28.2 Arc Length, Work, and Fluid Pressure: Additional Applications of the Definite Integral 871 amount of work to lift up the ith slice ≈ (weight of ith slice) · (distance to the top) The volume of the ith slice =π(4) 2 x = 16πx m 3 . The density of water is 1000 kg/m 3 . Therefore the mass of the ith slice = 1000 · 16πx kg. We calculate the weight of the slice by multiplying by g ≈9.8 m/sec 2 . The weight of the ith slice ≈ 1000 ·16πx · 9.8 newtons ≈ 156,800πx newtons amt. of work to lift up the ith slice ≈ 156,800πx · x i total work = lim n→∞ n i=1 156,800π · x i x = 1.5 0.1 156,800πx dx total work = 156,800π x 2 2 1.5 0.1 = 156,800π 1.125 − 0.005 = 175,616π The work done is 175,616π joules, or ≈ 551,713.9 joules. (b) To reduce the water level to 0.7 m, to pump half the water out, requires partitioning the interval [0.1, 0.8] into n equal pieces because the water slices must move between 0.1 and 0.8 m. 0.1 0.8 Figure 28.23 0.8 0.1 156,800πx dx = 156,800π x 2 2 0.8 0.1 = 156,800π[0.32 −0.005] = 49,392π joules, or ≈ 155,169.5 joules Notice that it takes substantially less work to reduce the water depth from 1.4 m to 0.7 m than it does to reduce it from 0.7 m to 0 m. ◆ Fluid Pressure and Force Water is an invaluable resource of the earth, and civilizations have devoted much scientific energy toward understanding it. We build bridges to travel over it, tunnels to travel under it, dams to contain it, plumbing, draining, and irrigation systems to direct it, and submarines to explore its depths. Understanding fluid pressure (force per unit area) is critical to the success of these projects. 872 CHAPTER 28 More Applications of Integration The fundamental principle is that fluid pressure is directly proportional to depth and, when a system is at equilibrium, it is the same in all directions. Pressure = fluid density · (depth). This means that given a submerged object in an equilbrium state, the fluid pressure down, the fluid pressure up, and the horizontal fluid pressures on the object are all the same. The hydrostatic force exerted by a fluid on a plane surface of area A situated horizontally at a depth d is given by force = (pressure) · (area) = ρd · A, where ρ is the density of the liquid, its weight per unit volume. d x x x x x d d d d The hydrostatic force on the base of every bottle of water is the same Figure 28.24 This means that if a bottle whose base has area A is filled with fluid to a height d, the fluid force on the base is unaffected by the shape of the bottle. (See Figure 28.24.) NOTE In the metric system, water density is given as 1000 kg/m 3 . To get weight per unit volume, multiply by g. water density = (9.8)(1000) = 9800 newtons/m 3 . In the English system, water density = 62.4 lbs/ft 3 . When building a dam or an underwater support for a bridge, the force of the water is more complicated to calculate because the structure is submerged vertically or at an angle as opposed to horizontally; the pressure increases with the depth. ◆ EXAMPLE 28.11 Suppose a particular dam has the shape of a parabola 40 feet wide at the top and 50 feet deep. Calculate the hydrostatic force on the dam when the water level is 5 feet from the top of the dam. 28.2 Arc Length, Work, and Fluid Pressure: Additional Applications of the Definite Integral 873 SOLUTION We establish a coordinate system as shown in Figure 28.25 and take horizontal slices so that the depth does not vary much within a slice. Partition [0, 45] into n equal pieces, each of height y = 45 n . Let y i = iy, for i = 0, 1 , n. the force on the ith strip ≈ (water density in lb/ft 3 ) · depth of ith strip · area of ith strip ≈ 62.4 lbs/ft 3 · ( 45 − y i ) ft · area of the ith strip 50 20 45 0 5 Water level Depth = 45 – y i y i y 0 y 1 y n y x (20, 50) Top of dam Face of dam . . . . Figure 28.25 We can approximate the area of the horizontal slices by treating them as rectangles of height y and width 2x i . To express the width in terms of y we find the equation of the parabola. (This is where the dimensions of the dam come into play.) y − kx 2 because we’ve put the vertex of the parabola at the origin. 50 =k(20) 2 ⇒ k = 50 400 = 1 8 y = 1 8 x 2 and x =± 8y Therefore, the area of the ith strip ≈ 2x i y = 2 √ 8y i y = 4 √ 2 √ y i y ft 2 the force on the ith strip ≈ (62.4 lbs/ft 3 ) · (45 − y i )ft · 4 √ 2 √ y i y ft 2 874 CHAPTER 28 More Applications of Integration total force = lim n→∞ n i=1 62.4(45 − y i )4 √ 2 √ y i y = 45 0 62.4 (45 − y ) 4 √ 2 √ ydy fluid density · depth · area = 249.6 √ 2 45 0 (45y 1/2 − y 3/2 )dy = 249.6 √ 2 45 · 2 3 y 3/2 − 2 5 y 5/2 45 0 = 249.6 √ 2 30 ·45 3/2 − 2 5 · 45 5/2 ≈ 1,278,673 lb ◆ There are many other applications of integration. Integrals can arise in physics when computing moments, centroids, or centers of mass. They can arise in economics when computing consumer surplus and the present value of an income stream. They arise in biology when computing blood flow through a blood vessel. Integration is used to calculate the surface area of a surface of revolution. It is used extensively in statistics and probability. This is not an exhaustive list! Pick an application of integration that interests you and learn about it. You have the basic tools you need in order to understand the exposition. You might try to read about it from a few sources—for instance, a calculus textbook and a book in the relevant discipline. PROBLEMS FOR SECTION 28.2 1. Using the arc length formula, find the length of the line f(x)= mx + b from x = 0to x = 3. Confirm that this answer agrees with the distance formula. In Problems 2 through 4, approximate each length with error less than 0.05. 2. The length of one arc of the cosine curve, say from x = −π 2 to x = π 2 3. The length of f(x)=x 2 from x =−1tox = 1 4. The length of y = ln x from x = 1tox = e 5. Find the work done in pushing a stroller 200 feet along a path by applying a constant force of 12 pounds in the direction of motion. 6. A force of 5 pounds will stretch a certain spring 3 inches beyond its natural length. (a) What is the value of the spring constant? (b) How much work is done in stretching the spring from its natural length to 3 inches beyond its natural length? 28.2 Arc Length, Work, and Fluid Pressure: Additional Applications of the Definite Integral 875 (c) How much work is done in stretching the spring from 3 inches beyond its natural length to 6 inches beyond its natural length? (d) Why is the answer to part (c) larger than the answer to part (b)? 7. Suppose 1.2 foot-pounds of work are required to compress a spring 2 inches. (a) How much work is required to stretch this spring 2 inches from its equilibrium position? (b) How much work is required to stretch this spring 4 inches from equilibrium? (c) How much work is required to stretch the spring 5 inches from its equilibrium position? 8. A window washer weighing 160 pounds is attached to a rope hanging from the roof of the building whose windows he is washing. The rope weighs 0.6 lb/ft. Right now he is working 50 feet down from the rooftop. (a) How much work is required to bring him to the windows that are 25 feet from the rooftop? (b) How much work will it take to bring him from where he is to the roof? 9. Amelia and Beulah are city dwellers who have set up pulley systems to get their groceries delivered without walking the stairs. Amelia pulls her basket filled with 12 pounds of groceries up to her 40-foot-high balcony. Beulah pulls her basket filled with 16 pounds of cleaning supplies up to her 30-foot-high window. Assuming both women use ropes weighing 0.2 lb/ft, whose task requires more work? How much more work? (Assume friction is negligible.) 10. A rectangular swimming pool is 4 feet deep, 75 feet long, and 25 feet wide. It is completely filled with water. (a) What is the hydrostatic force on the bottom of the pool? (b) What is the hydrostatic force on the shorter wall of the pool? (c) What is the hydrostatic force on the longer wall of the pool? (d) How much work is required to pump all the water out of the pool (i.e., up to the top of the pool)? 11. A hemispherical tank with a radius of 7 feet is filled to a height of 6 feet with gasoline. How much work is required to pump all the gasoline over the top? The weight-density of gasoline is 42 lb/ft 3 . 7 ft 6 ft 7 ft Hemispherical tank 12. As part of the pasteurization and homogenization process, milk is stored in a large tank as shown on the following page. 876 CHAPTER 28 More Applications of Integration 3 ft 5 ft Semicircular cross-section The next step in the processing requires the milk to be pumped out of this holding tank. If the tank is filled to the brim, how much work is required to empty it? The weight-density of milk is 64.5 lb/ft 3 . 13. Refer to Problem 12. What is the force exerted by the milk on the semicircular side wall of the holding tank? 14. A dam in a canyon has the shape of an isosceles triangle. Its depth is 30 feet and its width is 15 feet. The water level is 2 feet below the top of the dam. What is the total hydrostatic force on the face of the dam? 15 30 ft Face of the dam 15. A dam has the shape of an isosceles trapezoid. It is 40 feet deep. Its base is 30 feet long while on the surface it measures 60 feet across. (Its thickness is irrelevant to the problem.) If the water reaches the top of the dam, what is the total hydrostatic force on the face of the dam? 16. Project Pick an application of integration not explicitly discussed in this text and learn about it. Then either write an exposition of the application or prepare a short lesson on it. Include examples. 29 CHAPTER Computing Integrals In this chapter we introduce techniques of integration that expand our computational ability. Section 29.1 presents integration by parts, the integration analogue of the Product Rule. Sec- tions 29.2 and 29.3 introduce methods of transforming a difficult integrand into something more manageable. In the final section of this chapter we expand our notion of integration to encompass unbounded integrands and/or an unbounded interval of integration. 29.1 INTEGRATION BY PARTS—THE PRODUCT RULE IN REVERSE Integration by parts is a method of integration derived from the Product Rule for differ- entiation. This technique will help us evaluate integrals such as x sin xdx, xe x dx, x ln xdx, ln xdx, and tan −1 xdx. It is a useful technique for integrands involving the product of different types of functions as well as those involving a single intractable function, and it is a powerful tool for transforming an integral. The Product Rule tells us that if f and g are differentiable functions, then d dx ( f(x)g(x) ) = f(x)g (x) + g(x)f (x). Let’s denote f(x)by u and g(x) by v. Then, using the differential notation introduced in the substitution section, we can write du = f (x) dx and dv = g (x) dx. The Product Rule becomes 877 878 CHAPTER 29 Computing Integrals d(uv) = udv+ v du. d(uv) = udv+ v du uv = udv+ v du This can be rearranged to give what is referred to as the formula for integration by parts. Integration by parts: udv= uv − v du This formula expresses the original integral in terms of an expression involving a new integral. Our aim is to arrive at a new integral that is easier to compute than the original. Given an integral h(x) dx, the general idea is to choose u and dv such that h(x) dv = udv and we can find v = dv, du is simpler than u, and v du is simpler (or no more difficult) than udv. Indefinite Integrals Using Integration by Parts ◆ EXAMPLE 29.1 Find xe x dx. SOLUTION Let u = x, dv =e x dx Then du = dx, v= e x dx = e x . x u e x dx dv = x u e x v − e x v dx du = xe x − e x + C REMARKS 1. Notice that we wrote v = e x dx =e x , a particular antiderivative of e x , as opposed to e x +C, the family of antiderivatives. Try letting v =e x +C 1 and verify that the answer remains the same. In Exercise 29.1 you’ll verify that choosing any one antiderivative always suffices. 2. Suppose we try the following. Let u = e x , dv = xdx du = e x dx, v = x 2 2 . We obtain xe x dx = e x x 2 2 − 1 2 x 2 e x dx. 29.1 Integration by Parts—The Product Rule in Reverse 879 This new integral is more complex than the original integral, indicating a poor choice of u and dv. (Notice that these choices do not conform to the guidelines set out on the previous page.) 3. Given xe x dx, we would like to get rid of the “extra x.” Integration by parts is a mechanism that essentially allows us to do that. ◆ EXERCISE 29.1 Show that replacing v by v + C in the formula for integration by parts does not alter the result. (You will end up getting uC − uC.) ◆ EXAMPLE 29.2 Find x ln xdx. SOLUTION Let u = ln x, dv = xdx du = 1 x dx, v = x 2 2 . x ln xdx= ln x u · xdx dv = (ln x) · x 2 2 − 1 2 x 2 · 1 x dx = 1 2 x 2 ln x − 1 2 xdx = 1 2 x 2 ln x − 1 4 x 2 + C REMARK Both x and ln x become “simpler” when differentiated, but we choose u =ln x because x is much easier to integrate than ln x. ◆ ◆ EXAMPLE 29.3 Find ln xdx. SOLUTION Let u = ln x, dv = dx du = 1 x dx, v = x. ln xdx= x ln x − x · 1 x dx = x ln x − 1 dx = x ln x − x + C REMARK Notice that in this example the entire integrand is denoted by u. The approach when finding tan −1 dx or sin −1 dx or (ln x) 2 dx is similar. Integration by parts is amazingly effective in transforming these difficult integrals into simple ones. ◆ Repeated Use of Integration by Parts Sometimes we have to use integration by parts multiple times. As long as the new integrals that arise are not more difficult than the original we may be on the right track. 880 CHAPTER 29 Computing Integrals ◆ EXAMPLE 29.4 Find x 2 e x dx. SOLUTION Let u = x 2 , dv = e x dx du = 2xdx, v = e x . x 2 e x dx = x 2 e x − e x · 2xdx = x 2 e x − 2 xe x dx The integral xe x dx can be done by using integration by parts, as shown in Example 29.1. Therefore, x 2 e x dx = x 2 e x − 2(xe x − e x ) + C (by Example 29.1) = x 2 e x − 2xe x + 2e x + C = e x (x 2 − 2x + 2) + C. REMARKS 1. When doing integration by parts multiple times it is easy to make sign errors. Be cautious—use parentheses. 2. Sometimes you may end up with a constant of −2C, 1 3 C, etc. You may replace −2C by C 1 . Sometimes −2C is simply replaced by C because C is an arbitrary constant. 3. To find x 4 e x dx we would do integration by parts four times. ◆ EXERCISE 29.2 Use integration by parts to derive the reduction formula x n e ax dx = 1 a x n e ax − n a x n−1 e ax dx where a and n are non-zero constants. In the following example we use integration by parts twice, but the resulting integral is no simpler than the original. In fact, it is the same. You may at first think that we’ve gotten nowhere, but watch how we can solve algebraically for the integral in question. ◆ EXAMPLE 29.5 Find e 2x sin 3xdx. SOLUTION Let u = e 2x , dv = sin 3xdx du = 2e 2x dx, v =− 1 3 cos 3xdx. e 2x sin 3xdx=− 1 3 e 2x cos 3x − − 1 3 cos 3x · 2e 2x dx =− 1 3 e 2x cos 3x + 2 3 e 2x cos 3xdx . methods of transforming a difficult integrand into something more manageable. In the final section of this chapter we expand our notion of integration to encompass unbounded integrands and/ or an unbounded. scientific energy toward understanding it. We build bridges to travel over it, tunnels to travel under it, dams to contain it, plumbing, draining, and irrigation systems to direct it, and submarines to explore. substantially less work to reduce the water depth from 1.4 m to 0.7 m than it does to reduce it from 0.7 m to 0 m. ◆ Fluid Pressure and Force Water is an invaluable resource of the earth, and