28.1 Computing Volumes 861 ◆ EXAMPLE 28.5 Let’s model a bagel by revolving a disk of radius 1 centered at the origin about the vertical line x = 2. What is the volume of the bagel? x 2 + y 2 = 1 x = 2 x = 2 y x 1 1 1 –1 y y = √1 – x 2 T x Figure 28.13 SOLUTION We can slice the disk either along the x-axis or along the y-axis. The former will result in cylindrical shells and the latter in annuli or, more appropriately, bagel chips. We’ll opt for slicing along the x-axis. We can revolve the half-disk in the first two quadrants, obtaining the volume of the top half of the bagel, and doubling it to get the final answer. (Notice that if we revolved the half-disk in quadrants I and IV around the line x = 2 we would get less than half the total volume. Why?) Partition [−1, 1] into n equal pieces using a standard partition. volume of the ith cylindrical shell ≈ 2πr i h i x h i = 1 − x 2 i r i = (the distance between x = 2 and x i ) = 2 − x i Note that this holds regardless of the sign of x i . volume of ith cylindrical shell ≈ 2π(2 − x i ) 1 − x 2 i x volume of half of the bagel = 1 −1 2π(2 − x) 1 − x 2 dx = 2π 1 −1 2 1 − x 2 − x 1 − x 2 dx = 4π 1 −1 1 − x 2 dx − 2π 1 −1 x 1 − x 2 dx The first integral we can recognize as giving the area of a semicircle of radius 1. 862 CHAPTER 28 More Applications of Integration 4π 1 −1 1 − x 2 dx = 4π 1 2 π(1) 2 = 2π 2 The second integral has an integrand that is an odd function. (−x) 1(−x) 2 =−x 1 − x 2 Therefore, 1 −1 x 1 − x 2 dx = 0 Volume of the whole bagel = 2 · 2π 2 = 4π 2 . Observation. Suppose this was not a real bagel but only a model made of clay. Suppose further that we chopped it open with a cleaver as shown in Figure 28.14 and opened it up into a solid cylinder of radius 1. The length of the cylinder should be neither the outer circumference of the bagel, nor the inner circumference of the bagel, but the circumference corresponding to the dotted line shown. The volume of this cylinder is the area of the circle, π, times the length, 4π, giving 4π 2 . This is precisely the volume of the bagel. In other words, as you unroll the clay there will be cracking on one side and buckling on the other, and they exactly cancel out. 1 1 2 2π(2) Figure 28.14 ◆ EXERCISE 28.1 Show that this result holds more generally. Rotating the disk x 2 + y 2 ≤ a 2 around the line x = b, where b>a>0 generates a torus (the mathematical term for such a bagel-shape) whose volume is πa 2 (2πb)= 2π 2 a 2 b. PROBLEMS FOR SECTION 28.1 1. A tent has a base that is an isosceles triangle. The mouth of the tent measures 8 feet and the length is 12 feet. The tent is constructed so that the cross sections perpendicular to the base are all equilateral triangles. Find the volume of the tent. 28.1 Computing Volumes 863 2. A Wisconsin cheese factory makes its cheese in solid cylinders of radius 2 inches. A wedge of cheese is cut from the cylinder by chopping through the diameter of the base at an angle of 45 degrees with the base. Find the volume of the wedge of cheese. 2 45° (Hint: The base of the wedge is a semicircle of radius 2. Attach a coordinate system to it. The cross-sections are right isosceles triangles.) 3. Find the volume generated when the region in the first quadrant bounded by y = x 2 and y = 3x is rotated about (a) the x-axis, (b) the y-axis, (c) the line x =−1. 4. Find the volume generated when the region bounded by the y-axis, y = x 2 , and y = 4 is rotated about the x-axis. Do this in three ways. y = 4 y x y = x 2 (a) Chop the shaded region into vertical strips and rotate. (b) Chop the shaded region into horizontal strips and rotate. (c) Subtract volumes. Subtract the volume generated by rotating the region under y = x 2 from that generated by rotating the region under y = 4. (The latter is just a cylinder.) 5. Find the volume generated by rotating the region bounded by the x-axis and y = √ sin x for 0 ≤x ≤ π about the line y = 0. 6. The region bounded above by y = 1 x , below by the x-axis, and laterally by x = 1 2 and x = 5 is rotated about the x-axis. Find the volume of the funnel generated. 864 CHAPTER 28 More Applications of Integration 7. Let A be the region bounded by y =x 2 and y =4 − x 2 . Find the volume generated by rotating region A about (a) the y-axis, (b) the x-axis. 8. Suppose that a cantaloupe is a perfect sphere with a radius of 5 inches. You slice off a piece as indicated. At its thickest your piece is 2 inches. What’s the volume of your piece of cantaloupe? 5" 2" 9. Find the volume generated by revolving the region bounded by y = x 2 and y =4 about (a) the y-axis, (b) the vertical line x = 2, (c) the horizontal line y = 4, (d) the horizontal line y = 5. 10. A parfait cup is formed by revolving the curve y =x 3 ,0≤x ≤2, about the y-axis. The parfait cup is filled to the brim with hot chocolate. If you plan to drink exactly half the hot chocolate in the cup, at what height should the liquid be when you stop drinking? 11. The top 31 feet of the Great Pyramid of Cheops are now missing. What percentage of the structure remains? Refer to Example 28.1 for details. 12. We model a large, 9-inch-tall soup bowl by rotating a region A around the y-axis. A is the region bounded by the y-axis, y = 1 4 x 2 , and y = 9. (a) What is the capacity of the bowl? (b) If the bowl was filled to the brim when set out on the table and an hour later was filled only to a height of 4 inches, how much soup was ladled out in the hour? 13. Find the volume of the ellipsoid generated by revolving the ellipse x 2 a 2 + y 2 b 2 = 1 about the x-axis. y x a –a –b b 28.2 Arc Length, Work, and Fluid Pressure: Additional Applications of the Definite Integral 865 28.2 ARC LENGTH, WORK, AND FLUID PRESSURE: ADDITIONAL APPLICATIONS OF THE DEFINITE INTEGRAL The strategy of getting a grip on something by partitioning, approximating, summing and computing a limit is applicable in a variety of situations. If we arrive at the limit of a Riemann sum, then we can solve the problem using a definite integral. In this section we apply this strategy to the problems of computing the length of a plane curve, computing the work done by a variable force acting on an object, and computing the pressure exerted by a fluid. Arc Length Not only is the definite integral useful in computing areas and volumes, but it is a tool for computing arc length as well. Suppose f and f are continuous on [a, b]. Our goal is to find the length of the curve y = f(x)for a ≤x ≤b. f x y = f(x) x 1 x 2 a x 3 x n x 0 = b = Figure 28.15 We partition [a, b] into n equal pieces, each of length x = b−a n . Let x i = a + ix, for i = 0, , n and P i = (x i , y i ), where y i = f(x i ), i = 0, , n. x 1 x 2 x 3 ∆x x n–1 x n x 0 P 0 P 1 P 2 P 3 P n–1 P n . . . . . . Figure 28.16 We approximate the length of the curve on the ith interval by s i , the length of the line segment joining P i−1 and P i . 866 CHAPTER 28 More Applications of Integration ∆x i ∆y i = f(x i ) – f(x i –1 ) P i–1 P i s i Figure 28.17 s i = (x i ) 2 + (y i ) 2 , where x i = x and y i = y i − y i−1 . In order to arrive at a Riemann sum, we need an expression for s i of the form g(x i )x. s i = ( x i ) 2 + ( y i ) 2 x i x i = (x i ) 2 + (y i ) 2 (x i ) 2 x i = x x 2 + y i x i 2 x i = 1 + y i x i 2 x i Because we’ve made a uniform partition, all the x i ’s are equal; the y i ’s however are not. y i = f(x i ) − f(x i−1 ). As n →∞, x →0 and y i x i → f (x i ). For x very small, 3 s i ≈ 1 + [f (x i )] 2 x. arc length ≈ n i=1 s i ≈ n i=1 1 + f (x i ) 2 x Taking the limit as n →∞, we obtain arc length = b a 1 + f (x) 2 dx. Arc length: We conclude that if f is differentiable on [a, b] and f is continuous on [a, b], then the length of the graph of f from x = a to x = b is given by b a 1 + f (x) 2 dx. 3 Using the Mean Value Theorem it can be shown that s i = 1 +[f (x ∗ i )] 2 x for some x ∗ i in the ith interval even if x is not very tiny. 28.2 Arc Length, Work, and Fluid Pressure: Additional Applications of the Definite Integral 867 ◆ EXAMPLE 28.6 Find the length of the curve given by f(x)= 1 3 x 3/2 on the interval [0, 2]. SOLUTION f (x) = 1 2 x 1/2 Arc length = 2 0 1 + x 1/2 2 2 dx = 2 0 1 + x 4 dx = 1 2 2 0 √ 4 + xdx Let u = 4 + x when x = 0, u = 4 du = dx when x = 2, u = 6. = 1 2 6 4 u 1/2 du = 1 2 2u 3/2 3 6 4 = 1 3 6 3/2 − 4 3/2 = 1 3 6 3/2 − 8 ◆ Example 28.6 was carefully chosen to give a straightforward integral. Usually the integrals obtained are not easy to evaluate exactly. Numerical methods together with a calculator or computer allow us to approximate them. EXERCISE 28.2 Verify that the formula derived tells us that the length of y = x from x = 0tox = 1is √ 2. Work How much work is needed to pump all the water out of a swimming pool? How much work is needed to compress a heavy spring? How much work is needed to launch a rocket? In order to make such calculations we need to clarify the meaning of work from the framework of a physicist. If a constant force F acting in the direction of motion of an object causes a displacement d, then the work done by the force on the object is defined to be work = force · displacement W = F · d. Notice that the physicists’ definition of work differs from our colloquial use of the word. For example, if you wait at a bus stop for 15 minutes while holding your 25-pound nephew, you might feel tired but according to the physicist you’ve done no work. Your nephew hasn’t moved. And if you give up on the bus and walk to the grocery store, carrying your nephew the whole way, your arms have still not done any work. Your arms have exerted upward force on your nephew; his motion, however, has been horizontal. Even worse, if you get tired and set him down after a few minutes, you’ve done negative work. If you lift him from his position around your waist to a perch on your shoulders, you and the physicist both 868 CHAPTER 28 More Applications of Integration agree that you have done some work. If your shoulder is 1.5 feet above your waist, you’ve done 1.5 feet × 25 pounds = 37.5 foot-pounds of work. Units of Measure. Because work equals force times distance, in the English system it can be measured in foot-pounds, as above. Scientists generally measure force in newtons, 4 abbreviated N. A newton-meter is called a joule. To obtain newtons from kilograms (a unit of mass), apply Newton’s second law: force = mass · acceleration = m · g, where g is the acceleration due to gravity. Near the surface of the earth g is approximately 9.8 m/sec 2 . If a constant force is acting in the direction of motion, calculating work simply amounts to a multiplication problem. Consider a variable force acting in a straight line over a fixed distance. For example, consider compressing a spring. The applied force must increase as the spring becomes more compressed. Situate the x-axis so that the force is applied along it from x = a to x =b. Let F(x)be a continuous function giving the force applied at position x. To calculate the work done we partition [a, b] into n equal pieces, each of length x = b−a n and let x i = a +ix for i =0, 1, , n. On each subinterval we treat the force as approximately constant. This is reasonable because F is continuous. work done in ith subinterval ≈ F(x i )x a x b x 0 x 1 x n–1 x n x 2 . . . spring Figure 28.18 Total work done ≈ n i=1 F(x i )x. The approximation becomes better and better as n increases. lim n→∞ n i=1 F(x i )x = b a F(x)dx We conclude that if a continuous force F acts along the x-axis from x =a to x =b the work done by the force is given by b a F(x)dx. ◆ EXAMPLE 28.7 A wheelbarrow is being pushed along a straight forest trail with a force of F(x) = 10 + 9x − x 2 pounds, where x is the distance from the trailhead. How much work is done in moving the wheelbarrow the first 9 feet? 4 One newton corresponds to approximately 0.225 pound. 28.2 Arc Length, Work, and Fluid Pressure: Additional Applications of the Definite Integral 869 SOLUTION Work = 9 0 (10 +9x − x 2 )dx = 10x + 9x 2 2 − x 3 3 9 0 = 90 + 364.5 − 243 = 211.5 ft-lb. ◆ Let’s return to the spring problem. Hooke’s law, an experimentally derived law, says that the force necessary to compress or stretch a spring x units from its natural length is proportional to x. F = kx, where k is called the spring constant. Hooke’s law is followed provided the elastic limit of the spring is not exceeded. Example 28.8 applies this law. Natural length Compressed spring x Figure 28.19 ◆ EXAMPLE 28.8 An industrial-strength spring is being used to cushion the impact of packages as they come hurling off a conveyor belt. The spring has a spring constant of 1000 N/m. Find the work done by a package that compresses the spring 0.4 m. SOLUTION Work = 0.4 0 1000xdx = 1000x 2 2 0.4 0 = 500 · 0.16 = 80 N-m, or 80 joules. ◆ ◆ EXAMPLE 28.9 A pulley system is being used to lift granite out of a quarry. If the cable used weighs 2 pounds per foot length, how much work is done lifting a 300-pound block of granite 150 feet to the level of the pulley? Assume friction is negligible. x 150 ft Granite Not drawn to scale Figure 28.20 870 CHAPTER 28 More Applications of Integration SOLUTION Let x =the length of cable between the granite and the pulley. x will vary between 150 feet and 0 feet. The force that must be applied to lift the granite varies. force = (weight of granite) + (weight of cable) = 300 + 2x Work = 150 0 (300 +2x) dx = 300x + 2x 2 2 150 0 = 300(150) + (150) 2 = 45,000 + 22,500 = 67,500 67,500 foot-pounds of work are required. ◆ ◆ EXAMPLE 28.10 A cylindrical above-ground swimming pool has side walls 1.5 m tall, a radius of 4 m, and is filled to a height of 1.4 m. How much work is required to pump water over the top and (a) empty the pool? (b) reduce the water level to 0.7 m? SOLUTION Our basic strategy is to slice along the vertical axis, chopping the water into thin solid disks. Think of pumping the water out one disk at a time. We’ll calculate the work done to lift the ith disk to the top of the pool, sum over all the relevant slices, and, letting the number of slices increase without bound, get the limit of a Riemann sum. 4 m 1.5 m x i ∆x { Figure 28.21 Let x = the distance (in meters) between the top of the pool and the water to be lifted. (a) To empty the pool, x must vary between 0.1 m and 1.5 m. Partition [0.1, 1.5] into n equal pieces, where x = 1.4 n and x i = .1 +ix for i = 0, , n. The force required to lift a slice is the force acting against gravity; it is equal to the weight of the slice. { 0.1 m 1.5 m 1.5 m 1.4 m x i Figure 28.22 . 28.21 Let x = the distance (in meters) between the top of the pool and the water to be lifted. (a) To empty the pool, x must vary between 0.1 m and 1.5 m. Partition [0.1, 1.5] into n equal pieces,. calculate the work done to lift the ith disk to the top of the pool, sum over all the relevant slices, and, letting the number of slices increase without bound, get the limit of a Riemann sum. 4 m 1.5. top half of the bagel, and doubling it to get the final answer. (Notice that if we revolved the half-disk in quadrants I and IV around the line x = 2 we would get less than half the total volume.