Contents 6. Area between curves........................................ 6.. Area between the curve and the -ais.......................... 6.. Overview of Area of a Region Between Two Curves................... 3 6..3 Area of a Region Between Two Curves.......................... 4 6. Volume............................................... 9 6.. Volume B Slicing..................................... 9 6.. Volumes of Solids of Revolution - Disks & Washers................... 6.3 Volumes b Clindrical Shells................................... 5 6.3. Clindrical shells formula................................. 5 6.3. Shell Method........................................ 6 6.4 Work................................................. 6.4. Work Done b a Constant Force............................. 6.4. Work Done b a Variable Force.............................. 6.4.3 Hooke s Law (Robert Hooke 635-73):......................... 6.4.4 Pumping Problems..................................... 3 6.5 Average Value of a Function.................................... 6 6.5. Average Value of a Function............................... 6 6.5. Mean Value Theorem for Integral Calculus....................... 6
6. Area between curves 6.. Area between the curve and the -ais Definition 6.. Let f() be continuous on [a,b]. The area of the region between the graph of f() and the -ais is A = b a f()d Let f() be continuous on [a,b]. The area of the region between the graph of f() and the -ais is A = b a f()d Eample 6... Set up the integral(s) needed to find the area of the region bounded b ) = 3, the -ais and = and =. 8 6 4 = 3.5..5. ) = and the -ais. =
3) = and the -ais on [,]. = 3 4 6.. Overview of Area of a Region Between Two Curves With a few modifications the area under a curve represented b a definite integral can be etended to find the area between to curves. Observe the following graphs of f() and g(). Since both graphs lie above the -ais, we can geometricall interpret the area of the region between the graphs as the area of the region under f() minus the area of the region under g(). Area of the region between f() and g() 7 6 5 4 3 f() g() 3 Area of the region under g() Area of the region under f() 7 6 5 4 3 3 7 6 5 4 3 3 3
Although we have just shown one case, i.e. when f and g are both positive, this condition is not necessar. All we need in order to evaluate the integrand [f() g()] is that f() and g() both be continuous and g() f() on the interval [a,b]. 6..3 Area of a Region Between Two Curves Finding the Area Between the Curves Using Vertical Rectangles; Integrating with respect to. If f() and g() are continuous on [a,b] and g() f() for all in [a,b], then the area of the region bounded b the graphs of f() and g() and the vertical lines = a and = b is A = b a [f() g()]d where f() is the upper curve and g() is the lower curve. In the net picture, ou can see that we simpl partition the interval [a,b] and draw our representative rectangles. Let s drawn a few rectangles to illustrate the concept. Steps to Find the Area Between Two Curves integrating with restpect to. Graph the curves and draw a representative rectangle. This reveals which curve is f() (upper curve) and which is g()(lower curve). It also helps find the limits of integration if ou do not alread know them.. Find the limits of integration; ou ma need to find the pts of intersection. 3. Write a formula for f() g(). Simplif it if ou can. 4. Integrate [f() g()] from a to b. The number ou get is the area. NOTE: If ou have ver complicated functions ou can use a graphing calculator, Matlab or another computer algebra sstem to draw the functions, this will greatl speed up our work. The main reason for doing step is to see how man times the graphs cross (if ever) in the specified domain. It also is to determine which function is dominant in which intervals. If ou don t have a quick means to graph the functions, ou can set the two functions equal to each other and tr to solve for where the are equal. This will give ou the intersection points of the two graphs. Once ou know these, ou can determine which function is dominant in which subintervals. Then just proceed as before. 4
Eample 6... Find the indicated areas ) Find the area of the region bounded b = and = +. 5 4 3 ) Find the area of the region bounded b =, = ( ) and =. =, = ( ), = 5
3) Find the area of the region bounded b = 3, =, = and =. 8 7 6 5 4 = 3 3 = 6
Finding the Area Between the Curves Using Horizontal Rectangles; Integrating with respect to. The procedure of adding rectangles works no matter how we draw the rectangles. We can just of easil draw our rectangles horizontall and integrate with respect to, (our representative rectangles will now be parallel to the - ais instead of the - ais). Consider the following graphs, we will draw horizontal rectangles in one and vertical in the other: Horizontal rectangles Vertical rectangles 3 3 In the first picture, we onl need to integrate one time. In the second picture, we would have to integrate twice using different limits of integration and different functions. Finding the Area Between Two Curves integrating with respect to If h() and k() are continuous on [c,d] and k() h() for all in [c,d], then the area of the region bounded b the graphs of h() and k() and the horizontal lines = c and = d is A = d where h() is the right curve and k() is the left curve. c [h() k()]d; Steps to Finding the Area Between Two Curves integrating with restpect to. Graph the curves and draw a representative rectangle. This reveals which curve is h() (right curve) and which is k() (left curve). It also helps find the limits of integration if ou do not alread know them.. Find the limits of integration; ou ma need to find the points of intersection. 3. Write a formula for h() k(). Simplif it if ou can. 4. Integrate [h() k()] from c to d. The number ou get is the area. 7
Eample 6..3. Find the indicated areas ) Set up the integrals to find the area of the region bounded b + = 4 and =. 3 4 ) Set up the integrals to find the area of the region bounded b =, = ( ) and =. =, = ( ), = 8
6. Volume 6.. Volume B Slicing We will be tring to find the volume of a solid shaped using the sum of cross section areas times a width. We will be driving toward developing a Riemann Sum so that we can transform them into integrals. Definition 6.. 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 Steps to Find Volumes b the Method of Slicing. Sketch the solid and a tpical cross section. a A() d.. Find a formula for A(), the area of a tpical cross section. 3. Find the limits of integration. 4. Integrate A() to find the volume, V = b a A() d. Eample 6... Find the volumes of the solids if the solid lies between planes perpendicular to the -ais at = and = 4. The cross sections perpendicular to the -ais between these planes run from the parabola = to the parabola =. The cross sections are semi-circular disks with diameters in the -plane. Note: the area of a semicircular disk is given b A = πr. = 3 4 = Eample 6... Find the volume of the solid that lies between planes perpendicular to the -ais at = and =. In each case, the cross sections perpendicular to the -ais, between these planes, run from the semicircle = to the semicircle =. ) Assume the cross sections are squares with bases in the -plane. Y X 9
) The cross sections are squares with diagonals in the -plane. (The length of a square s diagonal is times the length of its sides) Y X Eample 6..3. The base of the solid is the region between the curve = sin() and the interval [,π] on the -ais. The cross sections perpendicular to the -ais are vertical equilateral triangles with bases running from the -ais to the curve. Note: the area of a triangle is given b A = bh or A = 3 4 s = sin() π Y X Eample 6..4. The solid lies between the planes perpendicular to the -ais at = π 3 and = π 3. The cross sections perpendicular to the -ais are circular disks with diameters running from the curve = tan to the curve = sec.
6.. Volumes of Solids of Revolution - Disks & Washers Introduction In this part of the section, we begin with a known planar shape and rotate that shape about a line. The resulting three dimensional shape is known as a solid of revolution. The line about which we rotate the plane shape, is called the ais of rotation. Certain solids of revolution that can be generated like a clinder, a cone, or a sphere, we can find their volumes using formulas from geometr. However, when the solid or revolution takes on a non-regular shape, like a spool, a bullet, or a blimp, etc... there are often times no eas geometric volume formulas. So we fall back on integral calculus to compute the volume. A. Disks Method: Rotation about the - ais:. If we revolve a continuous function = R(), on [a,b], and the -ais, about the ais as our ais of revolution, then we have a solid formed (with no hole); the cross section perpendicular to the ais of revolution will be a disk of radius (R() ) which will have area A() = π(radius) = π[r()].. The volume of the solid generated b revolving about the -ais the region between the -ais and the graph of the continuous function = R(), a b, is V = b 3. Steps to find the volume of the solid of revolution a π[r()] d a. Graph the region. b. Determine the ais of revolution. c. Determine radius R(). d. Determine the limits of integration (Determined b the region). e. Integrate. Eample 6..5. Find the volume of the solid generated b revolving the region defined b =, = about the -ais. =, =..5.5.
B. Disks Method: Rotation about the -ais: (a) Similarl, we can find the volume of the solid when the region is rotated about the -ais. (b) The volume of the solid generated b revolving about the -ais the region between the -ais and the graph of the continuous function = R(), c d, is V = b a π[r()] d Eample 6..6. Find the volume of the solid generated b revolving the region defined b = 3, = 8 and = about the -ais. 8 = 3, = 8 6 4 C. Disks Method: Rotation about a line other than one of the aes: If ou are revolving the planar object about one of the lines that bound the function, then ou proceed in prett much the same wa, since ou still have a solid with no hole. In this tpe of problem, the radius of the disk is still the distance between the curve and the ais of revolution (i.e., subtract radius R from the ais of revolution or vise versa, whichever gives a positive length). Eample 6..7. Find the volume of the solid generated b revolving the region defined b = and = about =. = =
D. Washer Method: (a) If the region we revolve to generate the solid does not border on or cross the ais of revolution, the solid will have a hole in it. The cross section perpendicular to the ais of revolution are washers instead of disk. (b) The dimensions of a tpical washer are given: Outer Radius: R() Inner Radius: r() (c) Thus the area of the washer will be given as: A() = π(outer radius) π(inner radius) = π[r()] π[r()] A() = π ( [R()] [r()] ) (d) The volume of such a solid of revolution would be given as: V() = b a π ( [R()] [r()] ) d (e) This formula can be adjusted to be integrated with respect to. (f) The disk formula for finding volume is just the washer formula with r() =. Eample 6..8. Find the volume of the solid generated b revolving the region defined b = and = about the -ais. = = 3
Eample 6..9. Find the volume of the solid generated b revolving the region defined b =, = and = 4 about the -ais. =, = 4 4 Eample 6... Find the volume of the solid generated b revolving the region defined b = +, = +, = and = about = 3. = +, = + 3 4
6.3 Volumes b Clindrical Shells 6.3. Clindrical shells formula. Consider a representative rectangle that has the following characteristics: w = width of the rectangle h = height of the rectangle p = distance between ais of revolution and center of the rectangle. When this rectangle is revolved about the ais of revolution it forms a clindrical shell or tube of thickness w. In order to find the volume of this shell, we ll consider two clinders, one that has radius equal to R() = p+ w the other has radius r() = p w, since p is the average radius of the shell. Recall that the volume of a clinder is V = π(radius) (height) = πr h. So the volume of the shell is: V = πr h πr h = π(p+w/) h π(p w/) h = π [ (p +pw+w /4) (p pw +w /4) ] h = π(pw)h = π(average radius)(width)(height). An alternative wa to remember formula is to consider a clindrical shell with outer radius r and inner radius r. Now imagine cutting and unrolling this clindrical shell to get a nearl flat rectangular solid. Using the fact that the volume of a rectangular solid is V =(length)(width)(height); the volume of this rectangular solid would be V = π(r)(r r )h = (πr)h r. If we now take a solid of revolution and want to approimate its volume b adding the volumes of the shells swept out b n rectangles and take the limit of this sum as r the volume of the solid is: V = lim r k= n (πr)h r = b a πrh dr. 5
6.3. Shell Method A. Rotation about the -ais. The volume of the solid generated b revolving the region between the -ais and the graph of the continuous function = f() >, a < < b about the -ais, is V = π(shell radius)(shell height) d V = b. Steps to find the volume of the solid of revolution a πf()d a. Graph the region. b. Determine the ais of revolution. c. Determine shell radius. d. Determine shell height. e. Determine the limits of integration (Determined b the region). f. Integrate. Eample 6.3.. Find the volume of the solid generated b revolving the region defined b = + 4, = from = to = about the -ais. = + 4 6
Eample 6.3.. Find the volume of the solid generated b revolving the region defined b =, = 4 from about the -ais. =, = 4 4 3 B. Rotation about the -ais The volume of the solid generated b revolving the region between the -ais and the graph of the continuous function = g(), c d about the -ais, is V = π(shell radius)(shell height) d V = d c πg() d Eample 6.3.3. Find the volume of the solid generated b revolving the region in the first quadrant bounded b = 3 and = about the -ais. = 3.5 7
C. Rotation about a line other than one of the aes: Eample 6.3.4. Find the volume of the solid generated b revolving the region in the first quadrant bounded b = 3, = 9 and = about: a. the line = 9 b. the line = = 3, = 9 8 6 4 8
Eample 6.3.5. Find the volume of the solid generated b revolving the region bounded b = and = + about the line = 3. 4 =, = + 3 Eample 6.3.6. Find the volume of the solid generated b revolving the region in the first quadrant bounded b =, the -ais and the line = about the line =. =, = 3 9
6.4 Work 6.4. Work Done b a Constant Force Definition 6.3. If an object is moved a distance D in the direction of an applied constant force F, then the work W done b the force is defined as W = FD.. Notice that b this definition, if the object ou are tring to move, does not move, then no work has been done, even though ou have eerted work in our efforts to move it.. There are man tpes of forces - centrifugal, electromotive, and gravitational, etc. We tpicall measure forces according to their abilit to accelerate a unit of mass over time. Units Sstem of Measurement Units for Force Units for Distance Units for Work U.S. pound (lb) foot (ft) foot-pounds (ftlb) inch (in) inch-pounds (in-lb) S.I. dne centimeters (cm) erg newton (N) meters (m) Newton-meter (N-m) Joule (J) NOTE: joule = ( newton)( meter) = 7 ergs Conversions pound = 4.448 newtons slug = 4.59 kilograms newton =.48 pounds kilogram =.6854 slugs dne =.48 pounds gram =.6854 slugs Eample 6.4.. A 5 lb man is climbing a ft pole. Calculate the work done b the man. 6.4. Work Done b a Variable Force Definition 6.4. If an object is moved along a straight line b a continuousl varing force F(), then the work W done b the force as the object is moved from = a to = b is given b: W = lim n n f( i) = i= b a f()d
Eample 6.4.. 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.634 N/m? Eample 6.4.3. A 5 lb bucket is lifted from the ground into the air b pulling in ft of rope, weighing.6 lbs, over a constant speed. The bucket contains gallons of H (6lbs). How much work was spent lifting the H, the bucket and the rope? Eample 6.4.4. A 75 ft cable weighing 6 lb/ft is connected to an elevator weighing 5 lb. Find the work done in lifting the elevator to a height of 5 ft.
6.4.3 Hooke s Law (Robert Hooke 635-73): Definition 6.5. The force F required to compress or to stretch a spring (within its elastic limits) is proportional to the distance d that the spring is compressed or stretched from its original length (natural length). Mathematicall speaking: F = k where the constant of proportionalit k (also known as the force constant or spring constant) depends on the specific nature of the spring and is the number of length units the spring is compressed/stretched from its natural length. Eample 6.4.5. A spring of natural length is stretched to 5 under a weight of lb. a) Find the work done to stretch it from to 3. b) Find the work done to stretch it from 5 to 8. Eample 6.4.6. It takes in-lb of work to stretch a spring from its natural length of to. a) Determine the force constant. b) Determine the work done to compress the spring from to 8.
6.4.4 Pumping Problems A. The strateg: When finding work done to pump liquids, we need to think of thin slabs of water being removed and then summing those slabs up. Eventuall, taking the limit of Riemann Sums to come up with an integral. B. The Weight of Water: Because of variations in the earth s gravitational field, the weight of a cubic foot of water at sea level can var from about 6.6 lbs at the equator to as much as 6.59 lbs near the poles, a variation of about.5%. We will use a tpical densit estimate of δ = 6.5 lbs ft 3 for our calculations. Using the S.I. units of measurement the densit of a cubic meter of water is kg m 3. Therefore, the estimated weight of the water is δ = 9.8 m s kg m 3 = 98 N m 3. C. Placement of Tank: Whether we place the tank above or below ground does not affect the amount of work done to pump the liquid out of the tank. We will sometimes place a tank underground in order to make the algebra easier. D. Steps in Finding Work Done During Pumping:. Draw a figure with a coordinate sstem.. Find the volume V of a thin horizontal slab/slice of liquid. 3. Find the weight/force F of a thin horizontal slab/slice of liquid. F = δ( V) 4. Find the distance d that the liquid has to travel. 5. Find the work W it takes to lift the slab to its destination. W = Fd = δ( V)d 6. Find the limits of integration. (defined b the region) 7. Integrate the work epression from the base to the surface of the liquid. Eample 6.4.7. Constant Shaped Tank. An aquarium with dimensions 3 ft is full of H O (weighing 6.5 lbs/ft 3 ). Find the work in lifting and pushing the H O over the top. 3
. What if the tank is full of H O and all the H O is pumped out 8 feet above the top. (Although the water is moved up a spigot above the top of the tank, for simplicit we disregard the adjustment needed for this change in the receptacle and assume that the entire slab is still being moved up the additional distance.) Eample 6.4.8. Variable Shaped Tank. Find the work in pumping kerosene, weighing 5. lbs/ft 3, over the rim of a trough which is ft long and has a semicircular end of diameter 6 ft if the tank is filled to /3 of its height. 4
. A tank in the form of a right circular cone, verte down, is filled with H O to a depth of / its height. If the height of the tank is ft and its diameter is 8 ft, find the work done pumping all the H O 5 ft above the top of the tank. 3. To design the interior of a huge stainless steel tank, ou revolve the curve =, 4, about the -ais. The container, with dimensions in meters, is to be filled with seawater which weighs, N/m 3. How much work will it take to empt the tank b pumping the seawater to the tanks top? 5
6.5 Average Value of a Function 6.5. Average Value of a Function Definition 6.6. If f() is integrable on [a,b], its average(mean) value on [a,b] is fave = b a b a f()d Eample 6.5.. Find the average value of f() = on [,4]. 6.5. Mean Value Theorem for Integral Calculus Definition 6.7. The Mean Value Theorem for Definite Integrals (MVT): If f() is continuous on [a,b], then at some point c in [a,b], f(c) = b a Eample 6.5.. Appl MVT to the eample above. b a f()d Geometric Interpretation of the MVT For positive functions f(), there is a number c such that the rectangle with base [a,b] and height f(c) has the same area as the region under the graph of f() from a to b. 3 4 3 3 4 6 9 6
Eample 6.5.3. Find the average value of = 9+ on [,4]. Appl the MVT to this problem. Eample 6.5.4. In a certain cit the temperature (in F) t hours after 9 a.m. was approimated b the function T(t) = 5+4sin ( πt ). Find the average temperature during the period from 9 a.m. to 9 p.m. 7