solving an application problem is to make the technique fit the problem at hand. In industr, do ou epect that all problems have been encountered before or that new problems arise each da?

7 38 CHAPTER 5.. Applications of the Definite Integral 5-8 EXERCISES 5. WRITING EXERCISES. Suppose the functions f and g satisf f () g() for all in the interval [a, b]. Eplain in terms of the areas A = b f () d and A a = b g() d wh the area between the curves = f () and = g() is given b a f () g() d. b a. Suppose the functions f and g satisf f () g() for all in the interval [a, b]. Eplain in terms of the areas A = b f () d and A a = b g() d wh the area between the curves = f () and = g() is given b a f () g() d. b a 3. Suppose that the speeds of racing cars A and B are v A (t) and v B (t) mph, respectivel. If v A (t) v B (t) for all t, v A () = v B () and the race lasts from t = tot = hours, eplain wh car A will win the race b [v A(t) v B (t)] dt miles.. Suppose that the speeds of racing cars A and B are v A (t) and v B (t) mph, respectivel. If v A (t) v B (t) for t.5 and. t.6 and v B (t) v A (t) for.5 t. and.6 t, describe the difference between v A(t) v B (t) dt and [v A(t) v B (t)] dt. Which integral will tell ou which car wins the race? In eercises, find the area between the curves on the given interval.. = 3, =, 3. = cos, = +, 3. = e, =,. = e, =, In eercises 5, sketch and find the area of the region determined b the intersections of the curves. 5. =, = 7 6. =, = 7. = 3, = =, = 9. = e, =. = +, =. = 5 +, =. = sin ( π), = cos In eercises 3 8, sketch and estimate the area determined b the intersections of the curves. 3. = e, =. =, = 5. = sin, = 6. = cos, = 7. =, = + 8. = ln, = In eercises 9 6, sketch and find the area of the region bounded b the given curves. Choose the variable of integration so that the area is written as a single integral. 9. =, =, =. = ( > ), = 3, =. =, =, =. = 3, = + 3. =, =, = 6, =. =, = 5. = e, = e, = 6. = ln, = +, = 7. The average value of a function f () onthe interval [a, b] is A= b f () d. Compute the average value of f () = b a a on [, 3] and show that the area above = A and below = f () equals the area below = A and above = f (). 8. Prove that the result of eercise 7 is alwas true b showing that b [ f () A] d =. a 9. The United States oil consumption for the ears was approimatel equal to f (t) = 6.e.7t million barrels per ear, where t = corresponds to 97. Following an oil shortage in 97, the countr s consumption changed and was better modeled b g(t) =.3e.(t ) million barrels per ear, for t. Show that f () g() and eplain what this number represents. Compute the area between f (t) and g(t) for t. Use this number to estimate the number of barrels of oil saved b Americans reduced oil consumption from 97 to 98. Million barrels per ear 3 f(t) g(t) 6 8 (97) (98) t

8 5-9 SECTION 5... Area between Curves Suppose that a nation s fuelwood consumption is given b 76e.3t m 3 /r and new tree growth is 5 6e.9t m 3 /r. Compute and interpret the area between the curves for t. 3. Suppose that the birthrate for a certain population is b(t) = e.t million people per ear and the death rate for the same population is d(t) = e.t million people per ear. Show that b(t) d(t) for t and eplain wh the area between the curves represents the increase in population. Compute the increase in population for t. 3. Suppose that the birthrate for a population is b(t) = e.t million people per ear and the death rate for the same population is d(t) = 3e.t million people per ear. Find the intersection T of the curves (T > ). Interpret the area between the curves for t T and the area between the curves for T t 3. Compute the net change in population for t In collisions between a ball and a striking object (e.g., a baseball bat or tennis racket), the ball changes shape, first compressing and then epanding. If represents the change in diameter of the ball (e.g., in inches) for m and f () represents the force between the ball and striking object (e.g., in pounds), then the area under the curve = f () isproportional to the energ transferred. Suppose that f c () isthe force during compression and f e () is the force during epansion. Eplain wh m [ f c() f e ()] d is proportional to the energ lost b the ball (due to friction) and thus m [ f c() f e ()] d/ m f c () d is the proportion of energ lost in the collision. For a baseball and bat, reasonable values are shown (see Adair s book The Phsics of Baseball): (in.)...3. f c () (lb) f e () (lb) 7 75 Use Simpson s Rule to estimate the proportion of energ retained b the baseball. 3. Using the same notation as in eercise 33, values for the force f c () during compression and force f e () during epansion of a golf ball are given b of its etension determines the loss of energ (see the chapter introduction). Suppose that is the etension of the tendon, f s () isthe force during stretching of the tendon and f r () is the force during recoil of the tendon. The data given are for a hind leg tendon of a wallab (see Aleander s book Eploring Biomechanics): (mm) f s () (N) f r () (N) 3 7 Use Simpson s Rule to estimate the proportion of energ returned b the tendon. 36. The arch of a human foot acts like a spring during walking and jumping, storing energ as the foot stretches (i.e., the arch flattens) and returning energ as the foot recoils. In the data, is the vertical displacement of the arch, f s ()isthe force on the foot during stretching and f r () isthe force during recoil (see Aleander s book Eploring Biomechanics): (mm) f s () (N) f r () (N) Use Simpson s Rule to estimate the proportion of energ returned b the arch. 37. The velocities of two runners are given b f (t) = mph and g(t) = sin t mph. Find and interpret the integrals π [ f (t) g(t)] dt and π [ f (t) g(t)] dt. 38. The velocities of two racing cars A and B are given b f (t) = ( e t ) mph and g(t) = t mph, respectivel. The cars start at the same place at time t =. Estimate (a) the largest lead for car A and (b) the time at which car B catches up. In eercises 39 and, the graph shows the rate of flow of water in gallons per hour into and out of a tank. Assuming that the tank starts with gallons, estimate the amount of water in the tank at hours,, 3, and 5 and sketch a graph of the amount of water in the tank. (in.) f c () (lb) 5 8 f e () (lb) Use Simpson s Rule to estimate the proportion of energ retained b the golf ball. 35. Much like the compression and epansion of a ball discussed in eercises 33 and 3, the force eerted b a tendon as a function 39. In 8 6 Out 3 5 t

9 CHAPTER 5.. Applications of the Definite Integral The graph shows the suppl and demand curves for a product. The point of intersection (q, p )gives the equilibrium quantit and equilibrium price for the product. The consumer surplus is defined to be CS = q D(q) dq p q. Shade in In Out the area of the graph that represents the consumer surplus, and compute this in the case where D(q) = q and S(t) = + q + q. p S(q) D(q) Repeat eercise for the producer surplus defined b PS = p q q S(q) dq. 3. Let C ()bethe marginal cost of producing thousand copies of an item and let R ()bethe marginal revenue from the sale of that item, with graphs as shown. Assume that R () = C () at = and = 5. Interpret the area between the curves for each interval: (a), (b) 5, (c) 5 and (d) q C'() R'() A basic principle of economics is that profit is maimized when marginal cost equals marginal revenue. At which intersection is profit maimized in eercise 3? Eplain our answer. In terms of profit, what does the other intersection point represent? 5. Suppose that the parabola = a + b + c and the line = m + n intersect at = A and = B with A < B. Show that the area between the curves equals a 6 (B A)3. (Hint: Use A and B to rewrite the integrand and then integrate.) t 6. Suppose that the cubic = a 3 + b + c + d and the parabola = k + m + n intersect at = A and = B with B repeated (that is, the curves are tangent at B; see the figure). Show that the area between the curves equals a (B A) A 7. Consider two parabolas, each of which has its verte at =, but with different concavities. Let h be the difference in -coordinates of the vertices, and let w be the difference in the -coordinates of the intersection points. Show that the area between the curves is 3 hw. w 6 h 8. Show that for an constant m, the area between = and = m is 6 (m + 8) 3/. Find the minimum such area. 9. For = as shown, find the value of L such that A = A A A B = L 5. For = and = k as shown, find k such that A = A A A = k

10 5- SECTION 5... Volume: Slicing, Disks and Washers EXPLORATORY EXERCISES. At this stage, ou can compute the area of an simple planar region. For a general figure bounded on the left b a function = l(), on the right b a function = r(), on top b a function = t() and on the bottom b a function = b(), write the area as a sum of integrals. (Hint: Divide the region into subregions whose area can be written as the integral of r() l() or t() b().). Find the area between = and = m for an constant m >. Without doing further calculations, use this area to find the area between = and = m. 3. For >, let f () bethe area between = and = sin t for t. Without calculating f (), find as man relationships as possible between the graphical features (zeros, etrema, inflection points) of = f () and the graphical features of = sin. 5. VOLUME: SLICING, DISKS AND WASHERS As we shall see throughout this chapter, the integral is an amazingl versatile tool. In section 5., we used definite integrals to compute area. In this section, we use integrals to compute the volume of a three-dimensional solid. We begin with a simple problem. When designing a building, architects must perform numerous detailed calculations. For instance, in order to analze a building s heating and cooling sstems, engineers must calculate the volume of air being processed. FIGURE 5.a FIGURE 5.b You alread know how to compute some volumes. For instance, the building shown in Figure 5.a is essentiall a rectangular bo, whose volume is given b lwh, where l is the length, w is the width and h is the height. The right circular clinders seen in the building in Figure 5.b have volume given b πr h, where h is the height and r is the radius of the circular cross section. Notice in each case that the building has a familiar cross section (a rectangle in Figure 5.a and a circle in Figure 5.b) that is etended verticall. We call an such solid a clinder (an solid whose cross sections perpendicular to some ais running through the solid are all the same). Now, notice the connection between the volume

11 5- SECTION 5... Volume: Slicing, Disks and Washers EXPLORATORY EXERCISES. At this stage, ou can compute the area of an simple planar region. For a general figure bounded on the left b a function = l(), on the right b a function = r(), on top b a function = t() and on the bottom b a function = b(), write the area as a sum of integrals. (Hint: Divide the region into subregions whose area can be written as the integral of r() l() or t() b().). Find the area between = and = m for an constant m >. Without doing further calculations, use this area to find the area between = and = m. 3. For >, let f () bethe area between = and = sin t for t. Without calculating f (), find as man relationships as possible between the graphical features (zeros, etrema, inflection points) of = f () and the graphical features of = sin. 5. VOLUME: SLICING, DISKS AND WASHERS As we shall see throughout this chapter, the integral is an amazingl versatile tool. In section 5., we used definite integrals to compute area. In this section, we use integrals to compute the volume of a three-dimensional solid. We begin with a simple problem. When designing a building, architects must perform numerous detailed calculations. For instance, in order to analze a building s heating and cooling sstems, engineers must calculate the volume of air being processed. FIGURE 5.a FIGURE 5.b You alread know how to compute some volumes. For instance, the building shown in Figure 5.a is essentiall a rectangular bo, whose volume is given b lwh, where l is the length, w is the width and h is the height. The right circular clinders seen in the building in Figure 5.b have volume given b πr h, where h is the height and r is the radius of the circular cross section. Notice in each case that the building has a familiar cross section (a rectangle in Figure 5.a and a circle in Figure 5.b) that is etended verticall. We call an such solid a clinder (an solid whose cross sections perpendicular to some ais running through the solid are all the same). Now, notice the connection between the volume

12 CHAPTER 5.. Applications of the Definite Integral 5- formulas for these two clinders. The volume of a right circular clinder is Likewise, in the case of a bo, we have V = (πr ) }{{} h cross-sectional area. }{{} height HISTORICAL NOTES Archimedes (ca. 87 B.C.) A Greek mathematician and scientist who was among the first to derive formulas for volumes and areas. Archimedes is known for discovering the basic laws of hdrostatics (he reportedl leapt from his bathtub, shouting Eureka! and ran into the streets to share his discover) and levers ( Give me a place to stand on and I can move the earth. ). An ingenious engineer, his catapults, grappling cranes and reflecting mirrors terrorized a massive Roman arm that eventuall conquered his hometown of Sracuse. On the da the Romans 3-ear siege ended, Archimedes was studing diagrams drawn in the dirt when a soldier tried to arrest him. Archimedes last words were reportedl, Do not disturb m circles. V = (length width) height. }{{} cross-sectional area In general, the volume of an clinder is found b Volumes b Slicing V = (cross-sectional area) (height). If either the cross-sectional area or width of a solid is not constant, we will need to modif our approach somewhat. For instance, pramids and domes do not have constant crosssectional area, as seen in Figures 5.3a and 5.3b. Since we don t know how to find the volume, we take the approach we ve used a number of times now: first approimate the volume and then improve the approimation. For an solid that etends from = a to = b, westart b partitioning the interval [a, b] onthe -ais into n subintervals, each of width = b a. Asusual, we denote n = a, = a + and so on, so that i = a + i, for i =,,,...,n. FIGURE 5.3a Pramid Arena in Memphis FIGURE 5.3b U.S. Capitol Building

13 5-3 SECTION 5... Volume: Slicing, Disks and Washers 3 i th slice of solid i th approimating clinder i i i i D FIGURE 5.a Sliced solid FIGURE 5.b ith slice of solid FIGURE 5.c ith approimating clinder We then slice the solid perpendicular to the -ais at each of the (n ) points,,..., n (see Figure 5.a). We net need to approimate the volume of each of the n slices. First, notice that if n is large, then each slice of the solid will be thin and so, the cross-sectional area is nearl constant throughout an given slice. Suppose that the area of the cross section corresponding to an particular value of is given b A(). Observe that the slice between = i and = i is nearl a clinder (see Figure 5.b). So, for an point c i in the interval [ i, i ], the area of the cross sections on that interval are all approimatel A(c i ). The volume V i of the ith slice is approimatel the volume of the clinder ling along the interval [ i, i ], with constant cross-sectional area A(c i ) (see Figure 5.c), so that V i A(c i ) }{{} cross-sectional area, }{{} width where is the width of the slice. Repeating this process for each of the n slices, we find that the total volume V of the solid is approimatel V n A(c i ). i= Notice that as the number of slices increases, the volume approimation should improve and we get the eact volume b computing V = lim n n A(c i ), assuming the limit eists. You should recognize this limit as the definite integral i= Volume of a solid with cross-sectional area A() V = b a A() d. (.)

14 CHAPTER 5.. Applications of the Definite Integral 5- REMARK. We use the same process followed here to derive man important formulas. In each case, we divide an object into n smaller pieces, approimate the quantit of interest for each of the small pieces, sum the approimations and then take a limit, ultimatel recognizing that we have derived a definite integral. For this reason, it is essential that ou understand the concept behind formula (.). Memorization will not do this for ou. However, if ou understand how the various pieces of this puzzle fit together, then the rest of this chapter should fall into place for ou nicel. 6 feet The Pramid Arena in Memphis EXAMPLE. Computing Volume from Cross-Sectional Areas The Pramid Arena in Memphis (pictured in the margin) has a square base of side approimatel 6 feet and a height of approimatel 3 feet. Find the volume of the pramid with these measurements. Solution To use (.), we need to have a formula for the cross-sectional area. The square horizontal cross sections of the pramid make this eas, but we need a formula for the size of the square at each height. Orient the -ais upward through the point at the top of the pramid. At =, the cross section is a square of side 6 feet. At = 3, the cross section can be thought of as a square of side feet. If f () represents the side length of the square cross section at height, weknow that f () = 6, f (3) = and f ()isalinear function. (Think about this; the sides of the pramid do not curve.) The slope of the line is m = 6 3 = 5 and we use the -intercept of 6 to get 8 f () = Since this is the length of a side of a square, the cross-sectional area is simpl the square of this quantit. Then from (.), we have 3 3 ( V = A() d = 5 ) d. Observe that we can evaluate this integral b substitution, b taking u = , so that du = 5 d. This gives us 8 3 ( V = 5 ) d = 8 u du 5 6 = 8 6 u du = 8 u = 38,, ft. In eample., we knew how to compute the cross-sectional area eactl. In man important applications, the cross-sectional area is not known eactl, but must be approimated using measurements. In such cases, we can still find the volume (approimatel), but we ll need to use numerical integration. EXAMPLE. Estimating Volume from Cross-Sectional Data In medical imaging, such as CT (computerized tomograph) and MRI (magnetic resonance imaging) processes, numerous measurements are taken and processed b a

15 5-5 SECTION 5... Volume: Slicing, Disks and Washers 5 computer to construct a three-dimensional image of the tissue the phsician wishes to stud. The process is similar to the slicing process we have used to find the volume of a solid. In this case, however, mathematical representations of various slices of the tissue are combined to produce a three-dimensional image that can be displaed and sliced back apart for a phsician to determine the health of the tissue. Suppose that an MRI scan indicates that the cross-sectional areas of adjacent slices of a tumor are given b the values in the table. (cm) A() (cm ) Estimate the volume of the tumor FIGURE 5.5a FIGURE 5.5b Solution To find the volume of the tumor, we would compute [following (.)] V = A() d, ecept that we onl know A() atafinite number of points. Does this ring an bells? Notice that we can use Simpson s Rule (see section.7) with =. toestimate the value of this integral: V = b a 3n A() d [ ] A() + A(.) + A(.) + A(.3) + A(.) + A(.5) + A(.6) + A(.7) + A(.8) + A(.9) + A() =. ( ) 3 =.9667 cm 3. We now turn to the problem of finding the volume of the dome in Figure 5.3b. Since the horizontal cross sections are circles, we need onl to determine the radius of each circle. Starting with the vertical cross section of the dome in Figure 5.5a, we rotate the dome to get Figure 5.5b. Here, observe that the radius equals the height of the function defining the outline of the dome. We use this insight in eample.3. EXAMPLE.3 Computing the Volume of a Dome 5 Suppose that a dome has circular cross sections, with outline =± (9 ) for 9 (in units of feet, this gives dimensions similar to the Capitol Dome in Figure 5.3b. A graph of this sidewas parabola is shown in Figure 5.5c). Find the volume of the dome. Solution Following our previous discussion, we know that the radius of a circular cross section corresponds to the distance from the -ais to the top half of the parabola 5 = (9 ). That is, the radius is given b r() = 5 (9 ). Each cross section is a circle with this radius, so the cross-sectional areas are given b 5 FIGURE 5.5c A() = π ( ) 5 (9 ),

16 6 CHAPTER 5.. Applications of the Definite Integral 5-6 for 9. The volume is then given b V = = π 9 A() d = 9 [ 5 5 ] 9 π ( ) 5 (9 ) d = = 9,5π 86,78 ft 3. 9 π (5 5 ) d Observe that an alternative wa of stating the problem in eample.3 is to sa: Find 5 the volume formed b revolving the region bounded b the curve = (9 ) and the -ais, for 9, about the -ais. Eample.3 can be generalized to the method of disks used to compute the volume of a solid formed b revolving a two-dimensional region about a vertical or horizontal line. We consider this general method net. The Method of Disks Suppose that f () and f is continuous on the interval [a, b]. Take the region bounded b the curve = f () and the -ais, for a b, and revolve it about the -ais, generating a solid (see Figures 5.6a and 5.6b). We can find the volume of this solid b slicing it perpendicular to the -ais and recognizing that each cross section is a circular disk of radius r = f () (see Figure 5.6b). From (.), we then have that the volume of the solid is Volume of a solid of revolution b (Method of disks) V = π[ f ()] d. (.) }{{} a cross-sectional area = πr f() f() a b a b Circular cross sections FIGURE 5.6a = f () FIGURE 5.6b Solid of revolution Since the cross sections of such a solid of revolution are all disks, we refer to this method of finding volume as the method of disks. EXAMPLE. Using the Method of Disks to Compute Volume Revolve the region under the curve = on the interval [, ] about the -ais and find the volume of the resulting solid of revolution.

17 5-7 SECTION 5... Volume: Slicing, Disks and Washers 7 r r 3 FIGURE 5.7a = FIGURE 5.7b Solid of revolution Solution It s critical to draw a picture of the region and the solid of revolution, so that ou get a clear idea of the radii of the circular cross sections. You can see from Figures 5.7a and 5.7b that the radius of each cross section is given b r =. From (.), we then get the volume: V = π [ ] d = π d= π }{{} = 8π. cross-sectional area = πr In a similar wa, suppose that g() and g is continuous on the interval [c, d]. Then, revolving the region bounded b the curve = g() and the -ais, for c d, about the -ais generates a solid (see Figures 5.8a and 5.8b). Once again, notice from Figure 5.8b that the cross sections of the resulting solid of revolution are circular disks of radius r = g(). All that has changed here is that we have interchanged the roles of the variables and. The volume of the solid is then given b Volume of a solid of revolution (Method of disks) V = d π[g()] d. (.3) }{{} c cross-sectional area = πr d d g() g() c c FIGURE 5.8a Revolve about the -ais FIGURE 5.8b Solid of revolution

18 8 CHAPTER 5.. Applications of the Definite Integral 5-8 REMARK. When using the method of disks, the variable of integration depends solel on the ais about which ou revolve the two-dimensional region: revolving about the -ais requires integration with respect to, while revolving about the -ais requires integration with respect to. This is easil determined b looking at a sketch of the solid. Don t make the mistake of simpl looking for what ou can plug in where. This is a recipe for disaster, for the rest of this chapter will require ou to make similar choices, each based on distinctive requirements of the problem at hand. EXAMPLE.5 Using the Method of Disks with as the Independent Variable Find the volume of the solid resulting from revolving the region bounded b the curves = and = from = to = 3 about the -ais. Solution You will find a graph of the curve in Figure 5.9a and of the solid in Figure 5.9b. 3 FIGURE 5.9a = FIGURE 5.9b Solid of revolution Notice from Figures 5.9a and 5.9b that the radius of an of the circular cross sections is given b. So, we must solve the equation = for,toget =. Since the surface etends from = to =, the volume is given b (.3) to be V = π ( ) d = }{{} πr = π [ ] = π π( ) d [ ( (6 8) )] = 9π. The Method of Washers One complication that occurs in computing volumes is that the solid ma have a cavit or hole in it. Another occurs when a region is revolved about a line other than the -ais

19 5-9 SECTION 5... Volume: Slicing, Disks and Washers 9 or the -ais. Neither case will present ou with an significant difficulties, if ou look carefull at the figures. We illustrate these ideas in eamples.6 and.7. EXAMPLE.6 Computing Volumes of Solids with and without Cavities Let R be the region bounded b the graphs of =, = and =. Compute the volume of the solid formed b revolving R about (a) the -ais, (b) the -ais and (c) the line =. Solution (a) The region R is shown in Figure 5.a and the solid formed b revolving it about the -ais is shown in Figure 5.b. Notice that this part of the problem is similar to eample.5. R Outer radius: R FIGURE 5.a = Inner radius: FIGURE 5.a = FIGURE 5.b Solid of revolution From (.3), the volume is given b ( ) V = π d = π }{{} = π. πr (b) Revolving the region R about the -ais produces a cavit in the middle of the solid. See Figure 5.a for a graph of the region R and Figure 5.b for a picture of the solid. Our strateg is to compute the volume of the outside of the object (as if it were filled in) and then subtract the volume of the cavit. Before diving into a computation, be sure to visualize the geometr behind this. For the present eample, the outside surface of the solid is formed b revolving the line = about the -ais. The cavit is formed b revolving the curve = about the -ais. Look carefull at Figures 5.a and 5.b and make certain that ou see this. The outer radius, r O,isthe distance from the -ais to the line =, or r O =. The inner radius, r I,isthe distance from the -ais to the curve =,orr I =. Appling (.) twice, we see that the volume is given b ( ) V = π() d π }{{} d π (outer radius) }{{} = π ( 6 π (inner radius) ) d = π ( 8 ) 5 ( = π 3 ) = π.

20 5 CHAPTER 5.. Applications of the Definite Integral 5- The solid in part (c) is formed b revolving the region R about the line =. This time, the rotation produces a washer-like solid with a clindrical hole in the middle. The region R is shown in Figure 5.a and the solid is shown in Figure 5.b. FIGURE 5.b Solid with cavit Washer-shaped cross sections Inner radius: R Outer radius: FIGURE 5.a Revolve about = FIGURE 5.b Solid of revolution The volume is computed in the same wa as in part (b), b subtracting the volume of the cavit from the volume of the outside solid. From Figures 5.a and 5.b, notice that the radius of the outer surface is the distance from the line = tothe curve =. This outer radius is then r O =. The radius of the inner hole is the distance from the line = tothe line =. This inner radius is then r I = =. From (.), the volume is given b V = = π = π π ( ) }{{} π(outer radius) [( + ( ) 8 d 6 π( ) }{{} π(inner radius) d ) ] [ d = π ] 8 5 = 56 5 π. In parts (b) and (c) of eample.6, the volume was computed b subtracting an inner volume from an outer volume in order to compensate for a cavit inside the solid. This technique is a slight generalization of the method of disks and is referred to as the method of washers, since the cross sections of the solids look like washers.

21 5- SECTION 5... Volume: Slicing, Disks and Washers 5 EXAMPLE.7 Revolving a Region about Different Lines Let R be the region bounded b = and =. Find the volume of the solids obtained b revolving R about each of the following: (a) the -ais, (b) the line = 3, (c) the line = 7 and (d) the line = 3. Solution For part (a), we draw the region R in Figure 5.3a and the solid of revolution in Figure 5.3b. Radius R R r O 3 ri 3 FIGURE 5.3a Revolve about -ais FIGURE 5.3b Solid of revolution FIGURE 5.a Revolve about = 3 From Figure 5.3b, notice that each cross section of the solid is a circular disk, whose radius is simpl. Solving for, weget =, where we have selected to be positive, since in this contet, represents a distance. From (.3), the volume of the solid of revolution is given b V = ( ) π d = π } {{ } π(radius) ( ) d = π [ ] = 8π. 3 For part (b), we have sketched the region R in Figure 5.a and the solid of revolution in Figure 5.b. Notice from Figure 5.b that the cross sections of the solid are shaped like washers and the outer radius r O is the distance from the ais of revolution = 3tothe curve =. That is, r O = ( 3) = ( ) ( 3) = 7, while the inner radius is the distance from the -ais to the line = 3. That is, From (.), the volume is r I = ( 3) = 3. FIGURE 5.b Solid of revolution V = π(7 ) }{{} d π(outer radius) π(3) }{{} π(inner radius) d = 7 5 π, where we have left the details of the computation as an eercise.

22 5 CHAPTER 5.. Applications of the Definite Integral r I r O R FIGURE 5.5a Revolve about = 7 FIGURE 5.5b Solid of revolution Part (c) (revolving about the line = 7) is ver similar to part (b). You can see the region R in Figure 5.5a and the solid in Figure 5.5b. The cross sections of the solid are again shaped like washers, but this time, the outer radius is the distance from the line = 7tothe-ais, that is, r O = 7. The inner radius is the distance from the line = 7tothe curve =, From (.), the volume of the solid is then V = r I = 7 ( ) = 3 +. π(7) }{{} π(outer radius) d π(3 + ) }{{} d = π(inner radius) π, where we again leave the details of the calculation as an eercise. Finall, for part (d) (revolving about the line = 3), we show the region R in Figure 5.6a and the solid of revolution in Figure 5.6b. In this case, the cross sections of the solid are washers, but the inner and outer radii are a bit trickier to determine than in the previous parts. The outer radius is the distance between the line = 3 and the left half of the parabola, while the inner radius is the distance between the line = 3 and the right half of the parabola. The parabola is given b =,sothat =±. Notice that = corresponds to the right half of the parabola, while = describes the left half of the parabola. This gives us r I = 3 and r O = 3 ( ) = 3 +.

23 5-3 SECTION 5... Volume: Slicing, Disks and Washers R r I r O FIGURE 5.6a Revolve about = 3 FIGURE 5.6b Solid of revolution Consequentl, we get the volume ( V = π 3 + ) d } {{ } π(outer radius) ( π 3 ) d = 6π, } {{ } π(inner radius) where we leave the details of this rather mess calculation to ou. In section 5.3, we present an alternative method for finding the volume of a solid of revolution that, for the present problem, will produce much simpler integrals. REMARK.3 You will be most successful in finding volumes of solids of revolution if ou draw reasonable figures and label them carefull. Don t simpl look for what to plug in where. You onl need to keep in mind how to find the area of a cross section of the solid. Integration does the rest. EXERCISES 5. WRITING EXERCISES. Discuss the relationships (e.g., perpendicular or parallel) to the -ais and -ais of the disks in eamples. and.5. Eplain how this relationship enables ou to correctl determine the variable of integration.. The methods of disks and washers were developed separatel in the tet, but each is a special case of the general volume formula. Discuss the advantages of learning separate formulas versus deriving each eample separatel from the general formula. For eample, would ou prefer to learn the etra formulas or have to work each problem from basic principles? How man formulas is too man to learn? 3. To find the area of a triangle of the form in section 5., eplain wh ou would use -integration. In this section, would it be easier to compute the volume of this triangle revolved about the -ais or -ais? Eplain our preference.

24 5 CHAPTER 5.. Applications of the Definite Integral 5-. In part (a) of eample.7, Figure 5.3a etends from = to =,but we used as the radius. Eplain wh this is the correct radius and not. In eercises, find the volume of the solid with cross-sectional area A().. A() = +, 3. A() = e., 3. A() = π( ),. A() = ( + ), In eercises 5, set up an integral and compute the volume. 5. The outline of a dome is given b = 6 6 for 6 6 (units of feet), with circular cross-sections perpendicular to the -ais. Find its volume. 6. Find the volume of a pramid of height 6 feet that has a square base of side 3 feet. These dimensions are half those of the pramid in eample.. How does the volume compare? 7. The great pramid at Gizeh is 5 feet high, rising from a square base of side 75 feet. Compute its volume using integration. Does our answer agree with the geometric formula?. A potter jar has circular cross sections of radius + sin inches for π. Sketch a picture of the jar and compute its volume.. A potter jar has circular cross sections of radius sin inches for π. Sketch a picture of the jar and compute its volume. 3. Suppose an MRI scan indicates that cross-sectional areas of adjacent slices of a tumor are as given in the table. Use Simpson s Rule to estimate the volume. (cm) A() (cm ) (cm) A() (cm ) Suppose an MRI scan indicates that cross-sectional areas of adjacent slices of a tumor are as given in the table. Use Simpson s Rule to estimate the volume. (cm) A() (cm ) Estimate the volume from the cross-sectional areas. (ft) A() (ft ) Estimate the volume from the cross-sectional areas. 5 feet 75 feet 8. Suppose that instead of completing a pramid, the builders at Gizeh had stopped at height 5 feet (with a square plateau top of side 375 feet). Compute the volume of this structure. Eplain wh the volume is greater than half the volume of the pramid in eercise A church steeple is 3 feet tall with square cross sections. The square at the base has side 3 feet, the square at the top has side 6 inches and the side varies linearl in between. Compute the volume.. A house attic has rectangular cross sections parallel to the ground and triangular cross sections perpendicular to the ground. The rectangle is 3 feet b 6 feet at the bottom of the attic and the triangles have base 3 feet and height feet. Compute the volume of the attic. (m) A() (m ) (m) A() (m ).... In eercises 7, compute the volume of the solid formed b revolving the given region about the given line. 7. Region bounded b =, = and = about (a) the -ais; (b) = 3 8. Region bounded b =, = and = about (a) the -ais; (b) = 9. Region bounded b =, = and = about (a) the -ais; (b) =. Region bounded b =, = and = about (a) the -ais; (b) = In eercises, a solid is formed b revolving the given region about the given line. Compute the volume eactl if possible and estimate if necessar.. Region bounded b = e, =, = and = about (a) the -ais; (b) =. Region bounded b = sec, =, = π/ and = π/ about (a) = ; (b) the -ais

25 5-5 SECTION 5... Volume: Slicing, Disks and Washers Region bounded b =, the -ais and = about + (a) the -ais; (b) = 3. Region bounded b = e and = about (a) the -ais; (b) = 5. Let R be the region bounded b = 3, the -ais and the -ais. Compute the volume of the solid formed b revolving R about the given line. (a) the -ais (b) the -ais (c) = 3 (d) = 3 (e) = 3 (f) = 3 3 R 3 6. Let R be the region bounded b = and =. Compute the volume of the solid formed b revolving R about the given line. (a) = (b) the -ais (c) = 6 (d) = (e) = (f) = 9. Let R be the region bounded b = a, = h and the -ais (where a and h are positive constants). Compute the volume of the solid formed b revolving this region about the -ais. Show that our answer equals half the volume of a clinder of height h and radius h/a. Sketch a picture to illustrate this. 3. Use the result of eercise 9 to immediatel write down the volume of the solid formed b revolving the region bounded b = a, = h/a and the -ais about the -ais. 3. Suppose that the square consisting of all points (, ) with and isrevolved about the -ais. Show that the volume of the resulting solid is π. 3. Suppose that the circle + = isrevolved about the -ais. Show that the volume of the resulting solid is 3 π. 33. Suppose that the triangle with vertices (, ), (, ) and (, ) is revolved about the -ais. Show that the volume of the resulting solid is 3 π. 3. Sketch the square, circle and triangle of eercises 3 33 on the same aes. Show that the relative volumes of the revolved regions (clinder, sphere and cone, respectivel) are 3::. 35. Verif the formula for the volume of a sphere b revolving the circle + = r about the -ais. 36. Verif the formula for the volume of a cone b revolving the line segment = h + h, r, about the -ais. r 37. Let A be a right circular cclinder with radius 3 and height 5. Let B be the tilted circular clinder with radius 3 and height 5. Determine whether A and B enclose the same volume R Determine whether the two indicated parallelograms have the same area. (Eercises 37 and 38 illustrate Cavalieri s Theorem.) 7. Let R be the region bounded b =, = and =. Compute the volume of the solid formed b revolving R about the given line. (a) the -ais (b) the -ais (c) = (d) = (e) = (f) = 8. Let R be the region bounded b =, = and =. Compute the volume of the solid formed b revolving R about the given line. (a) the -ais (b) the -ais (c) = (d) = The base of a solid V is the circle + =. Find the volume if V has (a) square cross sections and (b) semicircular cross sections perpendicular to the -ais.. The base of a solid V is the triangle with vertices (, ), (, ) and (, ). Find the volume if V has (a) square cross sections and (b) semicircular cross sections perpendicular to the -ais.

26 56 CHAPTER 5.. Applications of the Definite Integral 5-6. The base of a solid V is the region bounded b = and =. Find the volume if V has (a) square cross sections, (b) semicircular cross sections and (c) equilateral triangle cross sections perpendicular to the -ais.. The base of a solid V is the region bounded b = ln, = and =. Find the volume if V has (a) square cross sections, (b) semicircular cross sections and (c) equilateral triangle cross sections perpendicular to the -ais. 3. The base of a solid V is the region bounded b = e, =, = and = ln 5. Find the volume if V has (a) square cross sections and (b) semicircular cross sections perpendicular to the -ais.. The base of solid V is the region bounded b = and =. Find the volume if V has (a) square cross sections and (b) semicircular cross sections perpendicular to the -ais. 5. Use the given table of values to estimate the volume of the solid formed b revolving = f (), 3, about the -ais f () Use the given table of values to estimate the volume of the solid formed b revolving = f (),, about the -ais f () Water is poured at a constant rate into the vase with outline as shown and circular cross sections. Sketch a graph showing the height of the water in the vase as a function of time Sketch a graph of the rate of flow versus time if ou poured water into the vase of eercise 7 in such a wa that the height of the water in the vase increased at a constant rate. EXPLORATORY EXERCISES. Generalize the result of eercise 3 to an rectangle. That is, sketch the rectangle with a a and b b, the ellipse a + = and the triangle with vertices ( a, b), b (, b) and (a, b). Show that the relative volumes of the solid formed b revolving these regions about the -ais are 3::.. Take the circle ( ) + = and revolve it about the -ais. The resulting donut-shaped solid is called a torus. Compute its volume. Show that the volume equals the area of the circle times the distance travelled b the center of the circle. This is an eample of Pappus Theorem, dating from the fourth centur B.C. Verif that the result also holds for the triangle in eercise 5, parts (c) and (d). 5.3 VOLUMES BY CYLINDRICAL SHELLS R r O r I FIGURE 5.7 Revolve about = 3 3 In this section, we present an alternative to the method of washers discussed in section 5.. This method will help with solving some problems such as eample.7, part (d), where the method of washers led to a rather awkward integral. There, we let R be the region bounded b the graphs of = and = (see Figure 5.7). If R is revolved about the line = 3, as indicated in Figure 5.7, how would ou compute the volume of the resulting solid? The geometr of the region R makes it awkward to integrate with respect to, since the left-hand and right-hand boundaries of R are the left and right halves of the parabola, respectivel. On the other hand, since R is nicel defined on top b = and on bottom b =, it might be easier to integrate with respect to. Unfortunatel, in this case, the method of washers requires the -integration. The solution lies in an alternative method of computing volumes that uses the opposite variable of integration. Before returning to this eample, we consider the general case for a region revolved about the -ais. Let R denote the region bounded b the graph of = f () and the -ais

27 56 CHAPTER 5.. Applications of the Definite Integral 5-6. The base of a solid V is the region bounded b = and =. Find the volume if V has (a) square cross sections, (b) semicircular cross sections and (c) equilateral triangle cross sections perpendicular to the -ais.. The base of a solid V is the region bounded b = ln, = and =. Find the volume if V has (a) square cross sections, (b) semicircular cross sections and (c) equilateral triangle cross sections perpendicular to the -ais. 3. The base of a solid V is the region bounded b = e, =, = and = ln 5. Find the volume if V has (a) square cross sections and (b) semicircular cross sections perpendicular to the -ais.. The base of solid V is the region bounded b = and =. Find the volume if V has (a) square cross sections and (b) semicircular cross sections perpendicular to the -ais. 5. Use the given table of values to estimate the volume of the solid formed b revolving = f (), 3, about the -ais f () Use the given table of values to estimate the volume of the solid formed b revolving = f (),, about the -ais f () Water is poured at a constant rate into the vase with outline as shown and circular cross sections. Sketch a graph showing the height of the water in the vase as a function of time Sketch a graph of the rate of flow versus time if ou poured water into the vase of eercise 7 in such a wa that the height of the water in the vase increased at a constant rate. EXPLORATORY EXERCISES. Generalize the result of eercise 3 to an rectangle. That is, sketch the rectangle with a a and b b, the ellipse a + = and the triangle with vertices ( a, b), b (, b) and (a, b). Show that the relative volumes of the solid formed b revolving these regions about the -ais are 3::.. Take the circle ( ) + = and revolve it about the -ais. The resulting donut-shaped solid is called a torus. Compute its volume. Show that the volume equals the area of the circle times the distance travelled b the center of the circle. This is an eample of Pappus Theorem, dating from the fourth centur B.C. Verif that the result also holds for the triangle in eercise 5, parts (c) and (d). 5.3 VOLUMES BY CYLINDRICAL SHELLS R r O r I FIGURE 5.7 Revolve about = 3 3 In this section, we present an alternative to the method of washers discussed in section 5.. This method will help with solving some problems such as eample.7, part (d), where the method of washers led to a rather awkward integral. There, we let R be the region bounded b the graphs of = and = (see Figure 5.7). If R is revolved about the line = 3, as indicated in Figure 5.7, how would ou compute the volume of the resulting solid? The geometr of the region R makes it awkward to integrate with respect to, since the left-hand and right-hand boundaries of R are the left and right halves of the parabola, respectivel. On the other hand, since R is nicel defined on top b = and on bottom b =, it might be easier to integrate with respect to. Unfortunatel, in this case, the method of washers requires the -integration. The solution lies in an alternative method of computing volumes that uses the opposite variable of integration. Before returning to this eample, we consider the general case for a region revolved about the -ais. Let R denote the region bounded b the graph of = f () and the -ais

28 5-7 SECTION Volumes b Clindrical Shells 57 on the interval [a, b], where < a < b and f () on[a, b] (see Figure 5.8a). If we revolve this region about the -ais, we get the solid shown in Figure 5.8b. f() a R FIGURE 5.8a Revolve about -ais b a FIGURE 5.8b Solid of revolution b a c i i f() i b FIGURE 5.9a ith rectangle We first partition the interval [a, b] into n subintervals of equal width = b a. n On each subinterval [ i, i ], pick a point c i and construct the rectangle of height f (c i )as indicated in Figure 5.9a. Revolving this rectangle about the -ais forms a thin clindrical shell (i.e., a hollow clinder, like a pipe), as in Figure 5.9b. To find the volume of this thin clindrical shell, imagine cutting the clinder from top to bottom and then flattening out the shell. After doing this, ou should have essentiall a thin rectangular sheet, as seen in Figure 5.9c. Thickness Height a b Circumference of clindrical shell FIGURE 5.9b Clindrical shell FIGURE 5.9c Flattened clindrical shell Notice that the length of such a thin sheet corresponds to the circumference of the clindrical shell, which is π radius = πc i. So, the volume V i of the ith clindrical shell is approimatel V i length width height = (π radius) thickness height = (πc i ) f(c i ).

29 58 CHAPTER 5.. Applications of the Definite Integral 5-8 The total volume V of the solid can then be approimated b the sum of the volumes of the n clindrical shells: n V π c i f (c }{{} i ). }{{}}{{} i= radius height thickness As we have done man times now, we can get the eact volume of the solid b taking the limit as n and recognizing the resulting definite integral. We have Volume of a solid of revolution (clindrical shells) REMARK 3. Do not rel on simpl memorizing formula (3.). You must strive to understand the meaning of the components. It s simple to do if ou just think of how the correspond to the volume of a clindrical shell: π(radius) (height) (thickness). If ou think of volumes in this wa, ou will be able to solve an problem ou encounter. EXAMPLE 3. V = lim n n πc i f (c i ) = i= b a π }{{} radius f () }{{} height Using the Method of Clindrical Shells d }{{} thickness. (3.) Revolve the region bounded b the graphs of = and = in the first quadrant about the -ais. Solution From Figure 5.3a, notice that the region has an upper boundar of = and a lower boundar of = and runs from = to =. Here, we have drawn a sample rectangle that generates a clindrical shell. The resulting solid of revolution can be seen in Figure 5.3b. We can write down an integral for the volume b analzing the various components of the solid in Figures 5.3a and 5.3b. From (3.), we have V = π }{{} ( ) }{{}}{{} d radius height thickness ( = π ( 3 3 ) d = π 3 ) = π 6. Radius Height FIGURE 5.3a Sample rectangle generating aclindrical shell FIGURE 5.3b Solid of revolution

30 5-9 SECTION Volumes b Clindrical Shells 59 We can now generalize this method to solve the introductor eample. EXAMPLE 3. A Volume Where Shells Are Simpler Than Washers Find the volume of the solid formed b revolving the region bounded b the graph of = and the -ais about the line = 3. Solution Look carefull at Figure 5.3a, where we have drawn a sample rectangle that generates a clindrical shell and at the solid shown in Figure 5.3b. Notice that the radius of a clindrical shell is the distance from the line = 3tothe shell: This gives us the volume V = = π r = 3. π (3 ) ( }{{} ) }{{}}{{} d radius height thickness ( ) d = 6π, where the routine details of the calculation of the integral are left to the reader. 3 3 Radius 3 Height 3 8 FIGURE 5.3a Tpical rectangle generating a clindrical shell FIGURE 5.3b Solid of revolution Your first step in a volume calculation should be to analze the geometr of the solid to decide whether it s easier to integrate with respect to or. Note that for a given solid, the variable of integration in the method of shells is eactl opposite that of the method of washers. So, our choice of integration variable will determine which method ou use.

31 6 CHAPTER 5.. Applications of the Definite Integral 5-3 R FIGURE 5.3a = and = EXAMPLE 3.3 Computing Volumes Using Shells and Washers Let R be the region bounded b the graphs of =, = and =. Compute the volume of the solid formed b revolving R about the lines (a) =, (b) = and (c) = 3. Solution The region R is shown in Figure 5.3a. The geometr of the region suggests that we should consider as the variable of integration. Look carefull at the differences among the following three volumes. (a) Revolving R about the line =, observe that the radius of a clindrical shell is the distance from the line = tothe shell:, for (see Figure 5.3b). The height is the difference in the -values on the two curves: solving for, wehave = and =. Following (3.), we get the volume V = π ( ) [( ) ] d = }{{}}{{}}{{} 3 π, radius height thickness where we leave the routine details of the calculation to ou. (b) Revolving R about the line =, notice that the height of the clindrical shells is the same as in part (a), but the radius r is the distance from the line = to the shell: r = ( ) = + (see Figure 5.3c). This gives us the volume V = π [ ( )] [( ) ] d }{{}}{{}}{{} radius height thickness = 8 3 π. r O 3 r I 3 ( ) 3 Radius Radius ( ) r O r I 3 FIGURE 5.3b Revolve about = FIGURE 5.3c Revolve about = FIGURE 5.3d Solid of revolution (c) Finall, revolving R about the line = 3, notice that to find the volume using clindrical shells, we would need to break the calculation into two pieces, since the height of the clindrical shells would be different for [, ] than for [, ]. (Think about this some.) On the other hand, this is done easil b the method of washers. Observe that the outer radius is the distance from the line = 3tothe line = : r O = 3, while the inner radius is the distance from the line = 3tothe line = : r I = 3 ( ) (see Figure 5.3d). This gives us the volume V = π (3 ) [3 ( )] }{{}}{{} d = π. outer radius inner radius

32 5-3 SECTION Volumes b Clindrical Shells 6 Note the importance of sketching a picture and visualizing the solid. The most challenging aspect of these problems is to figure out how to set up the integral. Look carefull at our picture and determine which variable ou will use in the integration (i.e., determine whether ou will use washers or shells). Then, determine the components of the appropriate integral [i.e., the radius (or radii) and possibl the height], again b looking at the picture. Finall, do whatever it takes to evaluate the integral. If ou don t know how to evaluate it, ou can fall back on our CAS or approimate it numericall (e.g., b Simpson s Rule). EXAMPLE 3. Approimating Volumes Using Shells and Washers Let R be the region bounded b the graphs of = cos and =. Compute the volume of the solid formed b revolving R about the lines (a) = and (b) =. Solution First, we sketch the region R (see Figure 5.33a). Since the top and bottom of R are each defined b a curve of the form = f (), we will want to integrate with respect to.wenet look for the points of intersection of the two curves, b solving the equation cos =. Since we can t solve this eactl, we must use an approimate method (e.g., Newton s method) to obtain the approimate intersections at =±.83. Radius R cos cos R FIGURE 5.33a FIGURE 5.33b = cos, = Revolve about = (a) If we revolve the region about the line =, we should use clindrical shells (see Figure 5.33b). In this case, observe that the radius r of a clindrical shell is the distance from the line = tothe shell: r =, while the height of a shell is cos.weget the volume V π ( ) (cos }{{} ) d 3.757, }{{} radius height where we have approimated the value of the integral numericall. (We will see how to find an antiderivative for this integrand in Chapter 6.) (b) If we revolve the region about the line = (see Figure 5.33c on the following page), we use the method of washers. In this case, observe that the outer radius of a washer is the distance from the line = tothe curve = : r O =, while the inner radius is the distance from the line = tothe curve = cos : r I = cos

33 6 CHAPTER 5.. Applications of the Definite Integral 5-3 r I cos r O R cos FIGURE 5.33c Revolve about = (again, see Figure 5.33c). This gives us the volume.83 V π ( ) ( cos ).83 }{{}}{{} d.8, other radius inner radius where we have approimated the value of the integral numericall. We close this section with a summar of strategies for computing volumes of solids of revolution. VOLUME OF A SOLID OF REVOLUTION Sketch the region to be revolved. Determine the variable of integration ( if the region has a well-defined top and bottom, if the region has well-defined left and right boundaries). Based on the ais of revolution and the variable of integration, determine the method (disks or washers for -integration about a horizontal ais or -integration about a vertical ais, shells for -integration about a vertical ais or -integration about a horizontal ais). Label our picture with the inner and outer radii for disks or washers; label the radius and height for clindrical shells. Set up the integral(s) and evaluate. EXERCISES 5.3 WRITING EXERCISES. Eplain wh the method of clindrical shells produces an integral with as the variable of integration when revolving about avertical ais. (Describe where the shells are and which direction to move in to go from shell to shell.). Eplain wh the method of clindrical shells has the same form whether or not the solid has a hole or cavit. That is, there is no need for separate methods analogous to disks and washers.

34 5-33 SECTION Volumes b Clindrical Shells Suppose that the region bounded b = and = is revolved about the line =. Carefull eplain which method (disks, washers or shells) would be easiest to use to compute the volume.. Suppose that the region bounded b = 3 3 and =,, is revolved about = 3. Eplain what would be necessar to compute the volume using the method of washers and what would be necessar to use the method of clindrical shells. Which method would ou prefer and wh? 3 3 In eercises 8, sketch the region, draw in a tpical shell, identif the radius and height of each shell and compute the volume.. The region bounded b = and the -ais,, revolved about =. The region bounded b = and the -ais,, revolved about = 3. The region bounded b =, = and = revolved about the -ais. The region bounded b =, = and = revolved about = 5. The region bounded b =, = and = revolved about = 3 6. The region bounded b =, = and = revolved about = 7. The right half of + ( ) = revolved about the -ais 8. The right half of + ( ) = revolved about = In eercises 9 6, use clindrical shells to compute the volume. 9. The region bounded b = and =,revolved about =. The region bounded b = and =,revolved about =. The region bounded b = and = revolved about =. The region bounded b = and = revolved about = 3. The region bounded b = and = revolved about =. The region bounded b = and = revolved about = 3 5. The region bounded b = ( ) and = 9revolved about = 5 6. The region bounded b = ( ) and = 9revolved about = 3 In eercises 7 6, use the best method available to find each volume. 7. The region bounded b =, = and = revolved about (a) the -ais (b) the -ais (c) = (d) = 8. The region bounded b = +, = and = revolved about (a) = (b) = (c) the -ais (d) the -ais 9. The region bounded b = and = 6revolved about (a) = 3 (b) = 3 (c) = 3 (d) = 6. The region bounded b = and = + revolved about (a) = (b) = (c) = (d) =. The region bounded b = cos and = revolved about (a) = (b) = (c) the -ais (d) the -ais. The region bounded b = sin and = revolved about (a) = (b) = (c) the -ais (d) the -ais 3. The region bounded b = ( ), = and = revolved about (a) the -ais (b) the -ais (c) = (d) =. The region bounded b =, = ( > ) and the -ais revolved about (a) the -ais (b) the -ais (c) = (d) = 5. The region to the right of = and to the left of = and = revolved about (a) the -ais (b) the -ais

35 6 CHAPTER 5.. Applications of the Definite Integral The region bounded b = e, = and the -ais revolved about (a) -ais (b) -ais In eercises 7 3, the integral represents the volume of a solid. Sketch the region and ais of revolution that produce the solid π( ) d π[( + ) ( + ) ] d π[( ) ] d 3. π ( ) d 3. π( ) d π( )( + ) d 33. Use a method similar to our derivation of equation (3.) to derive the following fact about a circle of radius R. Area = π R = R c(r) dr, where c(r) = πr is the circumference of a circle of radius r. 3. You have probabl noticed that the circumference of a circle (πr) equals the derivative with respect to r of the area of the circle (πr ). Use eercise 33 to eplain wh this is not a coincidence. 35. Ajewelr bead is formed b drilling a -cm radius hole from the center of a -cm radius sphere. Eplain wh the volume is given b π / d.evaluate this integral or compute the volume in some easier wa. 36. Find the size of the hole in eercise 35 such that eactl half the volume is removed. 37. An anthill is in the shape formed b revolving the region bounded b = and the -ais about the -ais. A researcher removes a clindrical core from the center of the hill. What should the radius be to give the researcher % of the dirt? 38. The outline of a rugb ball has the shape of + =. The 3 6 ball itself is the revolution of this ellipse about either the -ais or -ais. Find the volume of the ball. EXPLORATORY EXERCISES. From a sphere of radius R, ahole of radius r is drilled out of the center. Compute the volume removed in terms of R and r. Compute the length L of the hole in terms of R and r. Rewrite the volume in terms of L.Isitreasonable to sa that the volume removed depends on L and not on R?. In each case, sketch the solid and find the volume formed b revolving the region about (i) the -ais and (ii) the -ais. Compute the volume eactl if possible and estimate numericall if necessar. (a) Region bounded b = sec tan +, =, = π and = π. (b) Region bounded b = +, =, = and =. (c) Region bounded b = sin, =, = π and =. (d) Region bounded b = and =. (e) Region bounded b = e and = ( ). 5. ARC LENGTH AND SURFACE AREA Length and area are quantities ou alread understand intuitivel. But, as ou have learned with area, the calculation of these quantities can be surprisingl challenging for man geometric shapes. In this section, we compute the length of a curve in two dimensions and the area of a surface in three dimensions. As alwas, pa particular attention to the derivations. As we have done a number of times now, we start with an approimation and then proceed to the eact solution, using the notion of limit..5 d q f p FIGURE 5.3a = sin Arc Length What could we mean b the length of the portion of the sine curve shown in Figure 5.3a? (We call the length of a curve its arc length.) Ifthe curve represented a road, ou could measure the length on our car s odometer b driving along that section of road. If the curve were actuall a piece of string, ou could straighten out the string and then measure its length with a ruler. Both of these ideas are ver helpful intuitivel. The both involve turning the problem of measuring length in two dimensions into the (much easier) problem of measuring the length in one dimension.

36 6 CHAPTER 5.. Applications of the Definite Integral The region bounded b = e, = and the -ais revolved about (a) -ais (b) -ais In eercises 7 3, the integral represents the volume of a solid. Sketch the region and ais of revolution that produce the solid π( ) d π[( + ) ( + ) ] d π[( ) ] d 3. π ( ) d 3. π( ) d π( )( + ) d 33. Use a method similar to our derivation of equation (3.) to derive the following fact about a circle of radius R. Area = π R = R c(r) dr, where c(r) = πr is the circumference of a circle of radius r. 3. You have probabl noticed that the circumference of a circle (πr) equals the derivative with respect to r of the area of the circle (πr ). Use eercise 33 to eplain wh this is not a coincidence. 35. Ajewelr bead is formed b drilling a -cm radius hole from the center of a -cm radius sphere. Eplain wh the volume is given b π / d.evaluate this integral or compute the volume in some easier wa. 36. Find the size of the hole in eercise 35 such that eactl half the volume is removed. 37. An anthill is in the shape formed b revolving the region bounded b = and the -ais about the -ais. A researcher removes a clindrical core from the center of the hill. What should the radius be to give the researcher % of the dirt? 38. The outline of a rugb ball has the shape of + =. The 3 6 ball itself is the revolution of this ellipse about either the -ais or -ais. Find the volume of the ball. EXPLORATORY EXERCISES. From a sphere of radius R, ahole of radius r is drilled out of the center. Compute the volume removed in terms of R and r. Compute the length L of the hole in terms of R and r. Rewrite the volume in terms of L.Isitreasonable to sa that the volume removed depends on L and not on R?. In each case, sketch the solid and find the volume formed b revolving the region about (i) the -ais and (ii) the -ais. Compute the volume eactl if possible and estimate numericall if necessar. (a) Region bounded b = sec tan +, =, = π and = π. (b) Region bounded b = +, =, = and =. (c) Region bounded b = sin, =, = π and =. (d) Region bounded b = and =. (e) Region bounded b = e and = ( ). 5. ARC LENGTH AND SURFACE AREA Length and area are quantities ou alread understand intuitivel. But, as ou have learned with area, the calculation of these quantities can be surprisingl challenging for man geometric shapes. In this section, we compute the length of a curve in two dimensions and the area of a surface in three dimensions. As alwas, pa particular attention to the derivations. As we have done a number of times now, we start with an approimation and then proceed to the eact solution, using the notion of limit..5 d q f p FIGURE 5.3a = sin Arc Length What could we mean b the length of the portion of the sine curve shown in Figure 5.3a? (We call the length of a curve its arc length.) Ifthe curve represented a road, ou could measure the length on our car s odometer b driving along that section of road. If the curve were actuall a piece of string, ou could straighten out the string and then measure its length with a ruler. Both of these ideas are ver helpful intuitivel. The both involve turning the problem of measuring length in two dimensions into the (much easier) problem of measuring the length in one dimension.

37 5-35 SECTION 5... Arc Length and Surface Area 65.5 n d q f p FIGURE 5.3b Four line segments approimating = sin Length f( i ) f( i ) i s i i FIGURE 5.35 Straight-line approimation of arc length To accomplish this mathematicall, we first approimate the curve with several line segments ( ) joined together. In Figure 5.3b, the line segments connect the points (, ), π, ( π ) ( ) 3π,,,, and (π, ) on the curve = sin. Anapproimation of the arc length s of the curve is given b the sum of the lengths of these line segments: s (π ) ( ) (π ) ( ) (π ) ( ) (π ) ( ) You might notice that this estimate is too small. (Wh is that?) We will improve our approimation b using more than four line segments. In the table at left, we show estimates of the length of the curve using n line segments for larger values of n.asou would epect, the approimation of length will get closer to the actual length of the curve, as the number of line segments increases. This general idea should sound familiar. We develop this notion further now for the more general problem of finding the arc length of the curve = f () onthe interval [a, b]. Here, we ll assume that f is continuous on [a, b] and differentiable on (a, b). (Where have ou seen hpotheses like these before?) As usual, we begin b partitioning the interval [a, b] into n equal pieces: a = < < < n = b, where i i = = b a, for each i =,,...,n. n Between each pair of adjacent points on the curve, ( i, f ( i )) and ( i, f ( i )), we approimate the arc length s i b the straight-line distance between the two points (see Figure 5.35). From the usual distance formula, we have s i d{( i, f ( i )), ( i, f ( i ))} = ( i i ) + [ f ( i ) f ( i )]. Since f is continuous on all of [a, b] and differentiable on (a, b), f is also continuous on the subinterval [ i, i ] and is differentiable on ( i, i ). B the Mean Value Theorem, we then have f ( i ) f ( i ) = f (c i )( i i ), for some number c i ( i, i ). This gives us the approimation s i ( i i ) + [ f ( i ) f ( i )] = ( i i ) + [ f (c i )( i i )] = + [ f (c i )] ( i i ) = + [ f }{{} (c i )]. Adding together the lengths of these n line segments, we get an approimation of the total arc length, n s + [ f (c i )]. i= Notice that as n gets larger, this approimation should approach the eact arc length, that is, n s = lim + [ f (c i )]. n i=

38 66 CHAPTER 5.. Applications of the Definite Integral 5-36 You should recognize this as the limit of a Riemann sum for + [ f ()],sothat the arc length is given eactl b the definite integral: REMARK. Arc length of = f () on the interval [a, b] The formula for arc length is ver simple. Unfortunatel, ver few functions produce arc length integrals that can be evaluated eactl. You should epect to use a numerical integration method on our calculator or computer to compute most arc lengths. whenever the limit eists. EXAMPLE. s = b a + [ f ()] d, (.) Using the Arc Length Formula Find the arc length of the portion of the curve = sin with π.(we estimated this as 3.79 in our introductor eample.) Solution From (.), the arc length is π s = + (cos ) d. Trtofind an antiderivative of + cos,but don t tr for too long. (The best our CAS can do is EllipticE[, ], which doesn t seem especiall helpful.) Using a numerical integration method, the arc length is π s = + (cos ) d 3.8. Even for ver simple curves, evaluating the arc length integral eactl can be quite challenging EXAMPLE. Estimating an Arc Length Find the arc length of the portion of the curve = with. Solution Using the arc length formula (.), we get s = + () d = + d.789, where we have again evaluated the integral numericall. (In this case, ou can find an antiderivative using a clever substitution or a CAS.) The graphs of = and = look surprisingl similar on the interval [, ] (see Figure 5.36). The both connect the points (, ) and (, ), are increasing and are concave up. If ou graph them simultaneousl, ou will note that = starts out flatter and then becomes steeper from about =.7 on. (Tr proving that this is true!) Arc length gives us one wa to quantif the difference between the two graphs. FIGURE 5.36 = and = EXAMPLE.3 A Comparison of Arc Lengths of Power Functions Find the arc length of the portion of the curve = with and compare to the arc length of the portion of the curve = on the same interval. Solution From (.), the arc length for = is given b + ( 3 ) d = d.6. Notice that this arc length is about 8% larger than that of =,asfound in eample..

39 5-37 SECTION 5... Arc Length and Surface Area 67 In the eercises, ou will be asked to eplore the trend in the lengths of the portion of the curves = 6, = 8 and so on, on the interval [, ]. Can ou guess now what happens to the arc length of the portion of = n,onthe interval [, ], as n? As with man words, everda usage of the word length can be ambiguous and misleading. For instance, when somebod sas that the length of a frisbee throw was ards, the length refers to the horizontal distance covered, not to the arc length of the frisbee s flight path. In this case, reporting the horizontal distance is more meaningful than the arc length (and is easier to measure). In man other cases, arc length is the quantit of interest. For eample, suppose ou need to hang a banner between two poles that are feet apart. If ou have onl feet of rope to work with, ou are going to be in trouble. The length of rope required is determined b the arc length, rather than the horizontal distance. EXAMPLE. Computing the Length of Cable Hanging between Two Poles FIGURE 5.37 = 5(e / + e / ) 5 A cable is to be hung between two poles of equal height that are feet apart. It can be shown that such a hanging cable assumes the shape of a catenar, the general form of which is = a cosh /a = a (e/a + e /a ). In this case, suppose that the cable takes the shape of = 5(e / + e / ), for, as seen in Figure How long is the cable? Solution From (.), the arc length of the curve is given b ( ) e / s = + e / d = + (e/5 + e /5 ) d = (e/5 + + e /5 ) d = (e/ + e / ) d = (e/ + e / ) d = 5(e / e / ) = = = (e e ) 3.5 feet, which corresponds to the horizontal distance of feet plus about 3 feet of slack. Surface Area Circular cross sections FIGURE 5.38 Surface of revolution In sections 5. and 5.3, we saw how to compute the volume of a solid formed b revolving a two-dimensional region about a fied ais. In addition, we often want to determine the area of the surface that is generated b the revolution. For instance, when revolving the line = +, for, about the -ais, the surface generated looks like a megaphone with two open ends, as shown in Figure 5.38.

40 68 CHAPTER 5.. Applications of the Definite Integral 5-38 Cut along a seam l u r l FIGURE 5.39a Right circular cone FIGURE 5.39b Flattened cone r r FIGURE 5. Frustum of a cone L Notice that this is the bottom portion of a right circular cone that has had its top cut off b a plane parallel to its base. Before we go an further, we pause to find the curved surface area of a right circular cone. In Figure 5.39a, we show a right circular cone of base radius r and slant height l. (As ou ll see later, it is more convenient in this contet to specif the slant height than the altitude.) If we cut the cone along a seam and flatten it out, we get the circular sector shown in Figure 5.39b. Notice that the curved surface area of the cone is the same as the area A of the circular sector. This is the area of a circle of radius l multiplied b the fraction of the circle included: θ out of a possible π radians, or A = π(radius) θ π = θ πl π = θ l. (.) The onl problem with this is that we don t know θ. However, notice that b the wa we constructed the sector (i.e., b flattening the cone), the circumference of the sector is the same as the circumference of the base of the cone. That is, Dividing b l gives us πr = πl θ π = lθ. θ = πr. l From (.), the curved surface area of the cone is now A = θ l = πr l = πrl. l Recall that we were originall interested in finding the surface area of onl a portion of a right circular cone (look back at Figure 5.38). For the frustum of a cone shown in Figure 5., the curved surface area is given b A = π(r + r )L. You can verif this b subtracting the curved surface area of two cones, where ou must use similar triangles to find the height of the larger cone from which the frustum is cut. We leave the details of this as an eercise. Returning to the original problem of revolving the line = + onthe interval [, ] about the -ais (seen in Figure 5.38), we have r =, r = and L = (from the Pthagorean Theorem). The curved surface area is then A = π( + ) = 3π For the general problem of finding the curved surface area of a surface of revolution, consider the case where f () and where f is continuous on the interval [a, b] and differentiable

41 5-39 SECTION 5... Arc Length and Surface Area 69 f() f() a b a b FIGURE 5.a Revolve about -ais FIGURE 5.b Surface of revolution i L i f( i ) i f( i ) on (a, b). If we revolve the graph of = f () about the -ais on the interval [a, b] (see Figure 5.a), we get the surface of revolution seen in Figure 5.b. As we have done man times now, we first partition the interval [a, b] into n pieces of equal size: a = < < < n = b, where i i = = b a, for each n i =,,...,n.oneach subinterval [ i, i ], we can approimate the curve b the straight line segment joining the points ( i, f ( i )) and ( i, f ( i )), as in Figure 5.. Notice that revolving this line segment around the -ais generates the frustum of a cone. The surface area of this frustum will give us an approimation to the actual surface area on the interval [ i, i ]. First, observe that the slant height of this frustum is L i = d{( i, f ( i )), ( i, f ( i ))} = ( i i ) + [ f ( i ) f ( i )], FIGURE 5. Revolve about -ais from the usual distance formula. Because of our assumptions on f, wecan appl the Mean Value Theorem, to obtain f ( i ) f ( i ) = f (c i )( i i ), for some number c i ( i, i ). This gives us L i = ( i i ) + [ f ( i ) f ( i )] = + [ f (c i )] ( i i ). }{{} The surface area S i of that portion of the surface on the interval [ i, i ]isapproimatel the surface area of the frustum of the cone, S i π[ f ( i ) + f ( i )] + [ f (c i )] π f (c i ) + [ f (c i )], since if is small, f ( i ) + f ( i ) f (c i ). Repeating this argument for each subinterval [ i, i ], i =,,...,n,gives us an approimation to the total surface area S, n S π f (c i ) + [ f (c i )]. i= As n gets larger, this approimation approaches the actual surface area, n S = lim π f (c i ) + [ f (c i )]. n i=

42 7 CHAPTER 5.. Applications of the Definite Integral 5- Recognizing this as the limit of a Riemann sum gives us the integral SURFACE AREA OF A SOLID OF REVOLUTION S = b a π f () + [ f ()] d, (.3) REMARK. There are eceptionall few functions f for which the integral in (.3) can be computed eactl. Don t worr; we have numerical integration for just such occasions. whenever the integral eists. You should notice that the factor of + [ f ()] d in the integrand in (.3) corresponds to the arc length of a small section of the curve = f (), while the factor π f () corresponds to the circumference of the solid of revolution. This should make sense to ou, as follows. For an small segment of the curve, if we approimate the surface area b revolving a small segment of the curve of radius f () around the -ais, the surface area generated is simpl the surface area of a clinder, S = πrh = π f () + [ f ()] d, since the radius of such a small clindrical segment is f () and the height of the clinder is h = + [ f ()] d.itisfar better to think about the surface area formula in this wa than to simpl memorize the formula. EXAMPLE.5 Using the Surface Area Formula Find the surface area of the surface generated b revolving =, for, about the -ais. Solution S = Using the surface area formula (.3), we have π + ( 3 ) d = π d 3.365, where we have used a numerical method to approimate the value of the integral. EXERCISES 5. WRITING EXERCISES. Eplain in words how the arc length integral is derived from the lengths of the approimating secant line segments.. Eplain wh the sum of the lengths of the line segments in Figure 5.3b is less than the arc length of the curve in Figure 5.3a. 3. Discuss whether the arc length integral is more accuratel called a formula or a definition (i.e., can ou precisel define the length of a curve without using the integral?).. Suppose ou graph the trapezoid bounded b = +, =, = and =, cut it out and roll it up. Eplain wh ou would not get Figure (Hint: Compare areas and carefull consider Figures 5.39a and 5.39b.) In eercises, approimate the length of the curve using n secant lines for n ; n.. =,. =, 3. = cos, π. = ln, 3 In eercises 5, compute the arc length eactl. 5. = +, 6. =, 7. = 3/ +, 8. = (e + e ), 9. = ln,

43 5- SECTION 5... Arc Length and Surface Area 7. = 3 +, 3 6. = +, 8. = e / + e /, 3. = 3/ /, 3 ( ). = 3 /3 /3, 8 In eercises 5, set up the integral for arc length and then approimate the integral with a numerical method. 5. = 3, 6. = 3, 7. =, 8. = tan, π/ 9. = cos, π. = ln, 3. =. = u sin udu, π e u sin udu, π 3. A rope is to be hung between two poles feet apart. If the rope assumes the shape of the catenar = 5(e / + e / ),, compute the length of the rope.. A rope is to be hung between two poles 6 feet apart. If the rope assumes the shape of the catenar = 5(e /3 + e /3 ), 3 3, compute the length of the rope. 5. In eample., compute the sag in the cable that is, the difference between the -values in the middle ( = ) and at the poles ( = ). Given this, is the arc length calculation surprising? 6. Sketch and compute the length of the astroid defined b /3 + /3 =. 7. A football punt follows the path = (6 ) ards. Sketch 5 a graph. How far did the punt go horizontall? How high did it go? Compute the arc length. If the ball was in the air for seconds, what was the ball s average velocit? 8. A baseball outfielder s throw follows the path = ( ) ards. Sketch a graph. How far did the ball 3 go horizontall? How high did it go? Compute the arc length. Eplain wh the baseball plaer would want a small arc length, while the football plaer in eercise 7 would want a large arc length. 9. In eample., show that the arc length is eactl sinh(), where sinh = (e e ). 3. Evaluate the integral in eample. with our CAS. Compare its answer to that of eercise The elliptic integral of the second kind is defined b EllipticE(φ,m) = φ m sin udu. Referring to eample., man CASs report EllipticE(, )asanantiderivative of + cos.verif that this is an antiderivative. 3. Man CASs report the antiderivative d = Verif that this is an antiderivative. 3/ d. 33. For eample., find an antiderivative with our CAS, evaluate the antiderivative at the endpoints and compare the difference in values to the value our CAS gives using numerical integration. Tr to do this for eample Briefl eplain what it means when our CAS returns f () d when asked to evaluate the indefinite integral f () d. In eercises 35, set up the integral for the surface area of the surface of revolution and approimate the integral with a numerical method. 35. =,, revolved about the -ais 36. = sin, π, revolved about the -ais 37. =,, revolved about the -ais 38. = 3,, revolved about the -ais 39. = e,, revolved about the -ais. = ln,, revolved about the -ais. = cos, π/, revolved about the -ais. =,, revolved about the -ais In eercises 3 6, compute the arc length L of the curve and the length L of the secant line connecting the endpoints of the curve. Compute the ratio L /L ; the closer this number is to, the straighter the curve is. 3. = sin, π 6 π 6. = cos, π 6 π 6 5. = e, = e, For = 6, = 8 and =, compute the arc length for. Using results from eamples. and.3, identif the pattern for the length of = n,, as n increases. Conjecture the limit as n. 8. To help understand the result of eercise 7, determine lim n n for each such that <. Compute the length of this limiting curve. Connecting this curve to the endpoint (, ), what is the total length? 9. Prove that = is flatter than = for < < / and steeper for > /. Compare the flatness and steepness of = 6 and =. 5. Suppose that the square consisting of all (, ) with and isrevolved about the -ais. Compute the surface area.

44 7 CHAPTER 5.. Applications of the Definite Integral 5-5. Suppose that the circle + = isrevolved about the -ais. Compute the surface area. 5. Suppose that the triangle with vertices (, ), (, ) and (, ) is revolved about the -ais. Compute the surface area. 53. Sketch the square, circle and triangle of eercises 5 5 on the same aes. Show that the relative surface areas of the solids of revolution (clinder, sphere and cone, respectivel) are 3::τ, where τ is the golden mean defined b τ = + 5. EXPLORATORY EXERCISES. In eercises 7 and 8, ou eplored the length of = n on the interval [, ]. In this eercise, ou will look at more general polnomials on the same interval. First, guess what the largest possible arc length would be for a polnomial on a given interval. Now, check the arc length for some selected parabolas. Eplain wh f c () = c( )isadownward-opening parabola (for c > ) with -intercepts and and verte at = /. What will happen to the arc length as c gets larger? Suppose we limit ourselves to polnomials that have function values ling between and. Verif that on the interval [, ], the functions g ( ) = ( ), g 3 () = (/ )( ) and g () = 8(/3 )(/3 )( ) all have this propert. Compare the arc lengths of g (), g 3 () and g (). What will be the limiting value of such arc lengths? Construct a function g 5 () that continues this pattern.. In this eercise, ou will eplore a famous parado (often called Gabriel s horn). Suppose that the curve = /, for R (where R is a large positive constant), is revolved about the -ais. Compute the enclosed volume and the surface area of the resulting surface. (In both cases, antiderivatives can be found, although ou ma need help from our CAS to get the surface area.) Determine the limit of the volume and surface area as R.Now for the parado. Based on our answers, ou should have a solid with finite volume, but infinite surface area. Thus, the three-dimensional solid could be completel filled with a finite amount of paint but the outside surface could never be completel painted. 3. Let C be the portion of the parabola = a inside the circle + = Find the value of a > that maimizes the arc length of C.. The figure shows an arc of a circle subtended b an angle θ, with a chord of length L and two chords of length s. Show that L s = cos(θ/). θ Start with a quarter-circle and use this formula repeatedl to derive the infinite product cos π cos π 8 cos π 6 cos π 3 = π where the left-hand side represents lim (cos π cos π...cos π ). n n n L s s 5.5 PROJECTILE MOTION In sections.,.3 and., we discussed aspects of the motion of an object moving in a straight line path (rectilinear motion). We saw that if we know a function describing the position of an object at an time t, then we can determine its velocit and acceleration b differentiation. A much more important problem is to go backward, that is, to find the

45 7 CHAPTER 5.. Applications of the Definite Integral 5-5. Suppose that the circle + = isrevolved about the -ais. Compute the surface area. 5. Suppose that the triangle with vertices (, ), (, ) and (, ) is revolved about the -ais. Compute the surface area. 53. Sketch the square, circle and triangle of eercises 5 5 on the same aes. Show that the relative surface areas of the solids of revolution (clinder, sphere and cone, respectivel) are 3::τ, where τ is the golden mean defined b τ = + 5. EXPLORATORY EXERCISES. In eercises 7 and 8, ou eplored the length of = n on the interval [, ]. In this eercise, ou will look at more general polnomials on the same interval. First, guess what the largest possible arc length would be for a polnomial on a given interval. Now, check the arc length for some selected parabolas. Eplain wh f c () = c( )isadownward-opening parabola (for c > ) with -intercepts and and verte at = /. What will happen to the arc length as c gets larger? Suppose we limit ourselves to polnomials that have function values ling between and. Verif that on the interval [, ], the functions g ( ) = ( ), g 3 () = (/ )( ) and g () = 8(/3 )(/3 )( ) all have this propert. Compare the arc lengths of g (), g 3 () and g (). What will be the limiting value of such arc lengths? Construct a function g 5 () that continues this pattern.. In this eercise, ou will eplore a famous parado (often called Gabriel s horn). Suppose that the curve = /, for R (where R is a large positive constant), is revolved about the -ais. Compute the enclosed volume and the surface area of the resulting surface. (In both cases, antiderivatives can be found, although ou ma need help from our CAS to get the surface area.) Determine the limit of the volume and surface area as R.Now for the parado. Based on our answers, ou should have a solid with finite volume, but infinite surface area. Thus, the three-dimensional solid could be completel filled with a finite amount of paint but the outside surface could never be completel painted. 3. Let C be the portion of the parabola = a inside the circle + = Find the value of a > that maimizes the arc length of C.. The figure shows an arc of a circle subtended b an angle θ, with a chord of length L and two chords of length s. Show that L s = cos(θ/). θ Start with a quarter-circle and use this formula repeatedl to derive the infinite product cos π cos π 8 cos π 6 cos π 3 = π where the left-hand side represents lim (cos π cos π...cos π ). n n n L s s 5.5 PROJECTILE MOTION In sections.,.3 and., we discussed aspects of the motion of an object moving in a straight line path (rectilinear motion). We saw that if we know a function describing the position of an object at an time t, then we can determine its velocit and acceleration b differentiation. A much more important problem is to go backward, that is, to find the

46 5-3 SECTION Projectile Motion 73 position and velocit of an object, given its acceleration. Mathematicall, this means that, starting with the derivative of a function, we must find the original function. Now that we have integration at our disposal, we can accomplish this with ease. You ma alread be familiar with Newton s second law of motion, which sas that F = ma, where F is the sum of the forces acting on an object, m is the mass of the object and a is the acceleration of the object. Start b imagining that ou are diving. The primar force acting on ou throughout the dive is gravit. The force due to gravit is our own weight, which is related to mass b W = mg, where g is the gravitational constant. (Common approimations of g, accurate near sea level, are 3 ft/s and 9.8 m/s.) To keep the problem simple mathematicall, we will ignore an other forces, such as air resistance. Let h(t) represent our height above the water t seconds after starting our dive. Then the force due to gravit is F = mg, where the minus sign indicates that the force is acting downward, in the negative direction. From our earlier work, we know that the acceleration is a(t) = h (t). Newton s second law then gives us mg = mh (t) or h (t) = g. Notice that the position function of an object (regardless of its mass) subject to gravit and no other forces will satisf the same equation. The onl differences from situation to situation are the initial conditions (the initial velocit and initial position) and the questions being asked. EXAMPLE 5. Finding the Velocit of a Diver at Impact If a diving board is 5 feet above the surface of the water and a diver starts with initial velocit 8 ft/s (in the upward direction), what is the diver s velocit at impact (assuming no air resistance)? Solution If the height (in feet) at time t is given b h(t), Newton s second law gives us h (t) = 3. Since the diver starts 5 feet above the water with initial velocit of 8 ft/s, we have the initial conditions h() = 5 and h () = 8. Finding h(t) nowtakes little more than elementar integration. We have h (t) dt = 3 dt or h (t) = 3t + c. From the initial velocit, we have so that the velocit at an time t is given b 8 = h () = 3() + c = c, h (t) = 3t + 8. To find the velocit at impact, ou first need to find the time of impact. Notice that the diver will hit the water when h(t) = (i.e., when the height above the water is ).

47 7 CHAPTER 5.. Applications of the Definite Integral 5- Integrating the velocit function gives us the height function: h (t) dt = ( 3t + 8) dt or h(t) = 6t + 8t + c. From the initial height, we have 5 = h() = 6() + 8() + c = c, so that the height above the water at an time t is given b Impact then occurs when h(t) = 6t + 8t + 5. = h(t) = 6t + 8t + 5 = (t + 3)(t 5), so that t = 5 is the time of impact. (Ignore the etraneous solution t = 3.) When t = 5, the velocit is h ( ( 5 ) = 3 5 ) + 8 = 3 ft/s (impact velocit). To put this in more familiar units of velocit, multipl b 36/58 to convert to miles per hour. In this case, the impact velocit is about mph. (You probabl don t want to come down in the wrong position at that speed!) In eample 5., the negative sign of the velocit indicated that the diver was coming down. In man situations, both upward and downward motions are important. EXAMPLE 5. An Equation for the Vertical Motion of a Ball A ball is propelled straight upward from the ground with initial velocit 6 ft/s. Ignoring air resistance, find an equation for the height of the ball at an time t. Also, determine the maimum height and the amount of time the ball spends in the air. Solution With gravit as the onl force, the height h(t) satisfies h (t) = 3. The initial conditions are h () = 6 and h() =. We then have h (t) dt = 3 dt or h (t) = 3t + c. From the initial velocit, we have 6 = h () = 3() + c = c and so, h (t) = 6 3t. Integrating one more time gives us h (t) dt = (6 3t) dt

48 5-5 SECTION Projectile Motion 75 TODAY IN MATHEMATICS Vladimir Arnold (937 ) A Russian mathematician with important contributions to numerous areas of mathematics, both in research and popular eposition. The esteem in which he is held b his colleagues can be measured b the international conference known as Arnoldfest held in Toronto in honor of his 6th birthda. Man of his books are widel used toda, including a collection of challenges titled Arnold s Problems. Areview of this book states that Arnold did not consider mathematics a game with deductive reasoning and smbols, but a part of natural science (especiall of phsics), i.e., an eperimental science. Height 6 or h(t) = 6t 6t + c. From the initial height we have = h() = 6() 6() + c = c, and so, h(t) = 6t 6t. Since the height function is quadratic, its maimum occurs at the one time when h (t) =. [You should also consider the phsics of the situation: what happens phsicall when h (t) =?] Solving 6 3t = givest = (the time at the maimum height) and the corresponding height is h() = 6() 6() = 6 feet. Again, the ball lands when h(t) =. Solving = h(t) = 6t 6t = 6t( t) gives t = (launch time) and t = (landing time). The time of flight is thus seconds. You can observe an interesting propert of projectile motion b graphing the height function from eample 5. along with the line = 8 (see Figure 5.3). Notice that the graphs intersect at t = and t = 3. Further, the time interval [, 3] corresponds to eactl half the time spent in the air. Notice that this sas that the ball stas in the top one-fourth of its height for half of its time in the air. You ma have marveled at how some athletes jump so high that the seem to hang in the air. As this calculation suggests, all objects tend to hang in the air. 3 FIGURE 5.3 Height of the ball at time t t EXAMPLE 5.3 Finding the Initial Velocit Required to Reach a Certain Height It has been reported that basketball star Michael Jordan has a vertical leap of 5. Ignoring air resistance, what is the initial velocit required to jump this high? Solution Once again, Newton s second law leads us to the equation h (t) = 3 for the height h(t). We call the initial velocit v,sothat h () = v and look for the value of v that will give a maimum altitude of 5. As before, we integrate to get h (t) = 3t + c. Using the initial velocit, we get v = h () = 3() + c = c. This gives us the velocit function h (t) = v 3t.

49 76 CHAPTER 5.. Applications of the Definite Integral 5-6 Integrating once again and using the initial position h() =, we get h(t) = v t 6t. The maimum height occurs when h (t) =. (Wh?) Setting = h (t) = v 3t, u v gives us t = v. The height at this time (i.e., the maimum altitude) is then 3 h ( v 3 ) ( v ) = v 6 3 ( v 3 ) = v 3 v 6 = v 6. FIGURE 5.a Path of projectile u v v cos u v sin u FIGURE 5.b Vertical and horizontal components of velocit So, a jump of 5 =.5 requires v 6 =.5 orv = 88, so that v = 88 7 ft/s (equivalent to roughl.6 mph). So far, we have onl considered projectiles moving verticall. In practice, we must also consider movement in the horizontal direction. Ignoring air resistance, these calculations are also relativel straightforward. The idea is to appl Newton s second law separatel to the horizontal and vertical components of the motion. If (t) represents the vertical position, then we have (t) = g, asbefore. Ignoring air resistance, there are no forces acting horizontall on the projectile. So, if (t) represents the horizontal position, Newton s second law gives us (t) =. The initial conditions are slightl more complicated here. In general, we want to consider projectiles that are launched with an initial speed v at an angle θ from the horizontal. Figure 5.a shows a projectile fired with θ>. Notice that an initial angle of θ<would mean a downward initial velocit. As shown in Figure 5.b, the initial velocit can be separated into horizontal and vertical components. From elementar trigonometr, the horizontal component of the initial velocit is v = v cos θ and the vertical component is v = v sin θ. 8 EXAMPLE 5. The Motion of a Projectile in Two Dimensions An object is launched at angle θ = π/6 from the horizontal with initial speed v = 98 m/s. Determine the time of flight and the (horizontal) range of the projectile. Solution Starting with the vertical component of the motion (and again ignoring air resistance), we have (t) = 9.8 (since the initial speed is given in terms of meters per second). Referring to Figure 5.b, notice that the vertical component of the initial velocit is () = 98 sin π/6 = 9 and the initial altitude is () =. A pair of simple integrations gives us the velocit function (t) = 9.8t + 9 and the position function (t) =.9t + 9t. The object hits the ground when (t) = (i.e., when its height above the ground is ). Solving = (t) =.9t + 9t = 9t(.t) 6 8 FIGURE 5.5 Path of ball gives t = (launch time) and t = (landing time). The time of flight is then seconds. The horizontal component of motion is determined from the equation (t) = with initial velocit () = 98 cos π/6 = 9 3 and initial position () =. Integration gives us (t) = 9 3 and (t) = (9 3)t. InFigure 5.5, we plot the path of the ball. [You can do this using the parametric plot mode on our

50 5-7 SECTION Projectile Motion 77 REMARK 5. You should resist the temptation to reduce this section to a few memorized formulas. It is true that if ou ignore air resistance, the vertical component of position will alwas turn out to be (t) = gt + (v sin θ)t + (). However, our understanding of the process and our chances of finding the correct answer will improve dramaticall if ou start each problem with Newton s second law and work through the integrations (which are not difficult). graphing calculator or CAS, b entering equations for (t) and (t) and setting the range of t-values to be t. Alternativel, ou can easil solve for t, toget t = 9,tosee that the curve is simpl a parabola.] The horizontal range is then the 3 value of (t) att = (the landing time), EXAMPLE 5.5 () = (9 3)() = meters. The Motion of a Tennis Serve Venus Williams has one of the fastest serves in women s tennis. Suppose that she hits a serve from a height of feet at an initial speed of mph and at an angle of 7 below the horizontal. The serve is in if the ball clears a 3 -high net that is 39 awa and hits the ground in front of the service line 6 awa. (We illustrate this situation in Figure 5.6.) Determine whether the serve is in or out. ft 7º 3 ft 39 ft 6 ft FIGURE 5.6 Height of tennis serve Solution As in eample 5., we start with the vertical motion of the ball. Since distance is given in feet, the equation of motion is (t) = 3. The initial speed must be converted to feet per second: mph = 58 ft/s = 76 ft/s. The vertical 36 component of the initial velocit is then () = 76 sin( 7 ).5 ft/s. Integration then gives us (t) = 3t.5. The initial height is () = ft, so another integration gives us (t) = 6t.5t + ft. The horizontal component of motion is determined from (t) =, with initial velocit () = 76 cos( 7 ) 7.69 ft/s and initial position () =. Integrations give us (t) = 7.69 ft/s and (t) = 7.69t ft. Summarizing, we have (t) = 7.69t, (t) = 6t.5t +. For the ball to clear the net, must be at least 3 when = 39. We have (t) = 39 when 7.69t = 39 or t.33. At this time, (.33)., showing that the ball is high enough to clear the net. The second requirement is that we need to have 6 when the ball lands ( = ). We have (t) = when 6t.5t + =. From the quadratic formula, we get t.7 and t.366. Ignoring the negative solution, we compute (.366) 63.97, so that the serve lands nearl four feet beond the service line. The serve is not in.

51 78 CHAPTER 5.. Applications of the Definite Integral 5-8 One reason ou should start each problem with Newton s second law is so that ou pause to consider the forces that are (and are not) being considered. For eample, we have thus far ignored air resistance, as a simplification of realit. Some calculations using such simplified equations are reasonabl accurate. Others, such as in eample 5.6, are not. EXAMPLE 5.6 An Eample Where Air Resistance Can t Be Ignored Suppose a raindrop falls from a cloud 3 feet above the ground. Ignoring air resistance, how fast would the raindrop be falling when it hits the ground? Solution If the height of the raindrop at time t is given b (t), Newton s second law of motion tells us that (t) = 3. Further, we have the initial velocit () = (since the drop falls as opposed to being thrown down) and the initial altitude () = 3. Integrating and using the initial conditions gives us (t) = 3t and (t) = 3 6t. The raindrop hits the ground when (t) =. Setting = (t) = 3 6t gives us t = 3/ seconds. The velocit at this time is then ( 3/6) = 3 3/ ft/s. This corresponds to nearl 3 mph! Fortunatel, air resistance does pla a significant role in the fall of a raindrop, which has an actual landing speed of about mph. FIGURE 5.7 Cross section of a wing Regulation baseball, showing stitching The obvious lesson from eample 5.6 is that it is not alwas reasonable to ignore air resistance. Some of the mathematical tools needed to more full analze projectile motion with air resistance are developed in Chapter 7. The air resistance (more precisel, air drag) that slows the raindrop down is onl one of the was in which air can affect the motion of an object. The Magnus force, produced b the spinning of an object or lack of smmetr in the shape of an object, can cause the object to change directions and curve. Perhaps the most common eample of a Magnus force occurs on an airplane. One side of an airplane wing is curved and the other side is comparativel flat (see Figure 5.7). The lack of smmetr causes the air to move over the top of the wing faster than it moves over the bottom. This produces a Magnus force in the upward direction (lift), lifting the airplane into the air. A more down-to-earth eample of a Magnus force occurs in an unusual baseball pitch called the knuckleball. To throw this pitch, the pitcher grips the ball with his fingertips and throws the ball with as little spin as possible. Baseball plaers claim that the knuckleball dances around unpredictabl and is eceptionall hard to hit or catch. There still is no complete agreement on wh the knuckleball moves so much, but we will present one current theor due to phsicists Robert Watts and Terr Bahill. The cover of the baseball is sewn on with stitches that are raised up slightl from the rest of the ball. These curved stitches act much like an airplane wing, creating a Magnus force that affects the ball. The direction of the Magnus force depends on the eact orientation of the ball s stitches. Measurements b Watts and Bahill indicate that the lateral force (left/right from the pitcher s perspective) is approimatel F m =. sin(θ) lb, where θ is the angle (in radians) of the ball s position rotated from a particular starting position. Since gravit does not affect the lateral motion of the ball, the onl force acting on the ball laterall is the Magnus force. Newton s second law applied to the lateral motion of the

52 5-9 SECTION Projectile Motion 79 knuckleball gives m (t) =. sin(θ). The mass of a baseball is about. slug. (Slugs are the units of measurement of mass in the English sstem. To get the more familiar weight in pounds, simpl multipl the mass b g = 3.) Wenowhave (t) = sin(θ). If the ball is spinning at the rate of ω radians per second, then θ = ωt + θ, where the initial angle θ depends on where the pitcher grips the ball. We then have (t) = sin(ωt + θ ), (5.) with initial conditions () = and () =. For a tpical knuckleball speed of 6 mph, it takes about.68 second for the pitch to reach home plate. EXAMPLE 5.7 An Equation for the Motion of a Knuckleball Foraspin rate of ω = radians per second and θ =, find an equation for the motion of the knuckleball and graph it for t.68. Repeat this for θ = π/. Solution For θ =, Newton s second law gives us (t) = sin 8t, from (5.). Integrating this and using the initial condition () = gives us...6 t FIGURE 5.8a Lateral motion of a knuckleball for θ =....6 t (t) = [ cos 8t ( cos )] =.5(cos 8t ). 8 Integrating once again and using the second initial condition () =, we get ( ) (t) =.5 (sin 8t ).5t =.565 sin 8t.5t. 8 A graph of this function shows the lateral motion of the ball (see Figure 5.8a). The graph shows the path of the pitch as it would look viewed from above. Notice that after starting out straight, this pitch breaks nearl a foot awa from the center of home plate! For the case where θ = π/, we have from (5.) that ( (t) = sin 8t + π ). Integrating this and using the first initial condition gives us (t) = { ( cos 8t + π ) [ ( cos + π )]} ( =.5 cos 8t + π ) FIGURE 5.8b Lateral motion of a knuckleball for θ = π Integrating a second time ields (t) =.5 ( ) [ ( sin 8t + π ) ( π )] [ ( sin =.565 sin 8t + π ) ]. 8 A graph of the lateral motion in this case is shown in Figure 5.8b. Notice that this pitch breaks nearl inches to the pitcher s right and then curves back over the plate for a strike! You can see that, in theor, the knuckleball is ver sensitive to spin and initial position and can be ver difficult to hit when thrown properl.

53 8 CHAPTER 5.. Applications of the Definite Integral 5-5 EXERCISES 5.5 WRITING EXERCISES. In eample 5.6, the assumption that air resistance can be ignored is obviousl invalid. Discuss the validit of this assumption in eamples 5. and In the discussion preceding eample 5.3, we showed that Michael Jordan (and an other human) spends half of his airtime in the top one-fourth of the height. Compare his velocities at various points in the jump to eplain wh relativel more time is spent at the top than at the bottom. 3. In eample 5., we derived separate equations for the horizontal and vertical components of position. To discover one consequence of this separation, consider the following situation. Two people are standing net to each other with arms raised to the same height. One person fires a bullet horizontall from a gun. At the same time, the other person drops a bullet. Eplain wh the bullets will hit the ground at the same time.. For the falling raindrop in eample 5.6, a more accurate model would be (t) = 3 + f (t), where f (t) represents the force due to air resistance (divided b the mass). If v(t)is the downward velocit of the raindrop, eplain wh this equation is equivalent to v (t) = 3 f (t). Eplain in phsical terms wh the larger v(t) is, the larger f (t) is. Thus, a model such as f (t) = v(t) or f (t) = [v(t)] would be reasonable. (In most situations, it turns out that [v(t)] matches the eperimental data better.) In eercises, identif the initial conditions () and ().. An object is dropped from a height of 8 feet.. An object is dropped from a height of feet. 3. An object is released from a height of 6 feet with an upward velocit of ft/s.. An object is released from a height of feet with a downward velocit of ft/s. In eercises 5 56, ignore air resistance. 5. Adiver drops from 3 feet above the water (about the height of an Olmpic platform dive). What is the diver s velocit at impact? 6. Adiver drops from feet above the water (about the height of divers at the Acapulco Cliff Diving competition). What is the diver s velocit at impact? 7. Compare the impact velocities of objects falling from 3 feet (eercise 5), feet (eercise 6) and 3 feet (eample 5.6). If height is increased b a factor of h, b what factor does the impact velocit increase? 8. The Washington Monument is 555 feet, 5 inches high. In a 8 famous eperiment, a baseball was dropped from the top of the monument to see if a plaer could catch it. How fast would the ball be going? 9. A certain not-so-wil coote discovers that he just stepped off the edge of a cliff. Four seconds later, he hits the ground in a puff of dust. How high was the cliff?. A large boulder dislodged b the falling coote in eercise 9 falls for 3 seconds before landing on the coote. How far did the boulder fall? What was its velocit when it flattened the coote?. The coote s net scheme involves launching himself into the air with an Acme catapult. If the coote is propelled verticall from the ground with initial velocit 6 ft/s, find an equation for the height of the coote at an time t. Find his maimum height, the amount of time spent in the air and his velocit when he smacks back into the catapult.. On the rebound, the coote in eercise is propelled to a height of 56 feet. What is the initial velocit required to reach this height? 3. One of the authors has a vertical jump of inches. What is the initial velocit required to jump this high? How does this compare to Michael Jordan s velocit, found in eample 5.3?. If the author underwent an eercise program and increased his initial velocit b %, b what percentage would he increase his vertical jump? 5. Show that an object dropped from a height of H feet will hit the ground at time T = H seconds with impact velocit V = 8 H ft/s. 6. Show that an object propelled from the ground with initial velocit v ft/s will reach a maimum height of v /6 ft. 7. You can measure our reaction time using a ruler. Hold our thumb and forefinger on either side of a ardstick. Have a friend drop the ardstick and grab it as fast as ou can. Take the distance d that the ardstick falls and compute how long the ruler fell. Show that if d is measured in cm, our reaction time is approimatel t.5 d.for comparison purposes, a top athlete has a reaction time of about.5 s. 8. The coefficient of restitution of a ball measures how livel the bounce is. B definition, the coefficient equals v v, where v is the (downward) speed of the ball when it hits the ground and v is the (upward) launch speed after it hits the ground. If a ball is dropped from a height of H feet and rebounds to a height of ch for some constant c with < c <, compute its coefficient of restitution.

54 5-5 SECTION Projectile Motion 8 9. According to legend, Galileo dropped two balls from the Leaning Tower of Pisa. When both the heav lead ball and the light wood ball hit the ground at the same time, Galileo knew that the acceleration due to gravit is the same for all objects. Unfortunatel, air resistance would affect such an eperiment. Taking into account air resistance, a 6 wood ball would fall f (t) = 75 ln [ cosh ( 6 t)] feet in t seconds, 8 85 while a 6 lead ball would fall g(t) =,8 ln [ cosh ( t)] feet, where cosh = (e + e ). From a height of 79 feet, find the height of the wood ball when the lead ball hits the ground.. If a theatrical production wanted to show the balls of eercise 9 landing at the same time, how much earlier would the wood ball need to be released? In eercises 3, sketch the parametric graphs as in eample 5. to indicate the flight path.. An object is launched at angle θ = π/3 radians from the horizontal with an initial speed of 98 m/s. Determine the time of flight and the horizontal range. Compare to eample m/s u. Find the time of flight and horizontal range of an object launched at angle 3 with initial speed ft/s. Repeat with an angle of Repeat eample 5.5 with an initial angle of 6.Btrial and error, find the smallest and largest angles for which the serve will be in.. Repeat eample 5.5 with an initial speed of 7 ft/s. B trial and error, find the smallest and largest initial speeds for which the serve will be in. 5. A baseball pitcher releases the ball horizontall from a height of 6 ft with an initial speed of 3 ft/s. Find the height of the ball when it reaches home plate 6 feet awa. (Hint: Determine the time of flight from the -equation, then use the -equation to determine the height.) 6 ft 3 ft/s 6 6. Repeat eercise 5 with an initial speed of 8 ft/s. (Hint: Carefull interpret the negative answer.) 7. A baseball plaer throws a ball toward first base feet awa. The ball is released from a height of 5 feet with an initial speed of ft/s at an angle of 5 above the horizontal. Find the height of the ball when it reaches first base. 8. B trial and error, find the angle at which the ball in eercise 7 will reach first base at the catchable height of 5 feet. At this angle, how far above the first baseman s head would the thrower be aiming? 9. A daredevil plans to jump over 5 cars. If the cars are all compact cars with a width of 5 feet and the ramp angle is 3, determine the initial velocit required to complete the jump successfull. Repeat with a takeoff angle of 5.Inspite of the reduced initial velocit requirement, wh might the daredevil prefer an angle of 3 to 5? 3. A plane at an altitude of 56 feet wants to drop supplies to a specific location on the ground. If the plane has a horizontal velocit of ft/s, how far awa from the target should the plane release the supplies in order to hit the target location? (Hint: Use the -equation to determine the time of flight, then use the -equation to determine how far the supplies will drift.) 3. Consider a knuckleball (see eample 5.7) with lateral motion satisfing the initial value problem (t) = 5 sin(ωt + θ ), () = () =. With ω =, find an equation for (t) and graph the solution for t.68 with (a) θ = and (b) θ = π/. 3. Repeat eercise 3 for θ = π/ and (a) ω = and (b) ω =. 33. For the Olmpic diver in eercise 5, what would be the average angular velocit (measured in radians per second) necessar to complete somersaults? 3. In the Fling Zucchini Circus human cannonball act, a performer is shot out of a cannon from a height of feet at an angle of 5 with an initial speed of 6 ft/s. If the safet net stands 5 feet above the ground, how far should the safet net be placed from the cannon? If the safet net can withstand an impact velocit of onl 6 ft/s, will the Fling Zucchini land safel or come down squash? 35. In a basketball free throw, a ball is shot from a height of h feet toward a basket feet above the ground at a horizontal distance of 5 feet. (See the figure on the following page.) If h = 6,θ = 5 and v = 5 ft/s, show that the free throw is good. Since the basket is larger than the ball, a free throw has a margin of error of several inches. If an shot that passes through height ft with is good, show that, for the given initial speed v, the margin of error is 8 θ 57. Sketch parametric graphs to show several of these free throws.

55 8 CHAPTER 5.. Applications of the Definite Integral 5-5 h v u 5 ft ft 36. For the basketball shot in eercise 35, fi h = 6 and θ = 5. Find the range of speeds v for which the free throw will be good. Given that it requires a force of F =.v pounds to launch a free throw with speed v,how much margin of error is there in the applied force? 37. Soccer plaer Roberto Carlos of Brazil is known for his curving kicks. Suppose that he has a free kick from 3 ards out. Orienting the - and -aes as shown in the figure, suppose the kick has initial speed ft/s at an angle of 5 from the positive -ais. Assume that the onl force on the ball is a Magnus force to the left caused b the spinning of the ball. With (t) = and (t) =, determine whether the ball goes in the goal at = 9 and. u 38. In eercise 37, a wall of plaers is lined up ards awa, etending from = to =. Determine whether the kick goes around the wall. 39. To train astronauts to operate in a weightless environment, NASA sends them up in a special plane (nicknamed the Vomit Comet). To allow the passengers to eperience weightlessness, the vertical acceleration of the plane must eactl match the acceleration due to gravit. If (t)isthe vertical acceleration of the plane, then (t) = g. Show that, for a constant horizontal velocit, the plane follows a parabolic path. NASA s plane flies parabolic paths of approimatel 5 feet in height (5 feet up and 5 down). The time to complete such a path is the amount of weightless time for the passengers. Compute this time.. In the 99 Summer Olmpics in Barcelona, Spain, an archer lit the Olmpic flame b shooting an arrow toward a cauldron at a distance of about 7 meters horizontall and 3 meters verticall. If the arrow reached the cauldron at the peak of its trajector, determine the initial speed and angle of the arrow. (Hint: Show that (t) = ift = (v sin θ)/9.8. For this t, show that (t) (t) = cot θ = 7 3 and solve for θ. Then solve for v.) In eercises 6, we eplore two aspects of juggling. More information can be found in The Mathematics of Juggling b Burkard Polster.. Professional jugglers generall agree that is the maimum number of balls that a human being can successfull maintain. To get an idea wh, suppose that it takes second to catch and toss a ball. (In other words, using both hands, the juggler can process balls per second.) To juggle balls, each ball would need to be in the air for.5 seconds. Neglecting air resistance, how high would the ball have to be tossed to sta in the air this long? How much higher would the balls need to be tossed to juggle balls?. Another aspect of juggling balls is accurac. A ball juggled from the right hand to the left hand must travel the correct horizontal distance to be catchable. Suppose that a ball is tossed with initial horizontal velocit v and initial vertical velocit v. Assume that the ball is caught at the height from which it is thrown. Show that the horizontal distance traveled is w = v v feet. (Hint: This is a basic projectile problem, 6 like eample 5..) 3. Referring to eercise, suppose that a ball is tossed at an angle of α from the vertical. Show that tan α = v. Combining v this result with eercises 6 and, show that w = h tan α, where h is the maimum height of the toss.. Find a linear approimation for tan at =. Use this approimation and eercise 3 to show that α w h.ifan angle of α produces a distance of w and an angle of α + α produces a distance of w + w, show that α w h. 5. Suppose that w is the difference between the ideal horizontal distance for a toss and the actual horizontal distance of a toss. For the average juggler, an error of w = foot is manageable. Let α be the corresponding error in the angle of toss. If h is the height needed to juggle balls (see eercise ), find the maimum error in tossing angle. 6. Repeat eercise 5 using the height needed to juggle balls. How much more accurate does the juggler need to be to juggle balls? 7. In a tpical golf tee shot, the ball is launched at an angle of 9.3 at an initial speed of ft/s. In the absence of air resistance, how far (horizontall) would this shot carr? The actual carr on such a shot is about ards (7 feet). In this case, a backspin of rpm gives the ball a huge upward Magnus force, which offsets most of the air resistance and gravit.

56 5-53 SECTION Projectile Motion Suppose that a firefighter holds a water hose at slope m and the water eits the hose with ( speed v ft/s. ) Show that the water + m follows the path = 6 m. Ifthe firefighter stands feet from a wall, what is the maimum height v that the water can reach? 9. Astronaut Alan Shepard modified some of his lunar equipment and became the onl person to hit a golf ball on the Moon. Assume that the ball was hit with speed 6 ft/s at an angle of 5 above the horizontal. Assuming no air resistance, find the distance the ball would have traveled on Earth. Then find how far itwould travel on the moon, where there reall is no air resistance (use g = 5. ft/s ). 55. How fast is a vert skateboarder like Ton Hawk going at the bottom of a ramp? Ignoring friction and air resistance, the answer comes from conservation of energ, which states that the kinetic energ mv plus the potential energ mg remains constant. Assume that the energ at the top of a trick at height H is all potential energ and the energ at the bottom of the ramp is all kinetic energ. (a) Find the speed at the bottom as a function of H. (b) Compute the speed if H = 6 feet. (c) Find the speed halfwa down ( = 8). (d) If the ramp has the shape = for, find the horizontal and vertical components of speed halfwa at = 8. h' 3' 5. The gravitational force of the Moon is about one-sith that of Earth. A simple guess might be that a golf ball would travel si times as high and si times as far on the Moon compared to on Earth. Determine whether this is correct. 5. Sprinters have been timed at speeds up to 8 mph. If this could be converted to a vertical velocit, how high would the person jump on Earth? Ignore air resistance. 5. One version of Murph s Law states that a piece of bread with butter on it will alwas fall off the table and land butter-side down. This is actuall more a result of phsics than of bad luck. An object knocked off of a table will tpicall rotate with an angular velocit ω rad/s. At a constant angular velocit ω, the object will rotate ωt radians in t seconds. Let θ = represent a flat piece of bread with butter-side up. If the bread hits the floor with π <θ< 3π,ithas landed butter-side down. Assume that the bread falls from a height of 3 feet and with an initial tilt of θ = π. Find the range of ω-values such that the bread falls butter-side down. For a falling piece of bread, ω is fairl small. Based on our calculation, if ω varies from revolution per second to revolution per second, how likel is the bread to land butter-side down? 53. Suppose a target is dropping verticall at a horizontal distance of feet from ou. If ou fire a paint ball directl at the target, show that ou will hit it (assuming no air resistance and assuming that the paint ball reaches the target before either hits the ground). 5. An object is dropped from a height of feet. Another object directl below the first is launched verticall from the ground with initial velocit ft/s. Determine when and how high up the objects collide. ' Eercise 55. Eercise A science class builds a ramp to roll a bowling ball out of a window that is 3 feet above the ground. Their goal is for the ball to land on a watermelon that is feet from the building. Assuming no friction or air resistance, determine how high the ramp should be to smash the watermelon. EXPLORATORY EXERCISES. In the tet and eercises 3 and 3, we discussed the differential equation (t) = 5 sin(ωt + θ ) for the lateral motion of a knuckleball. Integrate and appl the initial conditions () = and () = toderive the general equation (t) = 5 ( ) 5 6ω sin(ωt + θ ) ω cos θ t 5 6ω sin θ. If ou have access to three-dimensional graphics, graph (t,ω) for θ = with t.68 and. ω. (Note: Some plotters will have trouble with ω =.) Repeat with θ = π/,θ = π/ and two choices of our own for θ. A pitcher wants the ball to move as much as possible back and forth but end up near home plate ( = ). Based on these criteria, pick the combinations of θ and ω that produce the four best pitches. Graph these pitches in two dimensions with = (t) asinfigures 5.8a and 5.8b.. Although we have commented on some inadequacies of the gravit-onl model of projectile motion, we have not presented an alternatives. Such models tend to be somewhat more mathematicall comple. The model eplored in this eercise takes into account air resistance in a wa that is mathematicall tractable but still not ver realistic. Assume that the force of

57 8 CHAPTER 5.. Applications of the Definite Integral 5-5 air resistance is proportional to the speed and acts in the opposite direction of velocit. For a horizontal motion (no gravit), we have a(t) = F(t)/m = cv(t) for some constant c. Eplain what the minus sign indicates. Since a(t) = v (t), the model is v (t) = cv(t). Show that the function v(t) = v e ct satisfies the equation v (t) = cv(t) and the initial condition v() = v. If an object starts at () = a, integrate v(t) = v e ct to find its position at an time t. Show that the amount of time needed to reach = b (for a < b)isgivenbt = ( c ln c b a ). For abaseball (c =.5) thrown at 5 ft/s from a =, determine how long it takes to reach b = 6 and compute its velocit at that point. B what percentage has the velocit decreased? In baseball, two different tpes of radar guns are used to measure the velocit of a pitch. One measures the speed of the ball right as it leaves the pitcher s hand. The other measures the speed of the ball partwa to home plate. If the first gun registers 9 mph and the second registers 89 mph, how far from the plate does the second gun make its measurement? 3. The goal in the old computer game called Gorillas is to enter a speed and angle to launch an eplosive banana to tr v to hit a gorilla at some other location. Suppose that ou are located at the origin and our opponent at (, ). (a) Find two speed/angle combinations that will hit the gorilla. (b) Estimate the smallest speed that can be used to hit the target. (c) Repeat parts (a) and (b) if there is a building in the wa that occupies 3 and APPLICATIONS OF INTEGRATION TO PHYSICS AND ENGINEERING In this section, we eplore several applications of integration in phsics. In each case, we will define a basic concept to help solve a specific problem. We ll then use the definite integral to generalize the concept to solve a much wider range of problems. This use of integration is an ecellent eample of how the mathematical theor helps ou find solutions of practical problems. Imagine that ou are at the bottom of a snow-covered hill with a sled. To get a good ride, ou want to push the sled as far up the hill as ou can. A phsicist would sa that the higher up ou are, the more potential energ ou have. Sliding down the hill converts the potential energ into kinetic energ. (This is the fun part!) But pushing the sled up the hill requires ou to do some work: ou must eert a force over a long distance. Our first task is to quantif work. If ou push a sled up a hill, ou re doing work, but can ou give a measure of how much? Certainl, if ou push twice the weight (i.e., eert twice the force), ou re doing twice the work. Further, if ou push the sled twice as far, ou ve done twice the work. In view of these observations, for an constant force F applied overadistance d, wedefine the work W done as W = Fd. Unfortunatel, forces are generall not constant. We etend this notion of work to the case of a nonconstant force F() applied on the interval [a, b] asfollows. First, we divide the interval [a, b] into n equal subintervals, each of width = b a and consider the work n done on each subinterval. If is small, then the force F() applied on the subinterval [ i, i ] can be approimated b the constant force F(c i ) for some point c i [ i, i ]. The work done moving the object along the subinterval is then approimatel F(c i ).

58 8 CHAPTER 5.. Applications of the Definite Integral 5-5 air resistance is proportional to the speed and acts in the opposite direction of velocit. For a horizontal motion (no gravit), we have a(t) = F(t)/m = cv(t) for some constant c. Eplain what the minus sign indicates. Since a(t) = v (t), the model is v (t) = cv(t). Show that the function v(t) = v e ct satisfies the equation v (t) = cv(t) and the initial condition v() = v. If an object starts at () = a, integrate v(t) = v e ct to find its position at an time t. Show that the amount of time needed to reach = b (for a < b)isgivenbt = ( c ln c b a ). For abaseball (c =.5) thrown at 5 ft/s from a =, determine how long it takes to reach b = 6 and compute its velocit at that point. B what percentage has the velocit decreased? In baseball, two different tpes of radar guns are used to measure the velocit of a pitch. One measures the speed of the ball right as it leaves the pitcher s hand. The other measures the speed of the ball partwa to home plate. If the first gun registers 9 mph and the second registers 89 mph, how far from the plate does the second gun make its measurement? 3. The goal in the old computer game called Gorillas is to enter a speed and angle to launch an eplosive banana to tr v to hit a gorilla at some other location. Suppose that ou are located at the origin and our opponent at (, ). (a) Find two speed/angle combinations that will hit the gorilla. (b) Estimate the smallest speed that can be used to hit the target. (c) Repeat parts (a) and (b) if there is a building in the wa that occupies 3 and APPLICATIONS OF INTEGRATION TO PHYSICS AND ENGINEERING In this section, we eplore several applications of integration in phsics. In each case, we will define a basic concept to help solve a specific problem. We ll then use the definite integral to generalize the concept to solve a much wider range of problems. This use of integration is an ecellent eample of how the mathematical theor helps ou find solutions of practical problems. Imagine that ou are at the bottom of a snow-covered hill with a sled. To get a good ride, ou want to push the sled as far up the hill as ou can. A phsicist would sa that the higher up ou are, the more potential energ ou have. Sliding down the hill converts the potential energ into kinetic energ. (This is the fun part!) But pushing the sled up the hill requires ou to do some work: ou must eert a force over a long distance. Our first task is to quantif work. If ou push a sled up a hill, ou re doing work, but can ou give a measure of how much? Certainl, if ou push twice the weight (i.e., eert twice the force), ou re doing twice the work. Further, if ou push the sled twice as far, ou ve done twice the work. In view of these observations, for an constant force F applied overadistance d, wedefine the work W done as W = Fd. Unfortunatel, forces are generall not constant. We etend this notion of work to the case of a nonconstant force F() applied on the interval [a, b] asfollows. First, we divide the interval [a, b] into n equal subintervals, each of width = b a and consider the work n done on each subinterval. If is small, then the force F() applied on the subinterval [ i, i ] can be approimated b the constant force F(c i ) for some point c i [ i, i ]. The work done moving the object along the subinterval is then approimatel F(c i ).

59 5-55 SECTION Applications of Integration to Phsics and Engineering 85 The total work done W is then approimatel n W F(c i ). i= You should recognize this as a Riemann sum, which, as n gets larger, approaches the actual work, Work W = lim n n F(c i ) = i= b a F() d. (6.) We take (6.) as our definition of work. You ve probabl noticed that the farther a spring is compressed (or stretched) from its natural resting position, the more force is required to further compress (or stretch) the spring. According to Hooke s Law, the force required to maintain a spring in a given position is proportional to the distance it s compressed (or stretched). That is, if is the distance a spring is compressed (or stretched) from its natural length, the force F() eerted b the spring is given b for some constant k (the spring constant). F() = k, (6.) FIGURE 5.9 Stretched spring EXAMPLE 6. Computing the Work Done Stretching a Spring A force of 3 pounds stretches a spring foot from its natural length (see Figure 5.9). Find the work done in stretching the spring 6 inches beond its natural length. Solution First, we determine the value of the spring constant. From Hooke s Law (6.), we have that ( ) ( ) 3 = F = k, so that k = and F() =. From (6.), the work done in stretching the spring 6 inches (/ foot) is then W = / F() d = / d= 3 foot-pounds. In this case, notice that stretching the spring transfers potential energ to the spring. (If the spring is later released, it springs back toward its resting position, converting the potential energ to kinetic energ.) EXAMPLE 6. Computing the Work Done b a Weightlifter A weightlifter lifts a -pound barbell a distance of 3 feet. How much work was done? Also, determine the work done b the weightlifter if the weight is raised feet above the ground and then lowered back into place. Solution Since the force (the weight) is constant here, we simpl have W = Fd = 3 = 6 foot-pounds. On the other hand, find the amount of work done if the weightlifter lifts the same weight feet from the ground and then lowers it back into place. It ma seem strange, but since the barbell ends up in the same place as it started, the net distance covered is zero and the work done is zero. Of course, it would feel like work to the weightlifter, but this is where the mathematical notion of work differs from the usual use of the word. As we

60 86 CHAPTER 5.. Applications of the Definite Integral 5-56 have defined it, work accounts for the energ change in the object. Since the barbell has the same kinetic and potential energ that it started with, the total work done on it is zero. In eample 6.3, both the force and the distance are nonconstant. This presents some unique challenges and we ll need to first approimate the work and then recognize the definite integral that this approimation process generates. EXAMPLE 6.3 Computing the Work Required to Pump the Water Out of a Tank A spherical tank of radius feet is filled with water. Find the work done in pumping all of the water out through the top of the tank. Solution The basic formula W = Fd does not directl appl here, for several reasons. The most obvious of these is that the distance traveled b the water in each part of the tank is different, as the water toward the bottom of the tank must be pumped all the wa to the top, while the water near the top of the tank must be pumped onl a short distance. Let represent distance as measured from the bottom of the tank, as in Figure 5.5a. The entire tank corresponds to the interval, which we partition into = < < < n =, where i i = =, for each i =,,...,n. This partitions the tank into n n thin laers, each corresponding to an interval [ i, i ] (see Figure 5.5b). You can think of the water in the laer corresponding to [ i, i ]asbeing approimatel clindrical, of height. This laer must be pumped a distance of approimatel c i, for some c i [ i, i ]. Notice from Figure 5.5b that the radius of the ith laer depends on the value of. From Figure 5.5c (where we show a cross section of the tank), the radius r i corresponding to a depth of = c i is the base of a right triangle with hpotenuse and height c i. From the Pthagorean Theorem, we now have ( c i ) + r i =. r i c i c i r i c i ci FIGURE 5.5a Spherical tank FIGURE 5.5b The ith slice of water FIGURE 5.5c Cross section of tank

61 5-57 SECTION Applications of Integration to Phsics and Engineering 87 Solving this for ri,wehave ri = ( c i ) = ( c i + ci ) = c i ci. The force F i required to move the ith laer is then simpl the force eerted on the water b gravit (i.e., its weight). For this we will need to know the weight densit of water: 6. lb/ft 3.Wenowhave F i (Volume of clindrical slice) (weight of water per unit volume) = ( πri h) (6. lb/ft 3 ) = 6.π ( c i ci ). The work required to pump out the ith slice is then given approimatel b W i (Force) (distance) = 6.π ( c i ci ) ( ci ) = 6.πc i ( c i ). The work required to pump out all of the water is then the sum of the work required for each of the n slices: n W 6.πc i ( c i ). i= Finall, taking the limit as n gives the eact work, which ou should recognize as a definite integral: n W = lim 6.πc i ( c i ) = 6.π ( ) d n i= = 6.π ( + 3 ) d = 6.π [ ] ( ), = 6.π.6 6 foot-pounds. 3 Impulse is a phsical quantit closel related to work. Instead of relating force and distance to account for changes in energ, impulse relates force and time to account for changes in velocit. First, suppose that a constant force F is applied to an object from time t = totime t = T.Ifthe position of the object at time t is given b (t), then Newton s second law sas that F = ma = m (t). Integrating this equation once with respect to t gives us T T Fdt = m (t) dt, or F(T ) = m[ (T ) ()]. Recall that (t) isthe velocit v(t), so that FT = m[v(t ) v()]

62 88 CHAPTER 5.. Applications of the Definite Integral 5-58 or FT = m v, where v = v(t ) v() is the change in velocit. The quantit FT is called the impulse, mv(t)is the momentum at time t and the equation relating the impulse to the change in velocit is called the impulse-momentum equation. Since we defined the impulse for a constant force, we must generalize the notion to the case of a nonconstant force. Think about this and tr to guess what the definition should be. We define the impulse J of a force F(t) applied over the time interval [a, b] tobe Impulse J = b a F(t) dt. We leave the derivation of the impulse integral for the case of a nonconstant force as an eercise. The impulse-momentum equation also generalizes to the case of a nonconstant force: Impulse-momentum equation J = m[v(b) v(a)]. EXAMPLE 6. Estimating the Impulse for a Baseball Suppose that a baseball traveling at 3 ft/s (about 9 mph) collides with a bat. The following data (adapted from The Phsics of Baseball b Robert Adair) shows the force eerted b the bat on the ball at.-second intervals. t (s) F(t) (lb) d d m m FIGURE 5.5a Balancing two masses Estimate the impulse of the bat on the ball and (using m =. slug) the speed of the ball after impact. Solution In this case, the impulse J is given b.7 F(t) dt. Since we re given onl afied number of measurements of F(t), the best we can do is approimate the integral numericall (e.g., using Simpson s Rule). Recall that Simpson s Rule requires an even number n of subintervals, which means that ou need an odd number n + ofpoints in the partition. Using n = 8 and adding afunction value at t =.8 (wh is it fair to do this?), Simpson s Rule gives us J [ + (5) + (5) + (75) + (9) + (55) + (5) + () + ] In this case, the impulse-momentum equation J = m v becomes.867 =. v or v = 86.7 ft/s. Since the ball started out with a speed of 3 ft/s in one direction and it ended up moving in the opposite direction, the speed after impact is 56.7 ft/s. The concept of the first moment, like work, involves force and distance. Moments are used to solve problems of balance and rotation. Consider two children on a plaground seesaw (or teeter-totter). Suppose that the child on the left in Figure 5.5a is heavier (i.e., has larger mass) than the child on the right. If the children sit an equal distance from the pivot point, ou know what will happen: the left side will be pulled down. However, the

63 5-59 SECTION Applications of Integration to Phsics and Engineering 89 children can balance each other if the heavier child moves closer to the pivot point. That is, the balance is determined both b weight (force) and distance from the pivot point. If the children have masses m and m and are sitting at distances d and d, respectivel, from the pivot point, then the balance each other if and onl if m d = m d. (6.3) Let s turn the problem around slightl. Suppose there are two objects, of mass m and m, located at and, respectivel, with <.Weconsider the objects to be pointmasses. That is, each is treated as a single point, with all of the mass concentrated at that point (see Figure 5.5b). m m FIGURE 5.5b Two point-masses Suppose that ou want to find the center of mass, that is, the location at which ou could place the pivot of a seesaw and have the objects balance. From the balance equation (6.3), ou ll need m ( ) = m ( ). Solving this equation for gives us = m + m. m + m Notice that the denominator in this equation is the total mass of the sstem (i.e., the total mass of the two objects). The numerator of this epression is called the first moment of the sstem. More generall, for a sstem of n masses m, m,...,m n, located at =,,..., n, respectivel, the center of mass is given b the first moment divided b the total mass, that is, Center of mass = m + m + +m n n m + m + +m n. Now, suppose that we wish to find the mass and center of mass of an object of variable densit that etends from = a to = b. Here, we assume that the densit function ρ() (measured in units of mass per unit length) is known. Note that if the densit is a constant ρ, the mass of the object is simpl given b m = ρl, where L = b a is the length of the object. On the other hand, if the densit varies throughout the object, we can approimate the mass b dividing the interval [a, b] into n pieces of equal width = b a.on n each subinterval [ i, i ], the mass is approimatel ρ(c i ), where c i is a point in the subinterval. The total mass is then approimatel n m ρ(c i ). i= You should recognize this as a Riemann sum, which approaches the total mass as n, Mass m = lim n n ρ(c i ) = i= b a ρ() d. (6.)

64 9 CHAPTER 5.. Applications of the Definite Integral 5-6 EXAMPLE 6.5 Computing the Mass of a Baseball Bat A 3-inch baseball bat can be represented approimatel b an object etending from = to = 3 inches, with densit ρ() = ( ) slugs per inch. The densit takes into account the fact that a baseball bat is similar to an elongated cone. Find the mass of the object. Solution From (6.), the mass is given b 3 ( m = 6 + ) d 69 = 69 ( ) 3 [ 3 ( = ) 3 ( ) ] slug. To compute the weight (in ounces), multipl the mass b 3 6. The bat weighs roughl 3.5 ounces. To compute the first moment for an object of nonconstant densit ρ()etending from = a to = b,weagain divide the interval into n equal pieces. From our earlier argument, for each i =,,...,n, the mass of the ith slice of the object is approimatel ρ(c i ), for an choice of c i [ i, i ]. We then represent the ith slice of the object with a particle of mass m i = ρ(c i ) located at = c i.wecan now think of the original object as having been approimated b n distinct point-masses, as indicated in Figure 5.5. m m m 3 m m 5 c c c 3 c c 5 FIGURE 5.5 Si point-masses m 6 c 6 Notice that the first moment M n of this approimate sstem is M n = [ρ(c ) ]c + [ρ(c ) ]c + +[ρ(c n ) ]c n n = [c ρ(c ) + c ρ(c ) + +c n ρ(c n )] = c i ρ(c i ). Taking the limit as n, the sum approaches the first moment i= First moment M = lim n n c i ρ(c i ) = i= b a ρ() d. (6.5) The center of mass of the object is then given b Center of mass = M m = b a b a ρ() d. (6.6) ρ() d

65 5-6 SECTION Applications of Integration to Phsics and Engineering 9 Depth d Hoover Dam FIGURE 5.53 A plate of area A submerged to depth d A EXAMPLE 6.6 Finding the Center of Mass (Sweet Spot) of a Baseball Bat Find the center of mass of the baseball bat from eample 6.5. Solution From (6.5), the first moment is given b 3 ( M = 6 + ) [ d = ,6 +,9, ] 3.5. Recall that we had alread found the mass to be m 6. slug and so, from (6.6), the center of mass of the bat is = M m inches. 6. Note that for a baseball bat, the center of mass is one candidate for the so-called sweet spot of the bat, the best place to hit the ball. For our final application of integration in this section, we consider hdrostatic force. Imagine a dam holding back a lake full of water. What force must the dam withstand? As usual, we solve a simpler problem first. If ou have a flat rectangular plate oriented horizontall underwater, notice that the force eerted on the plate b the water (the hdrostatic force) issimpl the weight of the water ling above the plate. This is the product of the volume of the water ling above the plate and the weight densit of water (6. lb/ft 3 ). If the area of the plate is A ft and it lies d ft below the surface (see Figure 5.53), then the force on the plate is F = 6.Ad. According to Pascal s Principle, the pressure at a given depth d in a fluid is the same in all directions. This sas that if a flat plate is submerged in a fluid, then the pressure on one side of the plate at an given point is ρ d, where ρ is the weight densit of the fluid and d is the depth. In particular, this sas that it s irrelevant whether the plate is submerged verticall, horizontall or otherwise (see Figure 5.5). Consider now a verticall oriented wall (a dam) holding back a lake. It is convenient to orient the -ais verticall with = located at the surface of the water and the bottom of the wall at = a > (see Figure 5.55). In this wa, measures the depth of a section of the dam. Suppose w()isthe width of the wall at depth (where all distances are measured in feet). i Depth d b c w(c i ) i a a FIGURE 5.5 Pressure at a given depth is the same, regardless of the orientation FIGURE 5.55 Force acting on a dam