Applications of Definite Integrals

Transcription

1 58_Ch7_pp qd /3/6 :3 PM Page 378 Chapter 7 Applications of Definite Integrals T he art of potter developed independentl in man ancient civilizations and still eists in modern times. The desired shape of the side of a potter vase can be described b:! 5. " sin (/4) ( # # 8p) where is the height and is the radius at height (in inches). A base for the vase is preformed and placed on a potter s wheel. How much cla should be added to the base to form this vase if the inside radius is alwas inch less than the outside radius? Section 7.3 contains the needed mathematics. 378

2 Chapter 7 Overview Section 7. Integral as Net Change 379 B this point it should be apparent that finding the limits of Riemann sums is not just an intellectual eercise; it is a natural wa to calculate mathematical or phsical quantities that appear to be irregular when viewed as a whole, but which can be fragmented into regular pieces. We calculate values for the regular pieces using known formulas, then sum them to find a value for the irregular whole. This approach to problem solving was around for thousands of ears before calculus came along, but it was tedious work and the more accurate ou wanted to be the more tedious it became. With calculus it became possible to get eact answers for these problems with almost no effort, because in the limit these sums became definite integrals and definite integrals could be evaluated with antiderivatives. With calculus, the challenge became one of fitting an integrable function to the situation at hand (the modeling step) and then finding an antiderivative for it. Toda we can finesse the antidifferentiation step (occasionall an insurmountable hurdle for our predecessors) with programs like NINT, but the modeling step is no less crucial. Ironicall, it is the modeling step that is thousands of ears old. Before either calculus or technolog can be of assistance, we must still break down the irregular whole into regular parts and set up a function to be integrated. We have alread seen how the process works with area, volume, and average value, for eample. Now we will focus more closel on the underling modeling step: how to set up the function to be integrated. 7. What ou ll learn about Linear Motion Revisited General Strateg Consumption Over Time Net Change from Data Work... and wh The integral is a tool that can be used to calculate net change and total accumulation. Integral As Net Change Linear Motion Revisited In man applications, the integral is viewed as net change over time. The classic eample of this kind is distance traveled, a problem we discussed in Chapter 5. EXAMPLE Interpreting a Velocit Function Figure 7. shows the velocit d s v(t) t dt t 8 c m s ec of a particle moving along a horizontal s-ais for t 5. Describe the motion. Solve Graphicall The graph of v (Figure 7.) starts with v 8, which we interpret as saing that the particle has an initial velocit of 8 cm sec to the left. It slows to a halt at about t.5 sec, after which it moves to the right v with increasing speed, reaching a velocit of v cm sec at the end. Now tr Eercise (a). [, 5] b [, 3] Figure 7. The velocit function in Eample. EXAMPLE Finding Position from Displacement Suppose the initial position of the particle in Eample is s 9. What is the particle s position at (a) t sec? (b) t 5 sec? Solve Analticall (a) The position at t is the initial position s plus the displacement (the amount, s, that the position changed from t to t ). When velocit is continued

3 38 Chapter 7 Applications of Definite Integrals Reminder from Section 3.4 A change in position is a displacement. If s(t) is a bod s position at time t, the displacement over the time interval from t to t t is s(t t) s(t). The displacement ma be positive, negative, or zero, depending on the motion. constant during a motion, we can find the displacement (change in position) with the formula Displacement rate of change time. But in our case the velocit varies, so we resort instead to partitioning the time interval, into subintervals of length t so short that the velocit is effectivel constant on each subinterval. If t k is an time in the kth subinterval, the particle s velocit throughout that interval will be close to v t k. The change in the particle s position during the brief time this constant velocit applies is v t k t. rate of change time If v t k is negative, the displacement is negative and the particle will move left. If v t k is positive, the particle will move right. The sum v t k t of all these small position changes approimates the displacement for the time interval,. The sum v t k t is a Riemann sum for the continuous function v t over,. As the norms of the partitions go to zero, the approimations improve and the sums converge to the integral of v over,, giving Displacement v t dt ( t ) dt t 8 ] [ t t During the first second of motion, the particle moves 3 cm to the left. It starts at s 9, so its position at t is New position initial position displacement (b) If we model the displacement from t to t 5 in the same wa, we arrive at Displacement 5 5 v t dt [ t t The motion has the net effect of displacing the particle 35 cm to the right of its starting point. The particle s final position is ] Final position initial position displacement s Support Graphicall The position of the particle at an time t is given b s t t 8 u du 9, [ u ] because s t v t and s 9. Figure 7. shows the graph of s t given b the parametrization t NINT v u, u,, t 9, t t, t 5. continued

4 Section 7. Integral as Net Change 38 (a) Figure 7.a supports that the position of the particle at t is 6 3. (b) Figure 7.b shows the position of the particle is 44 at t 5. Therefore, the displacement is Now tr Eercise (b). T = X = Y = [, 5] b [, 6] (a) T = 5 X = 44 Y = 5 [, 5] b [, 6] (b) Figure 7. Using TRACE and the parametrization in Eample ou can see the left and right motion of the particle. The reason for our method in Eample was to illustrate the modeling step that will be used throughout this chapter. We can also solve Eample using the techniques of Chapter 6 as shown in Eploration. EXPLORATION Revisiting Eample The velocit of a particle moving along a horizontal s-ais for t 5 is d s t dt. t 8. Use the indefinite integral of ds dt to find the solution of the initial value problem d s t dt, t 8 s 9.. Determine the position of the particle at t. Compare our answer with the answer to Eample a. 3. Determine the position of the particle at t 5. Compare our answer with the answer to Eample b. We know now that the particle in Eample was at s 9 at the beginning of the motion and at s 5 44 at the end. But it did not travel from 9 to 44 directl it began its trip b moving to the left (Figure 7.). How much distance did the particle actuall travel? We find out in Eample 3. EXAMPLE 3 Calculating Total Distance Traveled Find the total distance traveled b the particle in Eample. Solve Analticall We partition the time interval as in Eample but record ever position shift as positive b taking absolute values. The Riemann sum approimating total distance traveled is and we are led to the integral v t k t, Total distance traveled 5 v t dt 5 8 t t dt. Evaluate Numericall We have NINT ( t t 8, t,, 5) Now tr Eercise (c).

5 38 Chapter 7 Applications of Definite Integrals What we learn from Eamples and 3 is this: Integrating velocit gives displacement (net area between the velocit curve and the time ais). Integrating the absolute value of velocit gives total distance traveled (total area between the velocit curve and the time ais). General Strateg The idea of fragmenting net effects into finite sums of easil estimated small changes is not new. We used it in Section 5. to estimate cardiac output, volume, and air pollution. What is new is that we can now identif man of these sums as Riemann sums and epress their limits as integrals. The advantages of doing so are twofold. First, we can evaluate one of these integrals to get an accurate result in less time than it takes to crank out even the crudest estimate from a finite sum. Second, the integral itself becomes a formula that enables us to solve similar problems without having to repeat the modeling step. The strateg that we began in Section 5. and have continued here is the following: Strateg for Modeling with Integrals. Approimate what ou want to find as a Riemann sum of values of a continuous function multiplied b interval lengths. If f is the function and a, b the interval, and ou partition the interval into subintervals of length, the approimating sums will have the form f c k with c k a point in the kth subinterval.. Write a definite integral, here b f d, to epress the limit of these sums as a the norms of the partitions go to zero. 3. Evaluate the integral numericall or with an antiderivative. EXAMPLE 4 Modeling the Effects of Acceleration A car moving with initial velocit of 5 mph accelerates at the rate of a t.4t mph per second for 8 seconds. (a) How fast is the car going when the 8 seconds are up? (b) How far did the car travel during those 8 seconds? (a) We first model the effect of the acceleration on the car s velocit. Step : Approimate the net change in velocit as a Riemann sum. When acceleration is constant, velocit change acceleration time applied. rate of change time To appl this formula, we partition, 8 into short subintervals of length t. On each subinterval the acceleration is nearl constant, so if t k is an point in the kth subinterval, the change in velocit imparted b the acceleration in the subinterval is approimatel a t k t mph. m ph sec sec The net change in velocit for t 8 is approimatel a t k t mph. Step : Write a definite integral. The limit of these sums as the norms of the partitions go to zero is 8 a t dt. continued

6 Section 7. Integral as Net Change 383 Step 3: Evaluate the integral. Using an antiderivative, we have Net velocit change ] 8 8.4tdt.t 76.8 mph. So, how fast is the car going when the 8 seconds are up? Its initial velocit is 5 mph and the acceleration adds another 76.8 mph for a total of 8.8 mph. (b) There is nothing special about the upper limit 8 in the preceding calculation. Appling the acceleration for an length of time t adds t.4u du mph u is just a dumm variable here. (b) to the car s velocit, giving v t 5 t The distance traveled from t to t 8 sec is 8.4u du 5.t mph. v t dt 8 5.t dt Etension of Eample 3 8 [ 5t.4t 3] 44.8 mph seconds. Miles-per-hour second is not a distance unit that we normall work with! To convert to miles we multipl b hours second 36, obtaining mile. 36 m i h sec mi h se c The car traveled.68 mi during the 8 seconds of acceleration. Now tr Eercise 9. Consumption Over Time The integral is a natural tool to calculate net change and total accumulation of more quantities than just distance and velocit. Integrals can be used to calculate growth, deca, and, as in the net eample, consumption. Whenever ou want to find the cumulative effect of a varing rate of change, integrate it. EXAMPLE 5 Potato Consumption From 97 to 98, the rate of potato consumption in a particular countr was C t.. t millions of bushels per ear, with t being ears since the beginning of 97. How man bushels were consumed from the beginning of 97 to the end of 973? We seek the cumulative effect of the consumption rate for t 4. Step : Riemann sum. We partition, 4 into subintervals of length t and let t k be a time in the kth subinterval. The amount consumed during this interval is approimatel C t k t million bushels. The consumption for t 4 is approimatel C t k t million bushels. continued

7 384 Chapter 7 Applications of Definite Integrals Step : Definite integral. The amount consumed from t to t 4 is the limit of these sums as the norms of the partitions go to zero. C t dt 4.. t dt million bushels 4 Step 3: Evaluate. Evaluating numericall, we obtain NINT.. t, t,, million bushels. Now tr Eercise. Table 7. Joules Time (min) Pumping Rates Rate (gal min) The joule, abbreviated J and pronounced jewel, is named after the English phsicist James Prescott Joule (88 889). The defining equation is joule ( newton)( meter). In smbols, J N m. It takes a force of about N to lift an apple from a table. If ou lift it m ou have done about J of work on the apple. If ou eat the apple, ou will have consumed about 8 food calories, the heat equivalent of nearl 335, joules. If this energ were directl useful for mechanical work (it s not), it would enable ou to lift 335, more apples up m. Net Change from Data Man real applications begin with data, not a full modeled function. In the net eample, we are given data on the rate at which a pump operates in consecutive 5-minute intervals and asked to find the total amount pumped. EXAMPLE 6 Finding Gallons Pumped from Rate Data A pump connected to a generator operates at a varing rate, depending on how much power is being drawn from the generator to operate other machiner. The rate (gallons per minute) at which the pump operates is recorded at 5-minute intervals for one hour as shown in Table 7.. How man gallons were pumped during that hour? Let R t, t 6, be the pumping rate as a continuous function of time for the hour. We can partition the hour into short subintervals of length t on which the rate is nearl constant and form the sum R t k t as an approimation to the amount pumped during the hour. This reveals the integral formula for the number of gallons pumped to be Gallons pumped 6 R t dt. We have no formula for R in this instance, but the 3 equall spaced values in Table 7. enable us to estimate the integral with the Trapezoidal Rule: 6 R t dt [ ] The total amount pumped during the hour is about 358 gal. Now tr Eercise 7. Work In everda life, work means an activit that requires muscular or mental effort. In science, the term refers specificall to a force acting on a bod and the bod s subsequent displacement. When a bod moves a distance d along a straight line as a result of the action of a force of constant magnitude F in the direction of motion, the work done b the force is W Fd. The equation W Fd is the constant-force formula for work. The units of work are force distance. In the metric sstem, the unit is the newtonmeter, which, for historical reasons, is called a joule (see margin note). In the U.S. customar sstem, the most common unit of work is the foot-pound.

8 Section 7. Integral as Net Change 385 Hooke s Law for springs sas that the force it takes to stretch or compress a spring units from its natural (unstressed) length is a constant times. In smbols, F k, where k, measured in force units per unit length,is a characteristic of the spring called the force constant. EXAMPLE 7 A Bit of Work It takes a force of N to stretch a spring m beond its natural length. How much work is done in stretching the spring 4 m from its natural length? We let F represent the force in newtons required to stretch the spring meters from its natural length. B Hooke s Law, F k for some constant k. We are told that F k, The force required to stretch the spring m is newtons. so k 5 N m and F 5 for this particular spring. We construct an integral for the work done in appling F over the interval from to 4. Step : Riemann sum. We partition the interval into subintervals on each of which F is so nearl constant that we can appl the constant-force formula for work. If k is an point in the kth subinterval, the value of F throughout the interval is approimatel F k 5 k. The work done b F across the interval is approimatel 5 k, where is the length of the interval. The sum Numericall, work is the area under the force graph. F k 5 k approimates the work done b F from to 4. Steps and 3: Integrate. The limit of these sums as the norms of the partitions go to zero is 4 F d 4 5d N m. ] Now tr Eercise 9. We will revisit work in Section 7.5. Quick Review 7. (For help, go to Section..) In Eercises, find all values of (if an) at which the function changes sign on the given interval. Sketch a number line graph of the interval, and indicate the sign of the function on each subinterval. Eample: f on, 3 Changes sign at. + + f () 3. sin on 3, Changes sign at p,, p. 3 on, 4 Changes sign at, 3. 3 on 4, Alwas positive on, Changes sign at 5. cos on, 4 Changes sign at p 4, 3 p 5p, e on, Alwas positive 7. on 5, 3 Changes sign at 8. on 3, 3 Changes sign at,,, 4 9. sec s in on,. sin on.,. Changes sign at 3 p, p 9. Changes sign at.9633 kp,.783 kp, where k is an integer

9 386 Chapter 7 Applications of Definite Integrals Section 7. Eercises In Eercises 8, the function v(t) is the velocit in m sec of a particle moving along the -ais. Use analtic methods to do each of the following: (a) Determine when the particle is moving to the right, to the left, and stopped. (b) Find the particle s displacement for the given time interval. If s() 3, what is the particle s final position? (c) Find the total distance traveled b the particle.. v t 5 cos t, t p See page v t 6 sin 3t, t p See page v t t, t See page v t 6t 8t, t See page v t 5 sin t cos t, t p See page v t 4 t, t 4 See page v t e sin t cos t, t p See page 389. t 8. v t, t 3 See page 389. t 9. An automobile accelerates from rest at 3 t mph sec for 9 seconds. (a) What is its velocit after 9 seconds? 63 mph (b) How far does it travel in those 9 seconds?. A particle travels with velocit for t 4 sec. v t t sin t m sec (a) What is the particle s displacement? (b) What is the total distance traveled? feet.4495 meters.94 meters. Projectile Recall that the acceleration due to Earth s gravit is 3 ft sec. From ground level, a projectile is fired straight upward with velocit 9 feet per second. (a) What is its velocit after 3 seconds? (b) When does it hit the ground? 5.65 sec 6 ft/sec (c) When it hits the ground, what is the net distance it has traveled? (d) When it hits the ground, what is the total distance it has traveled? 53.5 feet In Eercises 6, a particle moves along the -ais (units in cm). Its initial position at t sec is 5. The figure shows the graph of the particle s velocit v t. The numbers are the areas of the enclosed regions. 4 5 a b c 4. What is the particle s displacement between t and t c? 3 cm 3. What is the total distance traveled b the particle in the same time period? 33 cm a: b: 6 c: 8 4. Give the positions of the particle at times a, b, and c. 5. Approimatel where does the particle achieve its greatest positive acceleration on the interval, b? t a 6. Approimatel where does the particle achieve its greatest positive acceleration on the interval, c? t c In Eercises 7, the graph of the velocit of a particle moving on the -ais is given. The particle starts at when t (a) Find where the particle is at the end of the trip. (b) Find the total distance traveled b the particle. 3 v (m/sec) 3 v (m/sec) v (m/sec) v (m/sec) t (sec) t (sec) 8 9 t (sec) (a) t (sec) (a) 6 (a) 5 (b) 4 meters (b) 4 meters (b) 7 meters (a).5 (b) 9.5 meters. U.S. Oil Consumption The rate of consumption of oil in the United States during the 98s (in billions of barrels per ear) is modeled b the function C 7.8 e t 5,where t is the number of ears after Januar, 98. Find the total consumption of oil in the United States from Januar, 98 to Januar, billion barrels. Home Electricit Use The rate at which our home consumes electricit is measured in kilowatts. If our home consumes electricit at the rate of kilowatt for hour, ou will be charged

10 for kilowatt-hour of electricit. Suppose that the average consumption rate for a certain home is modeled b the function C t sin pt, where C t is measured in kilowatts and t is the number of hours past midnight. Find the average dail consumption for this home, measured in kilowatthours kilowatt-hours 3. Population Densit Population densit measures the number of people per square mile inhabiting a given living area. Washerton s population densit, which decreases as ou move awa from the cit center, can be approimated b the function, r at a distance r miles from the cit center. (a) If the population densit approaches zero at the edge of the cit, what is the cit s radius? miles (b) A thin ring around the center of the cit has thickness r and radius r. If ou straighten it out, it suggests a rectangular strip. Approimatel what is its area? pr r (c) Writing to Learn Eplain wh the population of the ring in part (b) is approimatel, r pr r. Population Population densit Area (d) Estimate the total population of Washerton b setting up and evaluating a definite integral. 83, Oil Flow Oil flows through a clindrical pipe of radius 3 inches, but friction from the pipe slows the flow toward the outer edge. The speed at which the oil flows at a distance r inches from the center is 8 r inches per second. (a) In a plane cross section of the pipe, a thin ring with thickness r at a distance r inches from the center approimates a rectangular strip when ou straighten it out. What is the area of the strip (and hence the approimate area of the ring)? pr r (b) Eplain wh we know that oil passes through this ring at approimatel 8 r pr r cubic inches per second. (c) Set up and evaluate a definite integral that will give the rate (in cubic inches per second) at which oil is flowing through the pipe. 396p in 3 /sec or 44.7 in 3 /sec 5. Group Activit Bagel Sales From 995 to 5, the Konigsberg Baker noticed a consistent increase in annual sales of its bagels. The annual sales (in thousands of bagels) are shown below. Year 6. (b) The answer in (a) corresponds to the area of midpoint rectangles. Part of each rectangle is above the curve and part is below. Sales (thousands) (b) 8( r ) in/sec (pr) r in flow in in 3 /sec Section 7. Integral as Net Change 387 (a) What was the total number of bagels sold over the -ear period? ( This is not a calculus question!) thousand (b) Use quadratic regression to model the annual bagel sales (in thousands) as a function B, where is the number of ears after 995. B() (c) Integrate B over the interval, to find total bagel sales for the -ear period. 94. (d) Eplain graphicall wh the answer in part (a) is smaller than the answer in part (c). See page Group Activit (Continuation of Eercise 5) (a) Integrate B over the interval.5,.5 to find total bagel sales for the -ear period thousand (b) Eplain graphicall wh the answer in part (a) is better than the answer in 5(c). 7. Filling Milk Cartons A machine fills milk cartons with milk at an approimatel constant rate, but backups along the assembl line cause some variation. The rates (in cases per hour) are recorded at hourl intervals during a -hour period, from 8: A.M. to 6: P.M. Time Rate (cases h) Use the Trapezoidal Rule with n to determine approimatel how man cases of milk were filled b the machine over the -hour period Writing to Learn As a school project, Anna accompanies her mother on a trip to the grocer store and keeps a log of the car s speed at -second intervals. Eplain how she can use the data to estimate the distance from her home to the store. What is the connection between this process and the definite integral? 9. Hooke s Law A certain spring requires a force of 6 N to stretch it 3 cm beond its natural length. (a) What force would be required to stretch the string 9 cm beond its natural length? 8 N (b) What would be the work done in stretching the string 9 cm beond its natural length? 8 N cm 3. Hooke s Law Hooke s Law also applies to compressing springs; that is, it requires a force of k to compress a spring a distance from its natural length. Suppose a,-lb force compressed a spring from its natural length of inches to a length of inches. How much work was done in compressing the spring (a) the first half-inch? (b) the second half-inch? 5 inch-pounds 375 inch-pounds See page 389.

11 388 Chapter 7 Applications of Definite Integrals Standardized Test Questions You ma use a graphing calculator to solve the following problems. 3. True or False The figure below shows the velocit for a particle moving along the -ais. The displacement for this particle is negative. Justif our answer. False. The displacement is the integral of the velocit from t to t 5 and is positive. 3. True or False If the velocit of a particle moving along the -ais is alwas positive, then the displacement is equal to the total distance traveled. Justif our answer. 33. Multiple Choice The graph below shows the rate at which water is pumped from a storage tank. Approimate the total gallons of water pumped from the tank in 4 hours. C (A) 6 (B) 4 (C) 36 (D) 4 (E) Multiple Choice The data for the acceleration a(t) of a car from to 5 seconds are given in the table below. If the velocit at t is 5 ft/sec, which of the following gives the approimate velocit at t 5 using the Trapezoidal Rule? D (A) 47 ft/sec (B) 5 ft/sec (C) ft/sec (D) 5 ft/sec 5 5 v (m/sec) 5 r (gal/hr) (E) 4 ft/sec t (sec) t (hr) t (sec) a(t) (ft sec ) Multiple Choice The rate at which customers arrive at a counter to be served is modeled b the function F defined b F(t) 6 cos p t for t 6, where F(t) is measured in customers per minute and t is measured in minutes. To the nearest whole number, how man customers arrive at the counter over the 6-minute period? B (A) 7 (B) 75 (C) 73 (D) 744 (E) Multiple Choice Pollution is being removed from a lake at a rate modeled b the function e.5t tons/r, where t is the number of ears since 995. Estimate the amount of pollution removed from the lake between 995 and 5. Round our answer to the nearest ton. A (A) 4 (B) 47 (C) 56 (D) 6 (E) 7 Etending the Ideas 37. Inflation Although the econom is continuousl changing, we analze it with discrete measurements. The following table records the annual inflation rate as measured each month for 3 consecutive months. Use the Trapezoidal Rule with n to find the overall inflation rate for the ear Month 38. Inflation Rate The table below shows the monthl inflation rate (in thousandths) for energ prices for thirteen consecutive months. Use the Trapezoidal Rule with n to approimate the annual inflation rate for the -month period running from the middle of the first month to the middle of the last month. 4 thousandths or.4 Month Annual Rate Januar.4 Februar.4 March.5 April.6 Ma.5 June.4 Jul.4 August.5 September.4 October.6 November.6 December.5 Januar.5 Monthl Rate (in thousandths) Januar 3.6 Februar 4. March 3. April.8 Ma.8 June 3. Jul 3.3 August 3. September 3. October 3.4 November 3.4 December 3.9 Januar True. Since the velocit is positive, the integral of the velocit is equal to the integral of its absolute value, which is the total distance traveled.

12 Section 7. Integral as Net Change Center of Mass Suppose we have a finite collection of masses in the coordinate plane, the mass m k located at the point k, k as shown in the figure. O Each mass m k has moment m k k with respect to the -ais and moment m k k about the -ais. The moments of the entire sstem with respect to the two aes are Moment about -ais: M m k k, Moment about -ais: M m k k. The center of mass is J, J where k J M m and M J M k m k k k. m k M m k k k k m k(k, k ) Suppose we have a thin, flat plate occuping a region in the plane. (a) Imagine the region cut into thin strips parallel to the -ais. Show that J dm, dm where dm d da, d densit (mass per unit area), and A area of the region. (b) Imagine the region cut into thin strips parallel to the -ais. Show that J dm, dm where dm d da, d densit, and A area of the region. In Eercises 4 and 4, use Eercise 39 to find the center of mass of the region with given densit. 4. the region bounded b the parabola and the line 4 with constant densit d, /5 4. the region bounded b the lines,, with constant densit d 4/3,. (a) Right: t p/, 3p/ t p Left: p/ t 3p/ Stopped: t p/, 3p/ (b) ; 3 (c). (a) Right: t p/3 Left: p/3 t p/ Stopped: t, p/3 (b) ; 5 (c) 6 3. (a) Right: t 5 Left: 5 t Stopped: t 5 (b) ; 3 (c) (a) Right: t Left: t Stopped: t, (b) 4; 7 (c) 6 5. (a) Right: t p/, 3p/ t p Left: p/ t p, p t 3p/ Stopped: t, p/, p, 3p/, p (b) ; 3 (c) /3 6. (a) Right: t 4 Left: never Stopped: t 4 (b) 6/3; 5/3 (c) 6/3 5. (d) The answer in (a) corresponds to the area of left hand rectangles. These rectangles lie under the curve B(). The answer in (c) corresponds to the area under the curve. This area is greater than the area of the rectangles. 8. One possible answer: Plot the speeds vs. time. Connect the points and find the area under the line graph. The definite integral also gives the area under the curve. 39. (a, b) Take dm d da as m k and letting da, k in the center of mass equations. 7. (a) Right: t p/, 3p/ t p Left: p/ t 3p/ Stopped: t p/, 3p/ (b) ; 3 (c) e (/e) (a) Right: t 3 Left: never Stopped: t (b) (ln )/.5; 4.5 (c) (ln )/.5

13 39 Chapter 7 Applications of Definite Integrals What ou ll learn about Area Between Curves 7. Area Enclosed b Intersecting Curves Boundaries with Changing Functions Integrating with Respect to Saving Time with Geometric Formulas... and wh The techniques of this section allow us to compute areas of comple regions of the plane. Areas in the Plane Area Between Curves We know how to find the area of a region between a curve and the -ais but man times we want to know the area of a region that is bounded above b one curve, f, and below b another, g (Figure 7.3). We find the area as an integral b appling the first two steps of the modeling strateg developed in Section 7... We partition the region into vertical strips of equal width and approimate each strip with a rectangle with base parallel to a, b (Figure 7.4). Each rectangle has area f c k g c k for some c k in its respective subinterval (Figure 7.5). This epression will be nonnegative even if the region lies below the -ais. We approimate the area of the region with the Riemann sum f c k g c k. Upper curve f() f() (c k, f (c k )) f (c k ) g(c k ) a b a b a c k b Lower curve g() g() (c k, g(c k )) Figure 7.3 The region between f () and g() and the lines a and b. Figure 7.4 We approimate the region with rectangles perpendicular to the -ais.. The limit of these sums as is b f g d. a Figure 7.5 The area of a tpical rectangle is f (c k ) g(c k ). This approach to finding area captures the properties of area, so it can serve as a definition. DEFINITION Area Between Curves If f and g are continuous with f g throughout a, b, then the area between the curves f () and g() from a to b is the integral of f g from a to b, A b a f g d.

14 Section 7. Areas in the Plane 39 sec EXAMPLE Appling the Definition Find the area of the region between sec and sin from to p 4. We graph the curves ( Figure 7.6) to find their relative positions in the plane, and see that sec lies above sin on, p 4. The area is therefore sin A p 4 sec sin d p 4 [ ] tan cos units squared. 4 Now tr Eercise. Figure 7.6 The region in Eample. = k k sin k = k sin k k = [, ] b [, 4] k = Figure 7.7 Two members of the famil of butterfl-shaped regions described in Eploration. EXPLORATION A Famil of Butterflies For each positive integer k, let A k denote the area of the butterfl-shaped region enclosed between the graphs of k sin k and k k sin k on the interval, p k. The regions for k and k are shown in Figure Find the areas of the two regions in Figure Make a conjecture about the areas A k for k Set up a definite integral that gives the area A k. Can ou make a simple u-substitution that will transform this integral into the definite integral that gives the area A? 4. What is lim k A k? 5. If P k denotes the perimeter of the kth butterfl-shaped region, what is lim k P k? (You can answer this question without an eplicit formula for P k.) = = [ 6, 6] b [ 4, 4] Figure 7.8 The region in Eample. Area Enclosed b Intersecting Curves When a region is enclosed b intersecting curves, the intersection points give the limits of integration. EXAMPLE Area of an Enclosed Region Find the area of the region enclosed b the parabola and the line. We graph the curves to view the region (Figure 7.8). The limits of integration are found b solving the equation either algebraicall or b calculator. The solutions are and. continued

15 39 Chapter 7 Applications of Definite Integrals Since the parabola lies above the line on,, the area integrand is. A d [ 3 3 ] 9 units squared Now tr Eercise 5. = cos = [ 3, 3] b [, 3] Figure 7.9 The region in Eample 3. Finding Intersections b Calculator The coordinates of the points of intersection of two curves are sometimes needed for other calculations. To take advantage of the accurac provided b calculators, use them to solve for the values and store the ones ou want. EXAMPLE 3 Using a Calculator Find the area of the region enclosed b the graphs of cos and. The region is shown in Figure 7.9. Using a calculator, we solve the equation cos to find the -coordinates of the points where the curves intersect. These are the limits of integration. The solutions are We store the negative value as A and the positive value as B. The area is NINT cos,, A, B This is the final calculation, so we are now free to round. The area is about Boundaries with Changing Functions Now tr Eercise 7. If a boundar of a region is defined b more than one function, we can partition the region into subregions that correspond to the function changes and proceed as usual. EXAMPLE 4 Finding Area Using Subregions Find the area of the region R in the first quadrant that is bounded above b and below b the -ais and the line. The region is shown in Figure Area d Area d (4, ) B A 4 Figure 7. Region R split into subregions A and B. (Eample 4) continued

16 Section 7. Areas in the Plane 393 While it appears that no single integral can give the area of R (the bottom boundar is defined b two different curves), we can split the region at into two regions A and B. The area of R can be found as the sum of the areas of A and B. Area of R d 4 d 3 ] area of A area of B 3 [ ] units squared 3 Now tr Eercise 9. Integrating with Respect to Sometimes the boundaries of a region are more easil described b functions of than b functions of. We can use approimating rectangles that are horizontal rather than vertical and the resulting basic formula has in place of. For regions like these d f() d f() d g() f() g() c g() c c use this formula d A = [f () g()]d. c (g(), ) f() g() (4, ) ( f (), ) 4 Figure 7. It takes two integrations to find the area of this region if we integrate with respect to. It takes onl one if we integrate with respect to. (Eample 5) EXAMPLE 5 Integrating with Respect to Find the area of the region in Eample 4 b integrating with respect to. We remarked in solving Eample 4 that it appears that no single integral can give the area of R, but notice how appearances change when we think of our rectangles being summed over. The interval of integration is,, and the rectangles run between the same two curves on the entire interval. There is no need to split the region (Figure 7.). We need to solve for in terms of in both equations: becomes, becomes,. continued

17 394 Chapter 7 Applications of Definite Integrals = 3, = +, 3 = + (c, d) We must still be careful to subtract the lower number from the higher number when forming the integrand. In this case, the higher numbers are the higher -values, which are on the line because the line lies to the right of the parabola. So, Area of R d [ 3 3 ] units squared. 3 Now tr Eercise. (a, b) [ 3, 3] b [ 3, 3] Figure 7. The region in Eample 6. (a, b) [ 3, 3] b [ 3, 3] (c, d) Figure 7.3 If we integrate with respect to in Eample 6, we must split the region at a. EXAMPLE 6 Making the Choice Find the area of the region enclosed b the graphs of 3 and. We can produce a graph of the region on a calculator b graphing the three curves 3,, and (Figure 7.). This convenientl gives us all of our bounding curves as functions of. If we integrate in terms of, however, we need to split the region at a (Figure 7.3). On the other hand, we can integrate from b to d and handle the entire region at once. We solve the cubic for in terms of : 3 becomes 3. To find the limits of integration, we solve 3. It is eas to see that the lower limit is b, but a calculator is needed to find that the upper limit d We store this value as D. The cubic lies to the right of the parabola, so Area NINT 3,,, D The area is about 4.. Now tr Eercise 7. Saving Time with Geometr Formulas Here is et another wa to handle Eample 4. Area 4 (4, ) Figure 7.4 The area of the blue region is the area under the parabola minus the area of the triangle. (Eample 7) EXAMPLE 7 Using Geometr Find the area of the region in Eample 4 b subtracting the area of the triangular region from the area under the square root curve. Figure 7.4 illustrates the strateg, which enables us to integrate with respect to without splitting the region. Area 3 ] 4 d 4 units squared 3 3 Now tr Eercise 35. The moral behind Eamples 4, 5, and 7 is that ou often have options for finding the area of a region, some of which ma be easier than others. You can integrate with respect to or with respect to, ou can partition the region into subregions,and sometimes ou can even use traditional geometr formulas. Sketch the region first and take a moment to determine the best wa to proceed.

18 Section 7. Areas in the Plane 395 Quick Review 7. (For help, go to Sections. and 5..) In Eercises 5, find the area between the -ais and the graph of the given function over the given interval.. sin over, p. e over, (e ) sec over p 4, p over, over 3, 3 9p/ In Eercises 6, find the - and -coordinates of all points where the graphs of the given functions intersect. If the curves never intersect, write NI and 6 (6, ); (, 5) 7. e and (, ) 8. p and sin (, ); (p, ) 9. and 3 (, ); (, ); (, ). sin and 3 (.986,.88); (, ); (.986,.88) Section 7. Eercises In Eercises 6, find the area of the shaded region analticall.. p/ 4. 4/3 3 cos 5. 8/5. 4p/3 sec t (, 8) 8 (, 8) 3 3 t 4 4 sin t 4 NOT TO SCALE 6. /5 3. / 3 (, ) 4

19 396 Chapter 7 Applications of Definite Integrals In Eercises 7 and 8, use a calculator to find the area of the region enclosed b the graphs of the two functions. 7. sin, cos(), In Eercises 9 and, find the area of the shaded region analticall. 9. 5/6. 5/6 In Eercises and, find the area enclosed b the graphs of the two curves b integrating with respect to.., 3. 3, In Eercises 3 and 4, find the total shaded area (, ) (, 4) (, ) (3, 5) In Eercises 5 34, find the area of the regions enclosed b the lines and curves. 5. and 3 6. and and and 8 9. a, a, and 3 a 3. and (3 points of intersection) (How man intersection points are there?). 4 and 4 3. and and and and 3 8/ and and sin and sin, p cos and sec, p 3 p cos p and p 3. sin p and 4 p.73 p 3. sec, tan, p 4, p 4 p 33. tan and tan, p 4 p sin co s and, p In Eercises 35 and 36, find the area of the region b subtracting the area of a triangular region from the area of a larger region. 35. The region on or above the -ais bounded b the curves 3 and The region on or above the -ais bounded b the curves 4 and Find the area of the propeller-shaped region enclosed b the curve 3 and the line. / 38. Find the area of the region in the first quadrant bounded b the line, the line, the curve, and the -ais. 39. Find the area of the triangular region in the first quadrant bounded on the left b the -ais and on the right b the curves sin and cos Find the area of the region between the curve 3 and the line b integrating with respect to (a), (b). 3/3 4. The region bounded below b the parabola and above b the line 4is to be partitioned into two subsections of equal area b cutting across it with the horizontal line c. (a) Sketch the region and draw a line c across it that looks about right. In terms of c, what are the coordinates of the points where the line and parabola intersect? Add them to our figure. ( c, c); ( c, c) (b) Find c b integrating with respect to. (This puts c in the limits of integration.) d 4 d c 4 3 c (c) Find c b integrating with respect to. (This puts c into the integrand as well.) 4. Find the area of the region in the first quadrant bounded on the left b the -ais, below b the line 4, above left b the curve, and above right b the curve. /3 4. (c) c (c ) d (4 c) c (4 ) d c 4 3 c c p.858

20 Section 7. Areas in the Plane The figure here shows triangle AOC inscribed in the region cut from the parabola b the line a. Find the limit of the ratio of the area of the triangle to the area of the parabolic region as a approaches zero. 3/4 44. Suppose the area of the region between the graph of a positive continuous function f and the -ais from a to b is 4 square units. Find the area between the curves f and f from a to b Writing to Learn Which of the following integrals, if either, calculates the area of the shaded region shown here? Give reasons for our answer. Neither; both are zero i. d d ii. d d A ( a, a ) a O C a (a, a ) a Standardized Test Questions You should solve the following problems without using a graphing calculator. 5. True or False The area of the region enclosed b the graph of and the line is 36. Justif our answer. True. 36 is the value of the appropriate integral. 5. True or False The area of the region in the first quadrant enclosed b the graphs of cos,, and the -ais is given b the definite integral.739 ( cos ) d. Justif our answer. False. It is.739 (cos ) d. 5. Multiple Choice Let R be the region in the first quadrant bounded b the -ais, the graph of, and the line 4. Which of the following integrals gives the area of R? (A) (C) [4 ( )]d (B) [( ) 4]d [4 ( )]d (D) [( ) 4]d 4 (E) [4 ( )]d 53. Multiple Choice Which of the following gives the area of the region between the graphs of and from to 3? E (A) (B) 9/ (C) 3/ (D) 3 (E) 7/ 54. Multiple Choice Let R be the shaded region enclosed b the graphs of e, sin(3), and the -ais as shown in the figure below. Which of the following gives the approimate area of the region R? B (A).39 (B).445 (C).869 (D).4 (E).34 A 46. Writing to Learn Is the following statement true, sometimes true, or never true? The area of the region between the graphs of the continuous functions f and g and the vertical lines a and b (a b) is b f g d. a Sometimes; If f () g() on (a, b), Give reasons for our answer. then true. 47. Find the area of the propeller-shaped region enclosed between the graphs of ln 4 (/).886 and Find the area of the propeller-shaped region enclosed between the graphs of sin and Find the positive value of k such that the area of the region enclosed between the graph of k cos and the graph of k is. k Multiple Choice Let f and g be the functions given b f () e and g() /. Which of the following gives the area of the region enclosed b the graphs of f and g between and? A (A) e e ln (B) ln e e (C) e (D) e e (E) ln e

21 398 Chapter 7 Applications of Definite Integrals Eploration 56. Group Activit Area of Ellipse An ellipse with major ais of length a and minor ais of length b can be coordinatized with its center at the origin and its major ais horizontal, in which case it is defined b the equation Etending the Ideas 57. Cavalieri s Theorem Bonaventura Cavalieri ( ) discovered that if two plane regions can be arranged to lie over the same interval of the -ais in such a wa that the have identical vertical cross sections at ever point (see figure), then the regions have the same area. Show that this theorem is true. a b. (a) Find the equations that define the upper and lower semiellipses as functions of. (b) Write an integral epression that gives the area of the ellipse. (c) With our group, use NINT to find the areas of ellipses for various lengths of a and b. (d) There is a simple formula for the area of an ellipse with major ais of length a and minor ais of length b. Can ou tell what it is from the areas ou and our group have found? (e) Work with our group to write a proof of this area formula b showing that it is the eact value of the integral epression in part (b). 56. (a) b a (b) a b a a d (c) Answers ma var. (d, e) abp a b Cross sections have the same length at ever point in [a, b]. 58. Find the area of the region enclosed b the curves and m, m. m ln (m) 57. Since f () g() is the same for each region where f () and g() represent the upper and lower edges, area b [ f () g()] d will be the same for a each.

22 Section 7.3 Volumes Volumes What ou ll learn about Volume As an Integral Square Cross Sections Circular Cross Sections Clindrical Shells Other Cross Sections... and wh The techniques of this section allow us to compute volumes of certain solids in three dimensions. S P Cross-section R() with area A() Volume As an Integral In Section 5., Eample 3, we estimated the volume of a sphere b partitioning it into thin slices that were nearl clindrical and summing the clinders volumes using MRAM. MRAM sums are Riemann sums, and had we known how at the time, we could have continued on to epress the volume of the sphere as a definite integral. Starting the same wa, we can now find the volumes of a great man solids b integration. Suppose we want to find the volume of a solid like the one in Figure 7.5. The cross section of the solid at each point in the interval a, b is a region R of area A. If A is a continuous function of, we can use it to define and calculate the volume of the solid as an integral in the following wa. We partition a, b into subintervals of length and slice the solid, as we would a loaf of bread, b planes perpendicular to the -ais at the partition points. The kth slice, the one between the planes at k and k,has approimatel the same volume as the clinder between the two planes based on the region R k (Figure 7.6). Plane at k Approimating clinder based on R( k ) has height k k k a k Plane at k Figure 7.5 The cross section of an arbitrar solid at point. b The clinder s base is the region R( k ) with area A( k ) NOT TO SCALE k Figure 7.6 Enlarged view of the slice of the solid between the planes at k and k. The volume of the clinder is The sum V k base area height A k. V k A k approimates the volume of the solid. This is a Riemann sum for A on a, b. We epect the approimations to improve as the norms of the partitions go to zero, so we define their limiting integral to be the volume of the solid. DEFINITION Volume of a Solid The volume of a solid of known integrable cross section area A from a to b is the integral of A from a to b, V b a A d.

23 4 Chapter 7 Applications of Definite Integrals To appl the formula in the previous definition, we proceed as follows. How to Find Volume b the Method of Slicing. Sketch the solid and a tpical cross section.. Find a formula for A. 3. Find the limits of integration. 4. Integrate A to find the volume. Tpical cross-section (m) Figure 7.7 A cross section of the pramid in Eample. 3 3 [ 3, 3] b [ 4, 4] Figure 7.8 The region in Eample. f() Figure 7.9 The region in Figure 7.8 is revolved about the -ais to generate a solid. A tpical cross section is circular, with radius f () cos. (Eample ) 3 Square Cross Sections Let us appl the volume formula to a solid with square cross sections. EXAMPLE A Square-Based Pramid A pramid 3 m high has congruent triangular sides and a square base that is 3 m on each side. Each cross section of the pramid parallel to the base is a square. Find the volume of the pramid. We follow the steps for the method of slicing.. Sketch. We draw the pramid with its verte at the origin and its altitude along the interval 3. We sketch a tpical cross section at a point between and 3 (Figure 7.7).. Find a formula for A. The cross section at is a square meters on a side, so A. 3. Find the limits of integration. The squares go from to Integrate to find the volume. V 3 Circular Cross Sections A d ] 3 9m 3 Now tr Eercise 3. The onl thing that changes when the cross sections of a solid are circular is the formula for A. Man such solids are solids of revolution, as in the net eample. EXAMPLE A Solid of Revolution The region between the graph of f cos and the -ais over the interval, is revolved about the -ais to generate a solid. Find the volume of the solid. Revolving the region (Figure 7.8) about the -ais generates the vase-shaped solid in Figure 7.9. The cross section at a tpical point is circular, with radius equal to f. Its area is A p f. continued

24 Section 7.3 Volumes 4 The volume of the solid is V A d NINT p cos,,, 5.43 units cubed. Now tr Eercise 7. [ /4, /] b [.5,.5] EXAMPLE 3 Washer Cross Sections The region in the first quadrant enclosed b the -ais and the graphs of cos and sin is revolved about the -ais to form a solid. Find its volume. The region is shown in Figure 7.. We revolve it about the -ais to generate a solid with a cone-shaped cavit in its center (Figure 7.). Figure 7. The region in Eample 3. Figure 7. The area of a washer is pr pr. (Eample 3) CAUTION! The area of a washer is pr pr, which ou can simplif to p(r r ), but not to p(r r). No matter how tempting it is to make the latter simplification, it s wrong. Don t do it. r R Figure 7. The solid generated b revolving the region in Figure 7. about the -ais. A tpical cross section is a washer: a circular region with a circular region cut out of its center. (Eample 3) This time each cross section perpendicular to the ais of revolution is a washer, a circular region with a circular region cut from its center. The area of a washer can be found b subtracting the inner area from the outer area (Figure 7.). In our region the cosine curve defines the outer radius, and the curves intersect at p 4. The volume is V p 4 p p 4 p[ sin p cos sin d cos d ] p 4 identit: cos sin cos p units cubed. Now tr Eercise 7. We could have done the integration in Eample 3 with NINT, but we wanted to demonstrate how a trigonometric identit can be useful under unepected circumstances in calculus. The double-angle identit turned a difficult integrand into an eas one and enabled us to get an eact answer b antidifferentiation. Clindrical Shells There is another wa to find volumes of solids of rotation that can be useful when the ais of revolution is perpendicular to the ais containing the natural interval of integration. Instead of summing volumes of thin slices, we sum volumes of thin clindrical shells that grow outward from the ais of revolution like tree rings.

25 4 Chapter 7 Applications of Definite Integrals [ 6, 4] b [ 3, 3] Figure 7.3 The graph of the region in Eploration, before revolution. Ais of revolution Figure 7.4 The region in Figure 7.3 is revolved about the line to form a solid cake. The natural interval of integration is along the -ais, perpendicular to the ais of revolution. (Eploration ) k k Figure 7.5 Cutting the cake into thin clindrical slices, working from the inside out. Each slice occurs at some k between and 3 and has thickness. (Eploration ) 3 3 EXPLORATION Volume b Clindrical Shells The region enclosed b the -ais and the parabola f 3 is revolved about the line to generate the shape of a cake (Figures 7.3, 7.4). (Such a cake is often called a bundt cake.) What is the volume of the cake? Integrating with respect to would be awkward here, as it is not eas to get the original parabola in terms of. (Tr finding the volume b washers and ou will soon see what we mean.) To integrate with respect to, ou can do the problem b clindrical shells, which requires that ou cut the cake in a rather unusual wa.. Instead of cutting the usual wedge shape, cut a clindrical slice b cutting straight down all the wa around close to the inside hole. Then cut another clindrical slice around the enlarged hole, then another, and so on. The radii of the clinders graduall increase, and the heights of the clinders follow the contour of the parabola: smaller to larger, then back to smaller (Figure 7.5). Each slice is sitting over a subinterval of the -ais of length. Its radius is approimatel k. What is its height?. If ou unroll the clinder at k and flatten it out, it becomes (essentiall) a rectangular slab with thickness. Show that the volume of the slab is approimatel p k 3 k k. 3. p k 3 k k is a Riemann sum. What is the limit of these Riemann sums as? 4. Evaluate the integral ou found in step 3 to find the volume of the cake! EXAMPLE 4 Finding Volumes Using Clindrical Shells The region bounded b the curve,the -ais, and the line 4 is revolved about the -ais to generate a solid. Find the volume of the solid.. Sketch the region and draw a line segment across it parallel to the ais of revolution (Figure 7.6). Label the segment s length (shell height) and distance from the ais of revolution (shell radius). The width of the segment is the shell thickness d. (We drew the shell in Figure 7.7, but ou need not do that.) Interval of integration 4 Shell height (4, ) Shell thickness d Shell radius 4 Figure 7.6 The region, shell dimensions, and interval of integration in Eample 4. Shell height 4 (4, ) Shell radius Figure 7.7 The shell swept out b the line segment in Figure 7.6.

26 Section 7.3 Volumes 43. Identif the limits of integration: runs from to. 3. Integrate to find the volume. V p( radius)( shell height) shell d p 4 d 8p Now tr Eercise 33(a). 4 Figure 7.8 The region and the height of a tpical shell in Eample 5. o k [, 3.5] b [.8,.] Figure 7.9 The base of the paperweight in Eample 6. The segment perpendicular to the -ais at k is the diameter of a semicircle that is perpendicular to the base. π = sin Figure 7.3 Cross sections perpendicular to the region in Figure 7.9 are semicircular. (Eample 6) EXAMPLE 5 Finding Volumes Using Clindrical Shells The region bounded b the curves 4,, and is revolved about the -ais to form a solid. Use clindrical shells to find the volume of the solid.. Sketch the region and draw a line segment across it parallel to the -ais (Figure 7.8). The segment s length (shell height) is 4. The distance of the segment from the ais of revolution (shell radius) is.. Identif the limits of integration: The -coordinate of the point of intersection of the curves 4 and in the first quadrant is about.56. So runs from to Integrate to find the volume. Other Cross Sections radius)( height) shell d.56 p()(4 ) d V p ( shell Now tr Eercise 35. The method of cross-section slicing can be used to find volumes of a wide variet of unusuall shaped solids, so long as the cross sections have areas that we can describe with some formula. Admittedl, it does take a special artistic talent to draw some of these solids, but a crude picture is usuall enough to suggest how to set up the integral. EXAMPLE 6 A Mathematician s Paperweight A mathematician has a paperweight made so that its base is the shape of the region between the -ais and one arch of the curve sin (linear units in inches). Each cross section cut perpendicular to the -ais (and hence to the -plane) is a semicircle whose diameter runs from the -ais to the curve. (Think of the cross section as a semicircular fin sticking up out of the plane.) Find the volume of the paperweight. The paperweight is not easil drawn, but we know what it looks like. Its base is the region in Figure 7.9, and the cross sections perpendicular to the base are semicircular fins like those in Figure 7.3. The semicircle at each point has radius s in sin and area A p sin. continued

27 44 Chapter 7 Applications of Definite Integrals The volume of the paperweight is V p A d p p sin d p NINT sin,,, p p Bonaventura Cavalieri ( ) Cavalieri, a student of Galileo, discovered that if two plane regions can be arranged to lie over the same interval of the -ais in such a wa that the have identical vertical cross sections at ever point, then the regions have the same area. This theorem and a letter of recommendation from Galileo were enough to win Cavalieri a chair at the Universit of Bologna in 69. The solid geometr version in Eample 7, which Cavalieri never proved, was named after him b later geometers. The number in parentheses looks like half of p, an observation that can be confirmed analticall, and which we support numericall b dividing b p to get.5. The volume of the paperweight is EXAMPLE 7 p p p.47 in 4 3. Cavalieri s Volume Theorem Now tr Eercise 39(a). Cavalieri s volume theorem sas that solids with equal altitudes and identical cross section areas at each height have the same volume ( Figure 7.3). This follows immediatel from the definition of volume, because the cross section area function A and the interval a, b are the same for both solids. b Same volume a Cross sections have the same length at ever point in [a, b]. Same cross-section area at ever level Figure 7.3 Cavalieri s volume theorem: These solids have the same volume. You can illustrate this ourself with stacks of coins. (Eample 7) a b Now tr Eercise 43.

28 Section 7.3 Volumes 45 EXPLORATION Surface Area We know how to find the volume of a solid of revolution, but how would we find the surface area? As before, we partition the solid into thin slices, but now we wish to form a Riemann sum of approimations to surface areas of slices (rather than of volumes of slices). = f() a b A tpical slice has a surface area that can be approimated b p f s, where s is the tin slant height of the slice. We will see in Section 7.4, when we stud arc length, that s, and that this can be written as s f k. Thus, the surface area is approimated b the Riemann sum n p f k f k. k. Write the limit of the Riemann sums as a definite integral from a to b. When will the limit eist?. Use the formula from part to find the surface area of the solid generated b revolving a single arch of the curve sin about the -ais. 3. The region enclosed b the graphs of and 4 is revolved about the -ais to form a solid. Find the surface area of the solid. Quick Review 7.3 (For help, go to Section..) In Eercises, give a formula for the area of the plane region in terms of the single variable.. a square with sides of length. a square with diagonals of length / 3. a semicircle of radius p / 4. a semicircle of diameter p /8 5. an equilateral triangle with sides of length ( 3 /4) 6. an isosceles right triangle with legs of length / 7. an isosceles right triangle with hpotenuse /4 8. an isosceles triangle with two sides of length and one side of length ( 5 /4) 9. a triangle with sides 3, 4, and 5 6. a regular heagon with sides of length (3 3 /)

29 46 Chapter 7 Applications of Definite Integrals Section 7.3 Eercises In Eercises and, find a formula for the area A() of the cross sections of the solid that are perpendicular to the -ais.. The solid lies between planes perpendicular to the -ais at and. The cross sections perpendicular to the -ais between these planes run from the semicircle to the semicircle. (a) The cross sections are circular disks with diameters in the -plane. p( ). The solid lies between planes perpendicular to the -ais at and 4. The cross sections perpendicular to the -ais between these planes run from to. (a) The cross sections are circular disks with diameters in the -plane. p 4 (b) The cross sections are squares with bases in the -plane. 4 (b) The cross sections are squares with bases in the -plane. 4( ) (c) The cross sections are squares with diagonals in the -plane. (The length of a square s diagonal is times the length of its sides.) ( ) (c) The cross sections are squares with diagonals in the -plane. (d) The cross sections are equilateral triangles with bases in the -plane. 3 In Eercises 3 6, find the volume of the solid analticall. 3. The solid lies between planes perpendicular to the -ais at and 4. The cross sections perpendicular to the ais on the interval 4 are squares whose diagonals run from to The solid lies between planes perpendicular to the -ais at and. The cross sections perpendicular to the -ais are circular disks whose diameters run from the parabola to the parabola. 6p/5 4 (d) The cross sections are equilateral triangles with bases in the -plane. 3 ( ) 5. The solid lies between planes perpendicular to the -ais at and. The cross sections perpendicular to the -ais between these planes are squares whose bases run from the semicircle to the semicircle. 6/3

30 Section 7.3 Volumes The solid lies between planes perpendicular to the -ais at and. The cross sections perpendicular to the -ais between these planes are squares whose diagonals run from the semicircle to the semicircle. 8/3 In Eercises 7, find the volume of the solid generated b revolving the shaded region about the given ais. 7. about the -ais p/3 8. about the -ais 6p 9. about the -ais 4 p. about the -ais p /6 tan 4 In Eercises, find the volume of the solid generated b revolving the region bounded b the lines and curves about the -ais..,, 3p/5. 3,, 3. 9, 36p 4., p/3 5.,, p/3 6.,, p.,, 8p In Eercises and, find the volume of the solid generated b revolving the region about the given line.. the region in the first quadrant bounded above b the line, below b the curve sec tan, and on the left b the -ais, about the line.3. the region in the first quadrant bounded above b the line, below b the curve sin, p, and on the left b the -ais, about the line p(3p 8) In Eercises 3 8, find the volume of the solid generated b revolving the region about the -ais. 3. the region enclosed b 5,,, p 4. the region enclosed b 3,, 4p 5. the region enclosed b the triangle with vertices,,,, and, 4p/3 6. the region enclosed b the triangle with vertices,,,, and, p/3 3 3/ sin cos 8p/7 7., , 8p/5 7p/5 9. sec,, p 4 p 4 p p 7. the region in the first quadrant bounded above b the parabola,below b the -ais, and on the right b the line 8p 8. the region bounded above b the curve and below b the line p/5 Group Activit In Eercises 9 3, find the volume of the solid described. 9. Find the volume of the solid generated b revolving the region bounded b and the lines and about (a) the -ais. 8p (b) the -ais. 3p/5 (c) the line. 8p/3 (d) the line 4. 4p/5 3. Find the volume of the solid generated b revolving the triangular region bounded b the lines,, and about (a) the line. p/3 (b) the line. 8p/3 3. Find the volume of the solid generated b revolving the region bounded b the parabola and the line about (a) the line. 6p/5 (b) the line. 56p/5 (c) the line. 64p/5 3. B integration, find the volume of the solid generated b revolving the triangular region with vertices,, b,,, h about (a) the -ais. (p/3)bh (b) the -ais. (p/3)b h In Eercises 33 and 34, use the clindrical shell method to find the volume of the solid generated b revolving the shaded region about the indicated ais. 33. (a) the -ais 6p/5 (b) the line 4p/5 (c) the line 8 5 p (d) the line 5 p ( 3 ) 34. (a) the -ais 8p/3 (b) the line 8p/5 (c) the line 5 8p (d) the line 5 8 4p 4 4 (, )

31 48 Chapter 7 Applications of Definite Integrals In Eercises 35 38, use the clindrical shell method to find the volume of the solid generated b revolving the region bounded b the curves about the -ais. 35.,, 8p 36.,,, for 5p/6 37.,, 4 8p/5 38.,, 7p/5 In Eercises 39 4, find the volume of the solid analticall. 39. The base of a solid is the region between the curve si n and the interval, p on the -ais. The cross sections perpendicular to the -ais are (a) equilateral triangles with bases running from the -ais to the curve as shown in the figure. 3 (b) squares with bases running from the -ais to the curve The solid lies between planes perpendicular to the -ais at p 3 and p 3. The cross sections perpendicular to the -ais are (a) circular disks with diameters running from the curve tan to the curve sec. p 3 (p /6) (b) squares whose bases run from the curve tan to the curve sec. 4 3 (p/3) 4. The solid lies between planes perpendicular to the -ais at and. The cross sections perpendicular to the -ais are circular disks with diameters running from the -ais to the parabola 5. 8p 4. The base of the solid is the disk. The cross sections b planes perpendicular to the -ais between and are isosceles right triangles with one leg in the disk. 8/3 sin 43. Writing to Learn A solid lies between planes perpendicular to the -ais at and. The cross sections b planes perpendicular to the -ais are circular disks whose diameters run from the line to the line as shown in the figure. Eplain wh the solid has the same volume as a right circular cone with base radius 3 and height. The volumes are equal b Cavalieri s Theorem. 44. A Twisted Solid A square of side length s lies in a plane perpendicular to a line L. One verte of the square lies on L. As this square moves a distance h along L, the square turns one revolution about L to generate a corkscrew-like column with square cross sections. (a) Find the volume of the column. (b) Writing to Learn What will the volume be if the square turns twice instead of once? Give reasons for our answer. s h 45. Find the volume of the solid generated b revolving the region in the first quadrant bounded b 3 and 4 about (a) the -ais, 5p/ (b) the line 8. 83p/ 46. Find the volume of the solid generated b revolving the region bounded b and about (a) the -ais, p/6 (b) the line. p/6 47. The region in the first quadrant that is bounded above b the curve, on the left b the line 4, and below b the line is revolved about the -ais to generate a solid. Find the volume of the solid b (a) the washer method and (b) the clindrical shell method. (a) p/48 (b) p/48 sin, p 48. Let f {,. s h (a) Show that f sin, p. (b) Find the volume of the solid generated b revolving the shaded region about the -ais. 4p sin,,

32 Section 7.3 Volumes Designing a Plumb Bob Having been asked to design a brass plumb bob that will weigh in the neighborhood of 9 g, ou decide to shape it like the solid of revolution shown here. (cm) , ; -ais , 3; -ais ,.5.5; -ais , 5; -ais 5.33 (a) Find the plumb bob s volume. 36p/5 cm 3 (b) If ou specif a brass that weighs 8.5 g cm 3, how much will the plumb bob weigh to the nearest gram? 9.3 g 5. Volume of a Bowl A bowl has a shape that can be generated b revolving the graph of between and 5 about the -ais. (a) Find the volume of the bowl. 5p (b) If we fill the bowl with water at a constant rate of 3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep? 3/(8p) 5. The Classical Bead Problem A round hole is drilled through the center of a spherical solid of radius r. The resulting clindrical hole has height 4 cm. (a) What is the volume of the solid that remains? 3p/3 (b) What is unusual about the answer? 5. Writing to Learn Eplain how ou could estimate the volume of a solid of revolution b measuring the shadow cast on a table parallel to its ais of revolution b a light shining directl above it. See page Same Volume about Each Ais The region in the first quadrant enclosed between the graph of a and the -ais generates the same volume whether it is revolved about the -ais or the -ais. Find the value of a (Continuation of Eploration ) Let g have a continuous first derivative on c, d. Show that the area of the surface generated b revolving the curve g about the -ais is See page 4. S d c p g g d. In Eercises 55 6, find the area of the surface generated b revolving the curve about the indicated ais. 55., ; -ais , ; -ais , 3; -ais , 5 8 ; -ais (cm The answer is independent of r. Standardized Test Questions You ma use a graphing calculator to solve the following problems. 63. True or False The volume of a solid of a known integrable cross section area A() from a to b is b A() d. Justif a our answer. True, b definition. 64. True or False If the region enclosed b the -ais, the line, and the curve is revolved about the -ais, the volume of the solid is given b the definite integral p d. Justif our answer. False. The volume is given b p 4 d. 65. Multiple Choice The base of a solid S is the region enclosed b the graph of ln, the line e, and the -ais. If the cross sections of S perpendicular to the -ais are squares, which of the following gives the best approimation of the volume of S? A (A).78 (B).78 (C).78 (D) 3.7 (E) Multiple Choice Let R be the region in the first quadrant bounded b the graph of 8 3/, the -ais, and the -ais. Which of the following gives the best approimation of the volume of the solid generated when R is revolved about the -ais? E (A) 6.3 (B) 5. (C) 5.4 (D) 39.7 (E) Multiple Choice Let R be the region enclosed b the graph of, the line 4, and the -ais. Which of the following gives the best approimation of the volume of the solid generated when R is revolved about the -ais? B (A) 64p (B) 8p (C) 56p (D) 36 (E) Multiple Choice Let R be the region enclosed b the graphs of e, e, and. Which of the following gives the volume of the solid generated when R is revolved about the -ais? D (A) (e e ) d (B) (e e ) d (C) (e e ) d (D) p (e e ) d (E) p (e e ) d

33 4 Chapter 7 Applications of Definite Integrals Eplorations 69. Ma-Min The arch sin, p, is revolved about the line c, c, to generate the solid in the figure. (a) Find the value of c that minimizes the volume of the solid. What is the minimum volume? p, p 8 (b) What value of c in, maimizes the volume of the solid? (c) Writing to Learn Graph the solid s volume as a function of c, first for c and then on a larger domain. What happens to the volume of the solid as c moves awa from,? Does this make sense phsicall? Give reasons for our answers. c 7. A Vase We wish to estimate the volume of a flower vase using onl a calculator, a string, and a ruler. We measure the height of the vase to be 6 inches. We then use the string and the ruler to find circumferences of the vase (in inches) at half-inch intervals. (We list them from the top down to correspond with the picture of the vase.) 6 sin Circumferences c (a) Find the areas of the cross sections that correspond to the given circumferences..3,.6,.5,., 3., 4.8, 7., 9.3,.7,.7, 9.3, 6.4, 3. (b) Epress the volume of the vase as an integral with respect to over the interval, 6. 4 p 6 C() d (c) Approimate the integral using the Trapezoidal Rule with n in 3 5. Partition the appropriate interval on the ais of revolution and measure the radius r() of the shadow region at these points. Then use an approimation such as the trapezoidal rule to estimate the integral b a pr () d. 54. For a tin horizontal slice, slant height s ( ) ( ) ()) (g. So the surface area is approimated b the Riemann sum n k p g( k ) ()) (g. The limit of that is the integral. V p(c p 8c p) Volume Etending the Ideas 7. Volume of a Hemisphere Derive the formula V 3 pr 3 for the volume of a hemisphere of radius R b comparing its cross sections with the cross sections of a solid right circular clinder of radius R and height R from which a solid right circular cone of base radius R and height R has been removed as suggested b the figure. 7. Volume of a Torus The disk a is revolved about the line b b a to generate a solid shaped like a doughnut, called a torus. Find its volume. (Hint: a a a d pa, since it is the area of a semicircle of radius a.) a bp 73. Filling a Bowl h R h R (a) Volume A hemispherical bowl of radius a contains water to a depth h. Find the volume of water in the bowl. ph (3a h)/3 (b) Related Rates Water runs into a sunken concrete hemispherical bowl of radius 5 m at a rate of. m 3 sec. How fast is the water level in the bowl rising when the water is 4 m deep? /(p) m/sec 74. Consistenc of Volume Definitions The volume formulas in calculus are consistent with the standard formulas from geometr in the sense that the agree on objects to which both appl. (a) As a case in point, show that if ou revolve the region enclosed b the semicircle a and the -ais about the -ais to generate a solid sphere, the calculus formula for volume at the beginning of the section will give 4 3 pa 3 for the volume just as it should. (b) Use calculus to find the volume of a right circular cone of height h and base radius r. 7. Hemisphere cross sectional area: p( R h ) A Right circular clinder with cone removed cross sectional area: pr ph A Since A A, the two volumes are equal b Cavalieri s theorem. Thus, volume of hemisphere volume of clinder volume of cone R h pr 3 3 pr 3 3 pr (a) A cross section has radius r a and area A() pr p(a ). V a p(a ) 4 a 3 pa3 (b) A cross section has radius r h and area A() p pr h h. V h pr h h d 3 pr h h

34 Section 7.3 Volumes 4 Quick Quiz for AP* Preparation: Sections You ma use a graphing calculator to solve the following problems.. Multiple Choice The base of a solid is the region in the first quadrant bounded b the -ais, the graph of sin, and the vertical line. For this solid, each cross section perpendicular to the -ais is a square. What is the volume? C (A).7 (B).85 (C).467 (D).57 (E).57. Multiple Choice Let R be the region in the first quadrant bounded b the graph of 3 and the -ais. A solid is generated when R is revolved about the vertical line. Set up, but do not evaluate, the definite integral that gives the volume of this solid. A 3 (A) p( )(3 ) d (B) 3 (C) 3 3 (D) (E) 3 p( )(3 ) d p()(3 ) d p(3 ) d (3 ) d 3. Multiple Choice A developing countr consumes oil at a rate given b r(t) e.t million barrels per ear, where t is time measured in ears, for t. Which of the following epressions gives the amount of oil consumed b the countr during the time interval t? D (A) r() (B) r() r() (C) r (t) dt (D) r(t) dt (E) r() 4. Free Response Let R be the region bounded b the graphs of, e, and the -ais. (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the horizontal line. (c) The region R is the base of a solid. For this solid, each cross section perpendicular to the -ais is a semicircle whose diameter runs from the graph of to the graph of e. Find the volume of this solid. 4. (a) The two graphs intersect where e, which a calculator shows to be Store this value as A. The area of R is A (e ) d.6. (b) Volume A p((e ) ( ) ) d.63. (c) Volume A p e d.35.

35 4 Chapter 7 Applications of Definite Integrals What ou ll learn about A Sine Wave 7.4 Length of a Smooth Curve Vertical Tangents, Corners, and Cusps... and wh The length of a smooth curve can be found using a definite integral. [, ] b [, ] Figure 7.3 One wave of a sine curve has to be longer than p. O k + k P k k Group Eploration Later in this section we will use an integral to find the length of the sine wave with great precision. But there are was to get good approimations without integrating. Take five minutes to come up with a written estimate of the curve s length. No fair looking ahead. Q k k = sin Figure 7.33 The line segment approimating the arc PQ of the sine curve above the subinterval k, k. (Eample ) Lengths of Curves A Sine Wave How long is a sine wave (Figure 7.3)? The usual meaning of wavelength refers to the fundamental period, which for sin is p. But how long is the curve itself? If ou straightened it out like a piece of string along the positive -ais with one end at, where would the other end be? EXAMPLE The Length of a Sine Wave What is the length of the curve sin from to p? We answer this question with integration, following our usual plan of breaking the whole into measurable parts. We partition, p into intervals so short that the pieces of curve (call them arcs ) ling directl above the intervals are nearl straight. That wa, each arc is nearl the same as the line segment joining its two ends and we can take the length of the segment as an approimation to the length of the arc. Figure 7.33 shows the segment approimating the arc above the subinterval k, k. The length of the segment is k. k The sum k k over the entire partition approimates the length of the curve. All we need now is to find the limit of this sum as the norms of the partitions go to zero. That s the usual plan, but this time there is a problem. Do ou see it? The problem is that the sums as written are not Riemann sums. The do not have the form f c k. We can rewrite them as Riemann sums if we multipl and divide each square root b k. k k k k k ( k k ) k This is better, but we still need to write the last square root as a function evaluated at some c k in the kth subinterval. For this, we call on the Mean Value Theorem for differentiable functions (Section 4.), which sas that since sin is continuous on k, k and is differentiable on k, k there is a point c k in k, k at which k k sin c k (Figure 7.34). That gives us which is a Riemann sum. si n c k k, Now we take the limit as the norms of the subdivisions go to zero and find that the length of one wave of the sine function is p si n d p c o s d Using NINT How close was our estimate? Now tr Eercise 9. k

36 Section 7.4 Lengths of Curves 43 P Slope sin' (c k ) k k c k k k Figure 7.34 The portion of the sine curve above k, k. At some c k in the interval, sin c k k k, the slope of segment PQ. (Eample ) d c (a, c) a ( k ) ( k ) f() k k Q k k (b, d) Figure 7.35 The graph of f, approimated b line segments. P Q b Length of a Smooth Curve We are almost read to define the length of a curve as a definite integral, using the procedure of Eample. We first call attention to two properties of the sine function that came into pla along the wa. We obviousl used differentiabilit when we invoked the Mean Value Theorem to replace k k b sin c k for some c k in the interval k, k. Less obviousl, we used the continuit of the derivative of sine in passing from si n c k k to the Riemann integral. The requirement for finding the length of a curve b this method, then, is that the function have a continuous first derivative. We call this propert smoothness. A function with a continuous first derivative is smooth and its graph is a smooth curve. Let us review the process, this time with a general smooth function f. Suppose the graph of f begins at the point a, c and ends at b, d, as shown in Figure We partition the interval a b into subintervals so short that the arcs of the curve above them are nearl straight. The length of the segment approimating the arc above the subinterval k, k is k. k The sum k k approimates the length of the curve. We appl the Mean Value Theorem to f on each subinterval to rewrite the sum as a Riemann sum, k k ( k ) k k f c k k. Passing to the limit as the norms of the subdivisions go to zero gives the length of the curve as L b a f d b a ( d d We could as easil have transformed k k into a Riemann sum b dividing and multipling b k, giving a formula that involves as a function of sa, g on the interval c, d : L k k k k ( ) k ) k d. k g c k k. For some point c k in ( k, k ) For some c k in ( k, k ) The limit of these sums, as the norms of the subdivisions go to zero, gives another reasonable wa to calculate the curve s length, L d g d d d. c c ( d d Putting these two formulas together, we have the following definition for the length of a smooth curve. DEFINITION ) Arc Length: Length of a Smooth Curve If a smooth curve begins at a, c and ends at b, d, a b, c d, then the length (arc length) of the curve is L b L d a ( d d c ( d d ) d ) d if is a smooth function of on a, b ; if is a smooth function of on c, d.

37 44 Chapter 7 Applications of Definite Integrals EXAMPLE Appling the Definition Find the eact length of the curve 4 3 for. which is continuous on,. Therefore, d 4 3 d 3, ) L ( d d d ( ) 8 d ] 3. 6 d (8, ) Now tr Eercise. We asked for an eact length in Eample to take advantage of the rare opportunit it afforded of taking the antiderivative of an arc length integrand. When ou add to the square of the derivative of an arbitrar smooth function and then take the square root of that sum, the result is rarel antidifferentiable b reasonable methods. We know a few more functions that give nice integrands, but we are saving those for the eercises. ( 8, ) Figure 7.36 The graph of 3 has a vertical tangent line at the origin where d d does not eist. (Eample 3) (, 8) (, 8) Figure 7.37 The curve in Figure 7.36 plotted with as a function of. The tangent at the origin is now horizontal. (Eample 3) Vertical Tangents, Corners, and Cusps Sometimes a curve has a vertical tangent, corner, or cusp where the derivative we need to work with is undefined. We can sometimes get around such a difficult in was illustrated b the following eamples. EXAMPLE 3 A Vertical Tangent Find the length of the curve 3 between 8, and 8,. The derivative d d is not defined at. Graphicall, there is a vertical tangent at where the derivative becomes infinite (Figure 7.36). If we change to as a function of, the tangent at the origin will be horizontal (Figure 7.37) and the derivative will be zero instead of undefined. Solving 3 for gives 3, and we have L ) d 3 d 7.6. Using NINT ( d d Now tr Eercise 5.

38 Section 7.4 Lengths of Curves 45 What happens if ou fail to notice that d d is undefined at and ask our calculator to compute NINT ( ( 3 3) 8),, 8,? This actuall depends on our calculator. If, in the process of its calculations, it tries to evaluate the function at, then some sort of domain error will result. If it tries to find convergent Riemann sums near, it might get into a long, futile loop of computations that ou will have to interrupt. Or it might actuall produce an answer in which case ou hope it would be sufficientl bizarre for ou to realize that it should not be trusted. EXAMPLE 4 Getting Around a Corner Find the length of the curve 4 from 4 to 4. We should alwas be alert for abrupt slope changes when absolute value is involved. We graph the function to check ( Figure 7.38). There is clearl a corner at where neither d d nor d d can eist. To find the length, we split the curve at to write the function without absolute values: [ 5, 5] b [ 7, 5] Figure 7.38 The graph of 4, 4 4, has a corner at where neither d/d nor d/d eists. We find the lengths of the two smooth pieces and add them together. (Eample 4) Then, L if, 4 { 5 if. 3 d 4 5 d B NINT Now tr Eercise 7. Finall, cusps are handled the same wa corners are: split the curve into smooth pieces and add the lengths of those pieces. Quick Review 7.4 (For help, go to Sections.3 and 3..) In Eercises 5, simplif the function.. on, 5. 4 on 3, 3. ta n on, p 3 sec 4. 4 on 4, c o s on, p cos In Eercises 6, identif all values of for which the function fails to be differentiable. 6. f f f f 4 4. f 3 si n kp, k an integer

39 46 Chapter 7 Applications of Definite Integrals Section 7.4 Eercises In Eercises, (a) set up an integral for the length of the curve; (b) graph the curve to see what it looks like; (c) use NINT to find the length of the curve..,. tan, p 3 3. sin, p 4., 5., from, to 7, 3 6. sin cos, p 7. tan tdt, p 6 8. se c t dt, 9. sec, p 3 p 3. e e, 3 3 p 3 p 4 In Eercises 8, find the eact length of the curve analticall b antidifferentiation. You will need to simplif the integrand algebraicall before finding an antiderivative from to 3. 3 from to 4 (8 8)/ from to 3 [Hint: d d is a perfect square.] 53/ from to [Hint: d d is a perfect square.] 3/ from to [Hint: d d is a perfect square.] 7/ , 53/6 7. se c 4 t dt, p 4 p t 4 dt, 7 3 /3 9. (a) Group Activit Find a curve through the point, whose length integral is L 4 4 d. (b) Writing to Learn How man such curves are there? Give reasons for our answer.. (a) Group Activit Find a curve through the point, whose length integral is /( ) L 4 d. (b) Writing to Learn How man such curves are there? Give reasons for our answer.. Find the length of the curve from to p 4. co s t dt. The Length of an Astroid The graph of the equation 3 3 is one of the famil of curves called astroids (not asteroids ) because of their starlike appearance (see figure). Find the length of this particular astroid b finding the length of half the first quadrant portion, 3 3, 4, and multipling b Fabricating Metal Sheets Your metal fabrication compan is bidding for a contract to make sheets of corrugated steel roofing like the one shown here. The cross sections of the corrugated sheets are to conform to the curve sin ( 3 p ), in. If the roofing is to be stamped from flat sheets b a process that does not stretch the material, how wide should the original material be? Give our answer to two decimal places..7 inches Original sheet in. Corrugated sheet 4. Tunnel Construction Your engineering firm is bidding for the contract to construct the tunnel shown on the net page. The tunnel is 3 ft long and 5 ft wide at the base. The cross section is shaped like one arch of the curve 5 cos p 5. Upon completion, the tunnel s inside surface (ecluding the roadwa) will be treated with a waterproof sealer that costs $.75 per square foot to appl. How much will it cost to appl the sealer? $38,4 O /3 /3 sin 3 (in.) 9. (b) Onl one. We know the derivative of the function and the value of the function at one value of.. (b) Onl one. We know the derivative of the function and the value of the function at one value of.

40 3. Because the limit of the sum k as the norm of the partition goes to zero will alwas be the length (b a) of the interval (a, b). 5 cos ( /5) Section 7.4 Lengths of Curves Multiple Choice Which of the following gives the best approimation of the length of the arc of cos() from to p/4? D (A).785 (B).955 (C). (D).38 (E) NOT TO SCALE 3 ft (ft) 35. Multiple Choice Which of the following epressions gives the length of the graph of 3 from to? C (A) ( 6 ) d (B) (C) 9 4 d (E) 4 d 6 d (D) d 36. Multiple Choice Find the length of the curve described b 3 3/ from to 8. B (A) 6 (B) 5 (C) (D) 5 8 (E) 96 5 In Eercises 5 and 6, find the length of the curve. 5. f 3 3, f 4,.89 In Eercises 7 9, find the length of the nonsmooth curve from to from to Writing to Learn Eplain geometricall wh it does not work to use short horizontal line segments to approimate the lengths of small arcs when we search for a Riemann sum that leads to the formula for arc length. 3. Writing to Learn A curve is totall contained inside the square with vertices,,,,,, and,. Is there an limit to the possible length of the curve? Eplain. Standardized Test Questions You should solve the following problems without using a graphing calculator. 3. True or False If a function f () is continuous on an interval [a, b], then the length of its curve is given b b d a d d. Justif our answer. False. The function must be differentiable. 33. True or False If a function f () is differentiable on an interval [a, b], then the length of its curve is given b b d. Justif our answer. True, b definition. a d d 3. No. Consider the curve 3 sin.5 for. 37. Multiple Choice Which of the following epressions should be used to find the length of the curve /3 from to? A (A) 9 4 (C) (E) 3 d 9/4 d Eploration d (B) 9 4 d (D) 6 d 38. Modeling Running Tracks Two lanes of a running track are modeled b the semiellipses as shown. The equation for lane is., and the equation for lane is 5.. The starting point for lane is at the negative -intercept 5,. The finish points for both lanes are the positive -intercepts. Where should the starting point be placed on lane so that the two lane lengths will be equal (running clockwise)? ( 9.99, 8.4) Start lane Start lane

41 48 Chapter 7 Applications of Definite Integrals Etending the Ideas 39. Using Tangent Fins to Find Arc Length Assume f is smooth on a, b and partition the interval a, b in the usual wa. In each subinterval k, k construct the tangent fin at the point k, f k as shown in the figure. ( k, f( k )) f() k Tangent fin with slope f'( k ) (a) Show that the length of the kth tangent fin over the interval k, k equals (b) Show that k f k k. lim n n (length of kth tangent fin) b f d, k a which is the length L of the curve f from a to b. 4. Is there a smooth curve f whose length over the interval a is alwas a? Give reasons for our answer. Yes. An curve of the form c, c a constant. k k 39. (a) The fin is the hpotenuse of a right triangle with leg lengths k and df d k k f ( k ) k. (b) lim n n ( k ) ( f ( k ) k ) k lim n n k ( ( f ) k ) k b a ()) ( f d

42 Section 7.5 Applications from Science and Statistics 49 What ou ll learn about Work Revisited 7.5 Fluid Force and Fluid Pressure Normal Probabilities... and wh It is important to see applications of integrals as various accumulation functions. Applications from Science and Statistics Our goal in this section is to hint at the diversit of was in which the definite integral can be used. The contets ma be new to ou, but we will eplain what ou need to know as we go along. Work Revisited Recall from Section 7. that work is defined as force (in the direction of motion) times displacement. A familiar eample is to move against the force of gravit to lift an object. The object has to move, incidentall, before work is done, no matter how tired ou get tring. If the force F() is not constant, then the work done in moving an object from a to b is the definite integral W b a F()d. 4.4 newtons lb ( newton)( meter) N m Joule EXAMPLE Finding the Work Done b a Force Find the work done b the force F() cos(p) newtons along the -ais from meters to meter. W cos(p) d p sin(p) / p sin p sin() p.38 Now tr Eercise. (N) Work 44 N. m EXAMPLE Work Done Lifting A leak bucket weighs newtons (N) empt. It is lifted from the ground at a constant rate to a point m above the ground b a rope weighing.4 N/m. The bucket starts with 7 N (approimatel 7. liters) of water, but it leaks at a constant rate and just finishes draining as the bucket reaches the top. Find the amount of work done (a) lifting the bucket alone; (b) lifting the water alone; (c) lifting the rope alone; (d) lifting the bucket, water, and rope together. (a) The bucket alone. This is eas because the bucket s weight is constant. To lift it, ou must eert a force of N through the entire -meter interval. (m) Figure 7.39 The work done b a constant -N force lifting a bucket m is 44 N m. (Eample ) Work N m 44 N m 44 J Figure 7.39 shows the graph of force vs. distance applied. The work corresponds to the area under the force graph. continued

43 4 Chapter 7 Applications of Definite Integrals 7 (N) F() = (m) Figure 7.4 The force required to lift the water varies with distance but the work still corresponds to the area under the force graph. (Eample ) 8 (N) Work (m) Figure 7.4 The work done lifting the rope to the top corresponds to the area of another triangle. (Eample ) (b) The water alone. The force needed to lift the water is equal to the water s weight, which decreases steadil from 7 N to N over the -m lift. When the bucket is m off the ground, the water weighs F 7 ( The work done is (Figure 7.4) W b a F d d [ ) ( 7 ) N. 7.75] J. (c) The rope alone. The force needed to lift the rope is also variable, starting at.4 8 N when the bucket is on the ground and ending at N when the bucket and rope are all at the top. As with the leak bucket, the rate of decrease is constant. At elevation meters, the meters of rope still there to lift weigh F.4 N. Figure 7.4 shows the graph of F. The work done lifting the rope is F d.4 d original weight of water proportion left at elevation [ 8.] N m 8 J. (d) The bucket, water, and rope together. The total work is J. Now tr Eercise 5. 8 or 5 (5, ) Figure 7.4 The conical tank in Eample 3. EXAMPLE 3 Work Done Pumping The conical tank in Figure 7.4 is filled to within ft of the top with olive oil weighing 57 lb ft 3. How much work does it take to pump the oil to the rim of the tank? We imagine the oil partitioned into thin slabs b planes perpendicular to the -ais at the points of a partition of the interval, 8. (The 8 represents the top of the oil, not the top of the tank.) The tpical slab between the planes at and has a volume of about V p(radius) (thickness) p ( ) The force F required to lift this slab is equal to its weight, F 57 V 57 p 4 lb. p 4 ft 3. weight per Weight ( unit volume ) volume The distance through which F must act to lift this slab to the level of the rim of the cone is about ft, so the work done lifting the slab is about W 57 p 4 ft lb. The work done lifting all the slabs from to 8 to the rim is approimatel W 57 p 4 ft lb. continued

44 Section 7.5 Applications from Science and Statistics 4 This is a Riemann sum for the function 57p 4 on the interval from to 8. The work of pumping the oil to the rim is the limit of these sums as the norms of the partitions go to zero. 57 p 4 d 57 p 4 3 d 57 p 4 [ ] 3,56 ft lb Now tr Eercise 7. W 8 8 Figure 7.43 To withstand the increasing pressure, dams are built thicker toward the bottom. Fluid Force and Fluid Pressure We make dams thicker at the bottom than at the top (Figure 7.43) because the pressure against them increases with depth. It is a remarkable fact that the pressure at an point on a dam depends onl on how far below the surface the point lies and not on how much water the dam is holding back. In an liquid, the fluid pressure p (force per unit area) at depth h is p wh, lb b Dimensions check: f t f lt3 ft, for eample where w is the weight-densit (weight per unit volume) of the liquid. EXAMPLE 4 The Great Molasses Flood of 99 Tpical Weight-densities (lb/ft 3 ) Gasoline 4 Mercur 849 Milk 64.5 Molasses Seawater 64 Water ft ft SHADED BAND NOT TO SCALE 9 ft Figure 7.44 The molasses tank of Eample 4. At : P.M.on Januar 5,99 (an unseasonabl warm da),a 9-ft-high,9-footdiameter clindrical metal tank in which the Puritan Distilling Compan stored molasses at the corner of Foster and Commercial streets in Boston s North End eploded. Molasses flooded the streets 3 feet deep, trapping pedestrians and horses, knocking down buildings, and oozing into homes. It was eventuall tracked all over town and even made its wa into the suburbs via trolle cars and people s shoes. It took weeks to clean up. (a) Given that the tank was full of molasses weighing lb ft 3, what was the total force eerted b the molasses on the bottom of the tank at the time it ruptured? (b) What was the total force against the bottom foot-wide band of the tank wall (Figure 7.44)? continued

45 4 Chapter 7 Applications of Definite Integrals (a) At the bottom of the tank, the molasses eerted a constant pressure of 9 k 45 Figure 7.45 The -ft band at the bottom of the tank wall can be partitioned into thin strips on which the pressure is approimatel constant. (Eample 4) p wh ( lb ft3 ) (9 ft) 9 lb ft. Since the area of the base was p 45, the total force on the base was ( lb 9 f t ) (5 p ft ) 57,5,56 lb. (b) We partition the band from depth 89 ft to depth 9 ft into narrower bands of width and choose a depth k in each one. The pressure at this depth k is p wh k lb ft (Figure 7.45). The force against each narrow band is approimatel pressure area k 9p 9p k lb. Adding the forces against all the bands in the partition and passing to the limit as the norms go to zero, we arrive at F pd 9p 9 d,53,553 lb 89 for the force against the bottom foot of tank wall. Now tr Eercise 5. Area = 4 5 Figure 7.46 The probabilit that the clock has stopped between : and 5: can be represented as an area of 4. The rectangle over the entire interval has area. Normal Probabilities Suppose ou find an old clock in the attic. What is the probabilit that it has stopped somewhere between : and 5:? If ou imagine time being measured continuousl over a -hour interval, it is eas to conclude that the answer is 4 (since the interval from : to 5: contains one-fourth of the time), and that is correct. Mathematicall, however, the situation is not quite that clear because both the -hour interval and the 3-hour interval contain an infinite number of times. In what sense does the ratio of one infinit to another infinit equal 4? The easiest wa to resolve that question is to look at area. We represent the total probabilit of the -hour interval as a rectangle of area sitting above the interval (Figure 7.46). Not onl does it make perfect sense to sa that the rectangle over the time interval, 5 has an area that is one-fourth the area of the total rectangle, the area actuall equals 4, since the total rectangle has area. That is wh mathematicians represent probabilities as areas, and that is where definite integrals enter the picture. Improper Integrals More information about improper integrals like f d can be found in Section 8.3. (You will not need that information here.) DEFINITION Probabilit Densit Function (pdf) A probabilit densit function is a function f with domain all reals such that f for all and f d. Then the probabilit associated with an interval a, b is b f d. a Probabilities of events, such as the clock stopping between : and 5:, are integrals of an appropriate pdf.

46 Section 7.5 Applications from Science and Statistics 43 EXAMPLE 5 Probabilit of the Clock Stopping Find the probabilit that the clock stopped between : and 5:. The pdf of the clock is, t f t {, otherwise. The probabilit that the clock stopped at some time t with t 5 is 5 f t dt 4. Now tr Eercise 7. f()d = Figure 7.47 A normal probabilit densit function. The probabilit associated with the interval a, b is the area under the curve, as shown. a b B far the most useful kind of pdf is the normal kind. ( Normal here is a technical term, referring to a curve with the shape in Figure 7.47.) The normal curve, often called the bell curve, is one of the most significant curves in applied mathematics because it enables us to describe entire populations based on the statistical measurements taken from a reasonabl-sized sample. The measurements needed are the mean m and the standard deviation s,which our calculators will approimate for ou from the data. The smbols on the calculator will probabl be J and s (see our Owner s Manual), but go ahead and use them as m and s,respectivel. Once ou have the numbers,ou can find the curve b using the following remarkable formula discovered b Karl Friedrich Gauss. DEFINITION Normal Probabilit Densit Function (pdf) The normal probabilit densit function (Gaussian curve) for a population with mean m and standard deviation s is f e m s. s p 68% of area 3 95% of the area 99.7% of the area 3 Figure 7.48 The rule for normal distributions. The mean m represents the average value of the variable. The standard deviation s measures the scatter around the mean. For a normal curve, the mean and standard deviation tell ou where most of the probabilit lies. The rule of thumb, illustrated in Figure 7.48, is this: The Rule for Normal Distributions =.5 = = Figure 7.49 Normal pdf curves with mean m and s.5,, and. Given a normal curve, 68% of the area will lie within s of the mean m, 95% of the area will lie within s of the mean m, 99.7% of the area will lie within 3s of the mean m. Even with the rule, the area under the curve can spread quite a bit, depending on the size of s. Figure 7.49 shows three normal pdfs with mean m and standard deviations equal to.5,, and.

47 44 Chapter 7 Applications of Definite Integrals EXAMPLE 6 A Telephone Help Line Suppose a telephone help line takes a mean of minutes to answer calls. If the standard deviation is s.5, then 68% of the calls are answered in the range of.5 to.5 minutes and 99.7% of the calls are answered in the range of.5 to 3.5 minutes. Now tr Eercise 9..3 [9,.5] b [,.5] Figure 7.5 The normal pdf for the spinach weights in Eample 7. The mean is at the center. EXAMPLE 7 Weights of Spinach Boes Suppose that frozen spinach boes marked as ounces of spinach have a mean weight of.3 ounces and a standard deviation of. ounce. (a) What percentage of all such spinach boes can be epected to weigh between and ounces? (b) What percentage would we epect to weigh less than ounces? (c) What is the probabilit that a bo weighs eactl ounces? Assuming that some person or machine is tring to pack ounces of spinach into these boes, we epect that most of the weights will be around, with probabilities tailing off for boes being heavier or lighter. We epect, in other words, that a normal pdf will model these probabilities. First, we define f using the formula: f e p The graph (Figure 7.5) has the look we are epecting. (a) For an arbitrar bo of this spinach, the probabilit that it weighs between and ounces is the area under the curve from to, which is NINT f,,,.933. So without doing an more measuring, we can predict that about 93.3% of all such spinach boes will weigh between and ounces. (b) For the probabilit that a bo weighs less than ounces, we use the entire area under the curve to the left of. The curve actuall approaches the -ais as an asmptote, but ou can see from the graph (Figure 7.5) that f approaches zero quite quickl. Indeed, f 9 is onl slightl larger than a billionth. So getting the area from 9 to should do it: NINT f,, 9,.67. We would epect onl about 6.7% of the boes to weigh less than ounces. (c) This would be the integral from to, which is zero. This zero probabilit might seem strange at first, but remember that we are assuming a continuous, unbroken interval of possible spinach weights, and is but one of an infinite number of them. Now tr Eercise 3. Quick Review 7.5 (For help, go to Section 5..) In Eercises 5, find the definite integral b (a) antiderivatives and (b) using NINT.. e d. e d a. e b..78 a. (/e) b..63 p a. / b sin d 4. 3 d 5 p d a. (/3) ln (9/) b..5

48 Section 7.5 Applications from Science and Statistics 45 In Eercises 6 find, but do not evaluate, the definite integral that is the limit as the norms of the partitions go to zero of the Riemann sums on the closed interval, p k sin k 7 7. k p k 7 p( ) sin d ( )(p) d 8. p cos k 7 9. k ) p( 3 4 p cos d k 7 p(/) ( ) d. sin k 7 ( 3 /4) sin d Section 7.5 Eercises In Eercises 4, find the work done b the force of F() newtons along the -ais from a meters to b meters.. F e 3, a, b J. F sin p 4, a, b J 3. F 9, a, b 3 9 J 4. F e sin cos, a, b J 5. Leak Bucket The workers in Eample changed to a larger bucket that held 5 L (49 N) of water, but the new bucket had an even larger leak so that it too was empt b the time it reached the top. Assuming the water leaked out at a stead rate, how much work was done lifting the water to a point meters above the ground? (Do not include the rope and bucket.) 49 J 6. Leak Bucket The bucket in Eercise 5 is hauled up more quickl so that there is still L (98 N) of water left when the bucket reaches the top. How much work is done lifting the water this time? (Do not include the rope and bucket.) 588 J 7. Leak Sand Bag A bag of sand originall weighing 44 lb was lifted at a constant rate. As it rose, sand leaked out at a constant rate. The sand was half gone b the time the bag had been lifted 8 ft. How much work was done lifting the sand this far? ( Neglect the weights of the bag and lifting equipment.) 944 ft-lb 8. Stretching a Spring A spring has a natural length of in. An 8-lb force stretches the spring to 4 in. (a) Find the force constant. lb/in. (b) How much work is done in stretching the spring from in. to in.? 4 in.-lb (c) How far beond its natural length will a 6-lb force stretch the spring? 8 in. 9. Subwa Car Springs It takes a force of,74 lb to compress a coil spring assembl on a New York Cit Transit Authorit subwa car from its free height of 8 in. to its full compressed height of 5 in. (a) What is the assembl s force constant? 738 lb/in. (b) How much work does it take to compress the assembl the first half inch? the second half inch? Answer to the nearest inchpound. 95 in.-lb and 74 in.-lb (Source: Data courtes of Bombardier, Inc., Mass Transit Division, for spring assemblies in subwa cars delivered to the New York Cit Transit Authorit from 985 to 987.). Bathroom Scale A bathroom scale is compressed 6 in. when a 5-lb person stands on it. Assuming the scale behaves like a spring that obes Hooke s Law, (a) how much does someone who compresses the scale 8 in. weigh? 3 lb 8.75 in.-lb (b) how much work is done in compressing the scale 8 in.?. Hauling a Rope A mountain climber is about to haul up a 5-m length of hanging rope. How much work will it take if the rope weighs.64 N m? 78 J. Compressing Gas Suppose that gas in a circular clinder of cross section area A is being compressed b a piston (see figure). (a) If p is the pressure of the gas in pounds per square inch and V is the volume in cubic inches, show that the work done in compressing the gas from state p, V to state p, V is given b the equation p, V Work p dv in. lb, p, V where the force against the piston is pa. (b) Find the work done in compressing the gas from V 43 in 3 to V 3 in 3 if p 5 lb in 3 and p and V obe the gas law pv.4 constant (for adiabatic processes). 37, in.-lb

49 46 Chapter 7 Applications of Definite Integrals Group Activit In Eercises 3 6, the vertical end of a tank containing water (blue shading) weighing 6.4 lb ft 3 has the given shape. (a) Writing to Learn Eplain how to approimate the force against the end of the tank b a Riemann sum. (b) Find the force as an integral and evaluate it. 3. semicircle (b) 3. lb 4. semiellipse (b) lb 5. triangle (b) 375 lb 6. parabola (b) 56. lb 3 3 ft 8 ft 6 ft 6 ft 8 ft 4 ft 4.5 ft 7. (d),494,4 ft-lb, min;,5, ft-lb, min (d) The Weight of Water Because of differences in the strength of Earth s gravitational field, the weight of a cubic foot of water at sea level can var from as little as 6.6 lb at the equator to as much as 6.59 lb near the poles, a variation of about.5%. A cubic foot of water that weighs 6.4 lb in Melbourne or New York Cit will weigh 6.5 lb in Juneau or Stockholm. What are the answers to parts (a) and (b) in a location where water weighs 6.6 lb ft 3? 6.5 lb ft 3? 8. Empting a Tank A vertical right clindrical tank measures 3 ft high and ft in diameter. It is full of kerosene weighing 5. lb ft 3. How much work does it take to pump the kerosene to the level of the top of the tank? 7,38,9 ft-lb 9. Writing to Learn The clindrical tank shown here is to be filled b pumping water from a lake 5 ft below the bottom of the tank. There are two was to go about this. One is to pump the water through a hose attached to a valve in the bottom of the tank. The other is to attach the hose to the rim of the tank and let the water pour in. Which wa will require less work? Give reasons for our answer. Through valve: 84,687.3 ft-lb Over the rim: 98,8.8 ft-lb Through a hose attached to a valve in the bottom is faster, because it takes more time to do more work. ft Open top 6 ft 6 ft 7. Pumping Water The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 6.4 lb ft 3. Ground level ft ft Valve at base. Drinking a Milkshake The truncated conical container shown here is full of strawberr milkshake that weighs 4 9 oz in 3. As ou can see, the container is 7 in. deep,.5 in. across at the base, and 3.5 in. across at the top (a standard size at Brigham s in Boston). The straw sticks up an inch above the top. About how much work does it take to drink the milkshake through the straw (neglecting friction)? Answer in inch-ounces in.-oz (.75, 7) (a) How much work does it take to empt the tank b pumping the water back to ground level once the tank is full?,497,6 ft-lb (b) If the water is pumped to ground level with a 5 -horsepower motor (work output 5 ft lb sec), how long will it take to empt the full tank (to the nearest minute)? min (c) Show that the pump in part (b) will lower the water level ft (halfwa) during the first 5 min of pumping..5 Dimensions in inches. Revisiting Eample 3 How much work will it take to pump the oil in Eample 3 to a level 3 ft above the cone s rim? 53,48.5 ft-lb

50 Section 7.5 Applications from Science and Statistics 47. Pumping Milk Suppose the conical tank in Eample 3 contains milk weighing 64.5 lb ft 3 instead of olive oil. How 6. Milk Carton A rectangular milk carton measures 3.75 in. b 3.75 in. at the base and is 7.75 in. tall. Find the force much work will it take to pump the contents to the rim? 34,58.65 ft-lb of the milk weighing 64.5 lb ft 3 on one side when the 3. Writing to Learn You are in charge of the evacuation and carton is full. 4. lb repair of the storage tank shown here. The tank is a hemisphere 7. Find the probabilit that a clock stopped between : and 5:. of radius ft and is full of benzene weighing 56 lb ft 3 /3. 8. Find the probabilit that a clock stopped between 3: and 6:. /4 9. Suppose a telephone help line takes a mean of minutes to answer calls. If the standard deviation is s, what percentage Outlet pipe of the calls are answered in the range of to 4 minutes? 68% ft z A firm ou contacted sas it can empt the tank for cent per foot-pound of work. Find the work required to empt the tank b pumping the benzene to an outlet ft above the tank. If ou have budgeted $5 for the job, can ou afford to hire the firm? 967,6 ft-lb, es 4. Water Tower Your town has decided to drill a well to increase its water suppl. As the town engineer, ou have determined that a water tower will be necessar to provide the pressure needed for distribution, and ou have designed the sstem shown here. The water is to be pumped from a 3-ft well through a vertical 4-in. pipe into the base of a clindrical tank ft in diameter and 5 ft high. The base of the tank will be 6 ft above ground. The pump is a 3-hp pump, rated at 65 ft lb sec. To the nearest hour, how long will it take to fill the tank the first time? (Include the time it takes to fill the pipe.) Assume water weighs 6.4 lb ft 3. 3 hr ft 5 ft 3. Test Scores The mean score on a national aptitude test is 498 with a standard deviation of points. (a) What percentage of the population has scores between 4 and 5?.34 (34%) (b) If we sample 3 test-takers at random, about how man should have scores above 7? Heights of Females The mean height of an adult female in New York Cit is estimated to be 63.4 inches with a standard deviation of 3. inches. What proportion of the adult females in New York Cit are (a) less than 63.4 inches tall?.5 (5%) (b) between 63 and 65 inches tall?.4 (4%) (c) taller than 6 feet?.36 (.36%) (d) eactl 5 feet tall? if we assume a continuous distribution;.7; 7.% between 59.5 in. and 6.5 in. 3. Writing to Learn Eercises 3 and 3 are subtl different, in that the heights in Eercise 3 are measured continuousl and the scores in Eercise 3 are measured discretel. The discrete probabilities determine rectangles above the individual test scores, so that there actuall is a nonzero probabilit of scoring, sa, 56. The rectangles would look like the figure below, and would have total area. Ground 4 in. Water surface NOT TO SCALE 6 ft 3 ft Submersible pump 5. Fish Tank A rectangular freshwater fish tank with base 4 ft and height ft (interior dimensions) is filled to within in. of the top. (a) Find the fluid force against each end of the tank lb (b) Suppose the tank is sealed and stood on end (without spilling) so that one of the square ends is the base. What does that do to the fluid forces on the rectangular sides? lb; the fluid force doubles Eplain wh integration gives a good estimate for the probabilit, even in the discrete case. Integration is a good approimation to the area. 33. Writing to Learn Suppose that f t is the probabilit densit function for the lifetime of a certain tpe of lightbulb where t is in hours. What is the meaning of the integral 8 The proportion of lightbulbs that last between and 8 f t dt? hours. Standardized Test Questions You ma use a graphing calculator to solve the following problems. 34. True or False A force is applied to compress a spring several inches. Assume the spring obes Hooke s Law. Twice as much work is required to compress the spring the second inch than is required to compress the spring the first inch. Justif our answer. False. Three times as much work is required.

51 48 Chapter 7 Applications of Definite Integrals 35. True. The force against each vertical side is 84.4 lb 35. True or False An aquarium contains water weighing 6.4 lb/ft 3. The aquarium is in the shape of a cube where the length of each edge is 3 ft. Each side of the aquarium is engineered to withstand pounds of force. This should be sufficient to withstand the force from water pressure. Justif our answer. 36. Multiple Choice A force of F() 35 newtons moves a particle along a line from m to 5 m. Which of the following gives the best approimation of the work done b the force? E (A) 75 J (B) 87.5 J (C) J (D) 38.5 J (E) 4375 J 37. Multiple Choice A leak bag of sand weighs 5 n. It is lifted from the ground at a constant rate, to a height of m above the ground. The sand leaks at a constant rate and just finishes draining as the bag reaches the top. Which of the following gives the work done to lift the sand to the top? (Neglect the bag.) D (A) 5 J (B) J (C) 5 J (D) 5 J (E) J 38. Multiple Choice A spring has a natural length of. m. A -n force stretches the spring to a length of.5 m. Which of the following gives the work done in stretching the spring from. m to.5 m? B (A).5 J (B) 5 J (C) J (D) J (E) 4 J 39. Multiple Choice A vertical right clindrical tank measures ft high and 6 ft in diameter. It is full of water weighing 6.4 lb/ft 3. How much work does it take to pump the water to the level of the top of the tank? Round our answer to the nearest ft-lb. E (A) 49,49 ft-lb (B) 85,696 ft-lb (C) 36,4 ft-lb (D) 448,776 ft-lb (E) 93,33 ft-lb Etending the Ideas 4. Putting a Satellite into Orbit The strength of Earth s gravitational field varies with the distance r from Earth s center, and the magnitude of the gravitational force eperienced b a satellite of mass m during and after launch is F r m MG. r Here, M kg is Earth s mass, G N m kg is the universal gravitational constant, and r is measured in meters. The work it takes to lift a -kg satellite from Earth s surface to a circular orbit 35,78 km above Earth s center is therefore given b the integral Work 35,78, MG r dr joules. 6,37, The lower limit of integration is Earth s radius in meters at the launch site. Evaluate the integral. (This calculation does not take into account energ spent lifting the launch vehicle or energ spent bringing the satellite to orbit velocit.) J 4. Forcing Electrons Together Two electrons r meters apart repel each other with a force of F 3 9 r newton. (a) Suppose one electron is held fied at the point, on the -ais (units in meters). How much work does it take to move a second electron along the -ais from the point, to the origin?.5 8 J (b) Suppose an electron is held fied at each of the points, and,. How much work does it take to move a third electron along the -ais from 5, to 3,? J 4. Kinetic Energ If a variable force of magnitude F moves a bod of mass m along the -ais from to,the bod s velocit v can be written as d dt (where t represents time). Use Newton s second law of motion, F m dv dt, and the Chain Rule d v dv d dv v dt d dt d to show that the net work done b the force in moving the bod from to is W F d mv mv, () where v and v are the bod s velocities at and. In phsics the epression mv is the kinetic energ of the bod moving with velocit v. Therefore, the work done b the force equals the change in the bod s kinetic energ, and we can find the work b calculating this change. Weight vs. Mass Weight is the force that results from gravit pulling on a mass. The two are related b the equation in Newton s second law, Thus, weight mass acceleration. newtons kilograms m sec, pounds slugs ft sec. See page 49. To convert mass to weight, multipl b the acceleration of gravit. To convert weight to mass, divide b the acceleration of gravit. In Eercises 43 49, use Equation from Eercise Tennis A -oz tennis ball was served at 6 ft sec (about 9 mph). How much work was done on the ball to make it go this fast? 5 ft-lb 44. Baseball How man foot-pounds of work does it take to throw a baseball 9 mph? A baseball weighs 5 oz.35 lb. 85. ft-lb 45. Golf A.6-oz golf ball is driven off the tee at a speed of 8 ft sec (about 9 mph). How man foot-pounds of work are done getting the ball into the air?.5 ft-lb 46. Tennis During the match in which Pete Sampras won the 99 U.S. Open men s tennis championship, Sampras hit a serve that was clocked at a phenomenal 4 mph. How much work did Sampras have to do on the -oz ball to get it to that speed? 64.6 ft-lb

52 Section 7.5 Applications from Science and Statistics Football A quarterback threw a 4.5-oz football 88 ft sec (6 mph). How man foot-pounds of work were done on the ball to get it to that speed? 9.7 ft-lb 48. Softball How much work has to be performed on a 6.5-oz softball to pitch it at 3 ft sec (9 mph)?.6 ft-lb 4. F m d v mv d v, so W F() d dt d mv d v d v mv dv d v mv mv Quick Quiz for AP* Preparation: Sections 7.4 and A Ball Bearing A -oz steel ball bearing is placed on a vertical spring whose force constant is k 8 lb ft. The spring is compressed 3 in. and released. About how high does the ball bearing go? (Hint: The kinetic (compression) energ, mgh, of a spring is ks,where s is the distance the spring is compressed, m is the mass, g is the acceleration of gravit, and h is the height.) 4.5 ft You should solve the following problems without using a graphing calculator.. Multiple Choice The length of a curve from to is given b 6 6 d. If the curve contains the point (, 4), which of the following could be an equation for this curve? A (A) 4 3 (B) 4 (C) 6 6 (D) (E) 7. Multiple Choice Which of the following gives the length of the path described b the parametric equations 4 t 4 and t 3, where t? D (A) t 6 9t 4 dt (B) t 6 dt (C) 9t 4 dt (D) t t dt (E) t 3 3t dt 3. Multiple Choice The base of a solid is a circle of radius inches. Each cross section perpendicular to a certain diameter is a square with one side ling in the circle. The volume of the solid in cubic inches is C (A) 6 (B) 6p (C) 8 (D) 8p 3 3 (E) 3p 4. Free Response The front of a fish tank is rectangular in shape and measures ft wide b.5 ft tall. The water in the tank eerts pressure on the front of the tank. The pressure at an point on the front of the tank depends onl on how far below the surface the point lies and is given b the equation p 6.4h, where h is depth below the surface measured in feet and p is pressure measured in pounds/ft. h ft The front of the tank can be partitioned into narrow horizontal bands of height h. The force eerted b the water on a band at depth h i is approimatel pressure area = 6.4h i h. (a) Write the Riemann sum that approimates the force eerted on the entire front of the tank. (b) Use the Riemann sum from part (a) to write and evaluate a definite integral that gives the force eerted on the front of the tank. Include correct units. (c) Find the total force eerted on the front of the tank if the front (and back) are semicircles with diameter ft. Include correct units. (a) n 6.4h i h i h ft.5 ft (b).5 6.4h dh 4.4 lbs (c).5 6.4h h dh 4.6 lbs

53 43 Chapter 7 Applications of Definite Integrals Calculus at Work I am working toward becoming an archeaoastronomer and ethnoastronomer of Africa. I have a Bachelor s degree in Phsics, a Master s degree in Astronom, and a Ph.D. in Astronom and Astrophsics. From 988 to 99 I was a member of the Peace Corps, and I taught mathematics to high school students in the Fiji Islands. Calculus is a required course in high schools there. For m Ph.D. dissertation, I investigated the possibilit of the birthrate of stars being related to the composition of star formation clouds. I collected data on the absorption of electromagnetic emissions emanating from these regions. The intensit of emissions graphed versus wavelength produces a flat curve with downward spikes at the characteristic wavelengths of the elements present. An estimate of the area between a spike and the flat curve results in a concentration in molecules/cm 3 of an element. This area is the difference in the integrals of the flat and spike curves. In particular, I was looking for a large concentration of water-ice, which increases the probabilit of planets forming in a region. Currentl, I am appling for two research grants. One will allow me to use the NASA infrared telescope on Mauna Kea to search for C 3 S in comets. The other will help me stud the histor of astronom in Tunisia. Jarita Holbrook Los Angeles, CA Chapter 7 Ke Terms arc length (p. 43) area between curves (p. 39) Cavalieri s theorems (p. 44) center of mass (p. 389) constant-force formula (p. 384) clindrical shells (p. 4) displacement (p. 38) fluid force (p. 4) fluid pressure (p. 4) foot-pound (p. 384) force constant (p. 385) Gaussian curve (p. 43) Hooke s Law (p. 385) inflation rate (p. 388) joule (p. 384) length of a curve (p. 43) mean (p. 43) moment (p. 389) net change (p. 379) newton (p. 384) normal curve (p. 43) normal pdf (p. 43) probabilit densit function (pdf) (p. 4) rule (p. 43) smooth curve (p. 43) smooth function (p. 43) solid of revolution (p. 4) standard deviation (p. 43) surface area (p. 45) total distance traveled (p. 38) universal gravitational constant (p. 48) volume b clindrical shells (p. 4) volume b slicing (p. 4) volume of a solid (p. 399) weight-densit (p. 4) work (p. 384) Chapter 7 Review Eercises The collection of eercises marked in red could be used as a chapter test. In Eercises 5, the application involves the accumulation of small changes over an interval to give the net change over that entire interval. Set up an integral to model the accumulation and evaluate it to answer the question.. A to car slides down a ramp and coasts to a stop after 5 sec. Its velocit from t to t 5 is modeled b v t t.t 3 ft sec. How far does it travel?.47 ft. The fuel consumption of a diesel motor between weekl maintenance periods is modeled b the function c t 4.t 4 gal da, t 7. How man gallons does it consume in a week? 3.36 gal 3. The number of billboards per mile along a -mile stretch of an interstate highwa approaching a certain cit is modeled b the function B e.3,where is the distance from the cit in miles. About how man billboards are along that stretch of highwa? 464