HOMEWORK SOLUTIONS MATH 9 Sections 6.4, 6.5, 7. Fall 6 Problem 6.4. Sketch the region enclosed by x = 4 y +, x = 4y, and y =. Use the Shell Method to calculate the volume of rotation about the x-axis SOLUTION. The first equation is a line with positive slope and x-intercept x =, the second is a line with negative slope and x-intercept x =. These lines intersect at 4 y + = 4 y or y = 4. Then, our picture looks as follows: Parallel to the x-axis we have shells with thickness dy, radius y, and height y 4 ( + y 4 ) = y. Thus, the volume of the solid is 4 π(radius)(height of shell)dy = π = π (y y 6 ) 4 4 y( y )dy = π 4 = π(6 64 64 ) = π( 6 ) = π y y dy 6.4. Problem 6.4.9 Use both the Shell and Disk Methods to calculate the volume obtained by rotating the region under the graph of f(x) = 8 x from x about the x-axis the y-axis SOLUTION. This region to be rotated is as follows:
About the x-axis, the disk method gives disks of thickness dx and radius y = 8 x. Thus the volume is π(8 x ) dx = π 64 6x + x 6 dx = π (64x 4x 4 + x7 7 ) = 576π 7 Using the shell method, we get shells with thickness dy, radius y, and height x = (8 y). Thus, the volume is 8 Substituting u = 8 y and du = dy we get πy(8 y) dy π (8 u)u du = π 8u 4 4 u 7 u du = π (6u 8 8 7 ) 8 = 576π 7 About the y-axis, disks have thickness dy and radius x = (8 y) so we get volume 8 π((8 y) ) dy = π 8 (8 y) dy = π ( 5 (8 y) 5 ) 8 = 96π 5 Using the shell method, we get shells with thickness dx, radius x and height y = 8 x. Thus the volume is πx(8 x )dx = π 8x x 4 dx = π (4x x5 5 ) = 96π 5 6.4.9 Problem 6.4.57 Use the Shell Method to find the volume of the torus obtained by rotating the circle (x a) + y = b about the y-axis (assume a > b) SOLUTION. Our picture is again:
Then we have cylinders with radius x going from x = a b to x = a+b. Each of these cylinders has height given by twice y = b (x a) so we have volume: a+b a b πx( b (x a) )dx = 4π Substituting u = x a and du = dx we get 4π b (u + a) b u du = 4π b a+b a b x b (x a) dx u b u du + 4πa b u du b Note that u b u is an odd function (u b u = (( u) b ( u) ) so it is integrated about the symmetric interval [-b,b] (or use substitution u = bsin(θ), du = bcos(θ)dθ to show this fact). Then the volume is just 4πa b b u du. Since the integral here is again half of a circle of radius b, we have that the volume is 4πa( πb ) = π ab. 6.4.57 Problem 6.4.6 The surface area of a sphere of radius r is 4πr. Use this to derive the formula for the volume V of a sphere of radius R in a new way Show that the volume of a thin spherical shell of inner radius r and thickness r is approximately 4πr r Approximate V by decomposing the sphere of radius R into N thin spherical shells of thickness r = R N Show that the approximation is a Riemann sum that converges to an integral. Evaluate the integral
SOLUTION. The volume of such a spherical shell is approximately the surface area at the inner radius times the thickness, 4πr r. This is not exact because as we move r away from the inner radius r, the surface area of these spheres actually involves a slightly larger radius than r. Now we estimate V by decomposing the sphere into N spherical shells. The boundary points of these shells are given by x k = R Nk, k =,.., N (e.g. the first shell goes from x = to x = R N and the last shell goes from x N = N N R to x N = R). Then, each shell has volume approximately equal to surface area at the outer radius times the thickness (e.g. the first shell has volume approximately 4π(x ) r = 4π( R N ) R N ). This will give us an overestimation, for the same reason that part a described an underestimation. Thus, we can describe the approximate volume of the whole sphere as: N 4π(x k ) r = k= N 4π( R R N k) N = 4π( R N N ) k= k= k N(N + )(N + ) = 4πR 6N = πr (N + N + N) N By definition, this is a Riemann sum partitioned by = x < x < x N = R and sampled at x,..., x N. Then, as we take the limit as N goes to infinity r becomes dr, x k becomes r, and we now take r everywhere in [, R]. Thus we obtain R 4πr dr = 4π ( r ) R = 4π R 6.4.6 Problem 6.5.9 Calculate the work (in joules) required to pump all of the water out of the full tank below. Distances are in meters, the density of water is kg m SOLUTION. The work against gravity to lift each layer of water of thickness dy is F(distance lifted) = mg(distance) = area dy density g (distance). If we define the origin at the center of the hemisphere and let the positive y-axis point downward then a layer at depth y has radius y by the Pythagorean theorem and thus area π( y ) m. This layer will 4
have to be lifted y + meters (because of the spout). π( y ) 9.8 (y + ) dy. Hence the work to lift this layer is So the total work to lift all layers is: y= 98π( y )(y+)dy = 98π = 98π( 5 y+ y y = 98π (5y + y y4 4 y ) ) = 7 π joules 6.5.9 Problem 6.5. Conical tank in Figure ; water exits through the spout. SOLUTION. Place the origin at the vertex of the inverted cone, and let the positive y-axis point upward. Consider a layer of water at a height of y meters. From similar triangles, the area of the layer is ( y ) π m, so the volume is ( y ) π δym. Thus the weight of one layer is ( y ) 98π δyn. The layer must be lifted y meters, so the total work needed to empty the tank is ( y ) 98π ( y)dy = π(.675 6 )J.55 7 J. 6.5. Problem 6.5.7 Calculate the work required to life a -m chain over the side of a building. Assume the chain has density 8kg/m. 5
SOLUTION. We again consider the work against gravity to lift a length dy segment of chain (y meters down the rope). This is given by mg(distance). Here, the mass of the dy meter portion of chain is 8 dy m(kg) m. This segment will be lifted y meters. Thus work to lift this segment is 8 dy 9.8 y. Then to lift all segments of the rope, starting from the top ( meters down the rope) and ending at the bottom ( meters down the rope), we have: 9.8 8 ydy = 78.4 ( y y= ) = 9 joules 6.5.7 Problem 6.5.5 The gravitational force between two objects of mass m and M, separated by a distance r, has magnitude GMm r where G = 6.67 m kg s. Show that if two objects of mass M and m are separated by a distance r, then the work required to increase the separation to a distance r is equal to W = GMm(r r ) SOLUTION. Using the equation We obtain W = r r W = a F(x)dx GMm r r dr = GMm r r dr = GMm ( r ) r r = GMm(r r ) 6.5.5 Problem 6.5.6 Use the result of Exercise 5 to calculate the work required to place a -kg satellite in an orbit km above the surface of the earth. Assume that the earth is a sphere of radius R e = 6.7 6 m and mass M e = 5.98 4 kg. Treat the satellite as a point mass. 6
SOLUTION. The satellite moves from the surface of the earth to * m above the earth. This is from distance r = R e m to r = R e + = R e + m. Thus the work is W = GMm(r r ) = GMm( R e R e + ) = (6.67 )(5.98 4 )()( 6.7 6 6.7 6 + ) joules 6.5.6 Problem 7..8 Find the equation of the tangent line at the point indicated y = e x, x = SOLUTION. Let f(x) = e x. Then f (x) = xe x and f () = e. At x =, f() = e, so the equation of the tangent line is y = e(x ) + e = ex e. 7..8 Problem 7..44 Calculate the derivative indicated d y dt ; y = e t sin t SOLUTION. Let Y = e t sin t. Then and dy dt = e t ( cos t) e t sin t = e t ( cos t sin t), d y dt = e t ( 9 sin t 6 cos t) e t ( cos t sin t) = e t ( 5 sin t cos t) Problem 7..5 f(x) = x e x 7..44 SOLUTION. Setting f (x) = (x + x)e x equal to zero and solving for x gives (x + x)e x = which is true if and only if x = or x =. Now, f (x) = (x + 4x + )e x. Because f () = >, x = corresponds to a local minimum. On the other hand, f ( ) = (4 8 + )e = /e <, so x = corresponds to a local maximum. Problem 7..8 e x e 4x e dx x 7..5 SOLUTION. ( e x e 4x e x ) dx = (e x e x )dx = e x ex + C. 7..8 7
Problem 7..9 Wind engineers have found that wind speed v (in m/s) at a given location follows a Rayleigh distribution of the type W(v) = ve v /64 This means that the probability that v lies between a and b is equal to the shaded area in the figure below. Show that the probability that v [, b] is e b /64.. Calculate the probability that v [, 5]. SOLUTION.. The probability that v [.b] is Let u = v /64. Then du = v dv and ve v /64 dv. b /64 ve v /64 dv = e u du = e b /64.. The probability that v [, 5] is the probability that v [, 5] minus the probability that v [, ] Using part (a), it follows that the probability that v [, 5] is Problem 7..94 ( e 5/64 ) ( e 4/64 ) = e /6 e 5/64.6. Recall the following property of integrals: If f(t) g(t) for all t, then for all x, f(t)dt The inequality e t holds for t because e >. Use () to prove that e x + x for x Then prove, by successive integration, the following inequalities (for x ): 7..9 g(t)dt () e x + x + x e x + x + x + 6 x SOLUTION. Integrating both sides of the inequality e t yields e t dt = e x x or e x + x. 8
Integrating both sides of this new inequality then gives Finally, integrating both sides again gives e t dt = e x x + x / or e x + x + x /. e t dt = e x x + x / + x /6 or e x + x + x / + x /6 as requested. 7..94 9