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Unformatted text preview: H OMEWORK S OLUTIONS MATH 1910 Fall 2017 Sections 6.3, 6.4, 6.5 Problem 6.3.11 Find the volume of revolution about the x-axis for f(x) = csc x over the interval [π/4, 3π/4]. S OLUTION. Z 3π/4 π(csc)2 dx Volume = π/4 3π/4 = π [− cot x]x=π/4 = −π(−1 − 1) = 2π 6.3.11 Problem 6.3.16 Sketch the region enclosed by the curves y = x2 and y = 2x + 3, describe the cross section perpendicular to the x-axis located at x, then find the volume of the solid obtained by rotating the region about the xaxis. S E C T I O N 6.3 Volumes of Revolution 685 two curves therefore intersectlooks at x = like −1 and x = 3. The region enclosed by the two curves is shown in S OLUTIONThe . The enclosed region this: the figure below. y 8 y = 2x + 3 6 4 y = x2 2 1 x 2 (b) When the region is rotated about the x-axis, each cross section is a washer with outer radius R = 2x + 3 and inner radius r = x2. The cross section perpendicular to the x-axis is a washer with outer radius 2x + 3 and inner (c) the The volume of theofsolid revolution radius x2 , so volume theofsolid of isrevolution is ! # Z3 ! 3 3 " Z3 2 2 2 2 4  1088π π 2 (2x + (4x 2 3) 2 − (x ) dx = π 2 + 12x + 9 − x )4dx π (2x + 3)−1 − (x ) dx = π 4x dx = . 6.3.16 −1 + 12x + 9 − x 15 %&3 −1 −1 $ & 4 3 1088π 1 x + 6x 2 + 9x − x 5 && = . 3 15 5 −1 =π 17. y = 16 − x, y = 3x + 12, x = −1 solution (a) Setting 16 − x = 3x + 12, we find that the two lines intersect at x = 1. The region enclosed by the two curves is shown in the figure below. y y = 16 − x y = 3x + 12 10 1 −1 −0.5 0.5 1 x (b) When the region is rotated about the x-axis, each cross section is a washer with outer radius R = 16 − x and inner radius r = 3x + 12. (c) The volume of the solid of revolution is ! 1 " # ! 1 55. Find the volume of the cone obtained by rotating the region under the segment joining (0, h) and (r, 0) about the y-axis. solution The segment joining (0, h) and (r, 0) has the equation h y =− x+h r or x = r (h − y). h Rotating the region under this segment about the y-axis produces a cone with volume "h ! " πr2 h πr2 (h − y)2 dx = − 2 (h − y)3 "" 2 h 3h 0 0 = Problem 6.3.56 1 2 π r h. 3 2 2 2 56. The torus (doughnut-shaped solid) in Figure 15 is obtained by rotating the circle (x − a) + y = 2b The torus (doughnut-shaped solid) shown below is obtained by rotating the circle (x − a) + y2 = b2 around the y-axis (assume that a > b). Show that it has volume 2π 2 ab2 . Hint:2After2simplifying it, evaluate around the y-axis (assuming that athe > area b).ofShow that it has volume 2π ab . the integral by interpreting it as a circle. y a a +b x FIGURE 15 Torus obtained by rotating a circle about the y-axis. 2 2 2 Rotating the region by the circle − a) + y = bwashers about the y-axis a torus S OLUTIONsolution . The cross sections ofenclosed this volume of(xrevolution withproduces outer # # radii given by whose cross sections are washers with outer radius R = a + b2 − y 2 and inner radius r = a − b2 − y 2 . the right half of the circle and inner radii given by the left half of the circle. Solving for x, we The volume of the torus is then get: '2 % ! b $% ! b& p & & p '2 ( 2 − y2 π 2 =b a 2+− by2 2− → y2 x − a− b2 − b y 22 − y dy2 =→ 4aπ (x − a) − a = ± x =−ba b±2 − y 2bdy. −b p remaining definite integral the is one-half the areafor of a the circlerighthand of radius b; therefore, the the volume of the(our outer Note that xNow, = the a+ b2 p − y2 gives equation side of circle torus is 2 2 radius), with x = a − b − y describing the lefthand side (our inner radius). Thus the 1 4aπ · π b2 = 2π 2 ab2 . volume of the solid is: 2 Z 57. b  Sketchp the hypocycloid x 2/3 + y 2/3p = 1 and find the  volume of the solid obtained by revolving it V = about π the x-axis. (a + b2 − y2 )2 − (a − b2 − y2 )2 dy −b solution A sketch of the hypocycloid is shown below. =π Zb   p p (a2 + 2a b2 − y2 + (b2 − y2y )) − (a2 − 2a b2 − y2 + (b2 − y2 )) dy 1 b Zb p 4a b2 − y2 dy =π −1 −b Zb p = 4aπ b2 − y2 dy. −b 1 x −1 ) * 2/3 3/2 For the y = ± 1−x . Rotating this region about the x-axis will produce a solid whose p Rbhypocycloid, ) *3/2 2− 2with Observe that b y dy describes the of half of a ofcircle ofofradius b,will sobewe have that cross sections are disks radius R = 1 − x 2/3area . Thus the volume the solid revolution −b 2 2 %2 3 1 +4aπ( πb ),= the volume of the torus! is desired. 2 2π ab −xas 9 7/3 9 2/32 3/2 π −1 (1 − x ) dx = π + x 3 7 2 '"1 " 32π − x 5/3 + x "" = . 5 105 −1 6.3.56 1 −a a −1 x FIGURE 16 The hyperbola with equation y 2 − x 2 = 1. solution Problem 6.3.59 √ Each cross section is a disk of radius R = 1 + x 2 , so the volume of the hyperboloid is % &'a % 3 & ! a ! a "# $2 ' 1 2a + 6a 1 + x 2 dx = π (1 + x 2 ) dx = π x + x 3 '' = π π 3 3 −a −a −a A ”bead” is59.formed byisremoving a cylinder of radius r rfrom center a sphere of Rradius A “bead” formed by removing a cylinder of radius from the the center of aofsphere of radius (FigureR. Find the volume of the with r =of1the and = 2. 17).bead Find the volume beadRwith r = 1 and R = 2. y y h r x R x FIGURE 17 A bead is a sphere with a cylinder removed. 2 2 2 The equation of the outer circle is x + y2 = 2 ,2and the2inner cylinder intersects the sphere S OLUTIONsolution . The equation of the outer circle is x + y = 2 = #4 so we are rotating the region √ when y =p ± 3. Each cross section of the bead is a washer with outer radius 4 − y 2 and inner radius 1, so 2 and x = 1 about the y-axis. These equations intersect at 1 = 4 − y2 enclosed√ by x = 4 − y the volume is given by √ √ √ ( * &2 ! √washers !to or y = ± 3. p Thus we have perpendicular from y = − 3 to y = 3 with " y-axis $ 3 %) 3the √ π inner 4 − y 2 1−so 12 the dy = π √ 3is: − y 2 dy = 4π 3. √ radius volume outer radius 4 − y2 and − 3 − 3  Z √3  p 2 2 Further Insights and Challenges 2 V=π √ 4 − y − 1 dy 60. Find the volume V of−the3bead (Figure 17) in terms of r and R. Then show that V = π6 h3 , where h is the height of the bead. ThisZ √ formula has a surprising consequence: Since V can be expressed in terms 3 of h alone, it follows that two beads of height 12cm, one formed from a sphere the size of an orange and the = π (3 − y )dy √ other from a sphere the size of the earth, would have the same volume! Can you explain intuitively how this − 3 is possible? √3  3 of the solution The equation for the outer circle bead is x 2 + y 2 = R 2 , and the inner cylinder intersects y # √ − cross section 2 . Each 2 2 the sphere when y = ± = R 2π− r3y √ of the bead is a washer with outer radius R − y 3 y=− 3 and inner radius r, so the volume is √ 3! * √ 3! √ 2 2 & ! √ 2 2 (%) √ R −r √ 3 (− 3) 2 (R3) −+ r 2 − y 2 ) dy 3− r32 −dy = π −√3(− ) 2 2 − 3 − R −r 3 √ &' R 2 −r 2 √ 2% √ 2 1 ! ' 4 √ = (R 2 − r 2 )3/2 π. =3π √ (R 2 − r 2 )y3 −√y 3 '' √ 3 3 − R 2 −r 2 3 =π 6 3− 3− 3 3 √ 1 Now, h = 2 R 2 − r 2 = 2(R 2 − √r 2 )1/2 , which gives h3 = π8(R 2 − r 2 )3/2 and finally (R 2 − r 2 )3/2 = 8 h3 . Substituting into the expression = 4π for3.the volume gives V = 6 h3 . The beads may have the same volume but π R −r √ R 2 −r 2 2 R 2=− π y2 clearly the wall of the earth-sized bead must be extremely thin while the orange-sized bead would be thicker. 3 6.3.59 18. f (x) = √ 1 x2 +1 [0, 2], , about x = 0 solution A sketch of the solid is shown below. Each shell has radius x and height √ of the solid is S E C T I O N 6.4 1 x2 +1 , so the volume The Method of Cylindrical Shells 705 # ! 2 " $% & ''2 √ 1 −2 2+ 2 11. y = (x 2 + 1)−2 , 2π y = 2x− (x 1) , x = 2 dx = 2π x + 1 '' = 2π( 5 − 1). √ x2 + 1 0 0 solution region enclosed by y = (x 2 + 1)−2 , y = 2 − (x 2 + 1)−2 and x = 2 is shown below. When ProblemThe 6.4.11 y −2 = rotating this region about the y-axis, each shell has radius x and height 2 − (x 2 + 1)−2 − (x 22 + 1)−2 Use the2 Shell−2 Method to compute the volume obtained by rotating the region enclosed by y = (x +1) , 1 2 − 2(x + 1) . The volume of the resulting solid is y = 2 − (x2 + 1)−2 and x = 2 around the y-axis. 0.8 # " #$2 ! 2 " 0.6 ! 2 $ 2x 1 2 −2 2 $ = 32 π. 2x − 2 dx = 2π x + 2 x(2 − 2(x + 1)looks ) dxlike = 2π 2π S OLUTION . The region this: 2 5 x + 1 $0 0 0 0.2 (x + 1) −2y 1 −1 y = 2 − (x 2 + 1)−2 2.0 19. f (x) = a − x 2 x [0, a],1.5 about x = −1 with a > 0, 1.0 below. Each shell has radius x − (−1) = x + 1 and height a − x, solution A sketch of the solid is shown y = (x 2 + 1) 2 0.5 so the volume of the solid is ! a ! a$ & x 2π (x + 1) (a − x) dx = 2π0.5 1.0a +1.5 (a −2.0 1)x − x 2 dx − 0 0 " #'a ' 12. y = 1 − |x − 1|, y = 0 a − 1 2 2 1 3 −2 Shells centered at the y-axis have radius 2 −x2(x− +x1) '' . Thus the volume is given = x2πandaxheight + 3 is shown 0 solution The region enclosed by y = 1 − |x − 1| and2the x-axis below. When rotating this by # [0, 1],2 the shell has radius x and region about the y-axis,Z two different shells are " generated. For each x ∈ 2 3  − 1) a a (a + 3) aZ 2(a 2 2  and 2xvolume = 2π +height = π. height x; for each x ∈ [1, 2], the shell has 2 − x.−The of the resulting solid is 2 radius −2x a 2πx 2 − 2(x + 1) = 2π dx3 22x − 23 2 (x + 1) ! 1 ! 2 !0 ! 0  1 2 2 2π x(x) dx + 2π x(2 − x) dx = 2π a 2(x 2 ) dx + 2π (2x − x 2 ) dx 1 = 2π x0 + 2 0 1 1 x + 1 x=0 " #$ " #$ 1 3 $$1 1 3 $$2 2 32 = 2π + 2π x − x x π. = $ $ = 2π. 6.4.11 3 0 1 5 3 −2 − ay 20.Problem f (x) = 1 6.4.20 − x 2 , [−1, 1], x=c a −2 −1 1.0 with cy => x 1 y=2−x 0.5 below. Each shell has radius c − x and height 1 − x 2 , so the solution A sketch of the solid is shown Sketch the solid obtained by rotating the region underneath the graph of f(x) = 1 − x2 over [−1, 1] about volume of the solid is x method. the axis x = c for c > 1, then calculate its volume using the Shell " #'1 ! 1 ! 1$ 0 0.5 1.0 1.5 $ & & 2.0 ' 1 4 c 3 1 2 2 3 2 ' = 8cπ . 2πS OLUTION (c − x). A1 sketch − x dx = 2π x − cx − x + c dx = 2π − − + cx x x x ' of the solid is shown below. 4 3 2 3 −1 −1 −1 In Exercises 13 and 14, use a graphing utility to find the points of intersection of the curves numerically and y then compute the volume of rotation of the enclosed region about the y-axis. 1 13. y = 12 x 2 , y = sin(x 2 ), x≥0 solution The region enclosed by y = 12 x 2 and y = sin x 2 is shown below. When rotating this region about the y-axis, each shell has radius x and height sin x 2 − 12 x 2 .x Using a computer algebra system, we find −1 1 on cthe 2c − 1 2cis+ 1x = 1.376769504. Thus, the volume of the that the x-coordinate of the point of intersection right resulting solid of revolution is " # 2 Method to calculate the volume of rotation InEach Exercises 21–26, shell sketchhas the!radius enclosed region use 1the Shell 1.376769504 cylindrical c− x andand height 1− x , so the volume of the solid is 2 2 x sin x − x dx = 1.321975576. 2π about the x-axis. Z1 Z 1 2 0  8cπ 21. x = y, y = 0, x2π(c = 1− x)(1 − x2 ) dx = 2π x3 − cx2 − x + c dx = . 6.4.20 y 3 −1 −1 solution When the region shown below is rotated about the x-axis, each shell has radius y and height 1 1 − y. The volume of the resulting solid is #' " ! 1 y = sin x 2 ! 1 1 2 1 3 ''1 π x2 = = 2π y(1 − y) dy = 2π (y − y 2 ) ydy − y y 2π 2 ' = 3. 2 3 0 0 0 x 4 1 0 14. y = 1 − x4, y = x, x≥0 solution The region enclosed by y = 1 − x 4 , y = x, and the y-axis is shown below. Using the shell method, each shell has radius x and height 1 − x 4 − x. Using a computer algebra system, we find that the 1 0 26. y = x 1/3 − 2, y = 0, 0.5 1.0 1.5 2.0 x x = 27 solution When the region shown below is rotated about the x-axis, each shell has radius y and height 27 − (y + 2)3 . The volume of the resulting solid is Problem 6.4.26 ! ! 1# $ $ # 1 2 3 2π enclosed dy y · 27 − 2)3 dy y 4 calculate Sketch the region by(y the+curves y= = 2π x1/3 − 19y 2, y −=12y 0, x −=6y27,−and its volume of 0 revolution about the x-axis using the shell method. 0 % &' 19 2 38π 3 4 1 5 ''1 3 = 2π = y − 4y − . y − y ' S OLUTION. The region looks like this: 2 5 2 5 0 y 1.0 0.8 0.6 y= 3 5 10 15 20 25 30 x−2 0.4 0.2 0 x 3 27. Determine which ofatthethe following is the appropriate needed to 2) determine ofis the solid Each shell centered x-axis has radius y andintegrand height 27 − (y + . Thus the thevolume volume obtained by rotating around the vertical axis given by x = −1 the area that is between the curves y = f (x) Z 1 over the interval [a, b], where aZ ≥ 1 0 and f (x) ≥ g(x) over that and y = g(x)   interval. 38π . 6.4.26 2πy 27 − (y + 2)3 dy = 2π 19y − 12y2 − 6y3 − y4 dy = (a) x(f (x) 0− g(x)) 5 0 (b) (x + 1)(f (x) − g(x)) (c) x((f (x) − 1) − (g(x) − 1)) (d) (x − 1)(f (x) − g(x)) Problem 6.4.57 (e) x(f (x + 1) − g(x + 1)) Use the Shell Method to find the volume of the torus obtained by rotating the circle (x − a)2 + y2 = b2 solution When(assume the region = fproblem (x) and6.3.56.) y = g(x) is rotated around the vertical axis about the y-axis a >between b). (Seethe thecurves picturey for x = −1, each shell has radius x − (−1) = x + 1 and height f (x) − g(x). The appropriate integrand is therefore (b): .(xWe + 1)(f − g(x)).with radius x going from x = a − b to x = a + b. Each of these S OLUTION have(x) cylinders p (b cylinders has height 2 b2 − (x − a)2 so we have volume: 28. Let y = f (x) be a decreasing function on [0, b], such that f (b) = 0. Explain why 2π 0 xf (x) dx = ( f (0) Z a+b h denotes Z a+b q q the inverse of f . π 0 (h(x))2 dx, where 2πx(2 b2 − (x − a)2 )dx = 4π x b2 − (x − a)2 dx a−b a−b Substituting u = x − a and du = dx we get Zb 4π −b √ Zb p Zb p p u b2 − u2 du + 4πa b2 − u2 du (u + a) b2 − u2 du = 4π −b −b √ p Note that u b2 − u2 is an odd function (u b2 − u2 = −((−u) b2 − (−u)2 ) so it is 0 integrated about the symmetric interval [-b,b] (or use substitution u = bsin(θ), du = bcos(θ)dθ to show this fact). Rb √ Thus the volume is 4πa −b b2 − u2 du. The integral here is equal to half of the area of a circle 2 2 2 of radius b, we have that the volume is 4πa( πb 2 ) = 2π ab . 5 6.4.57 π "y m . 2 Thus the weight of one layer is 9800π ! y "2 2 "y N. The layer must be lifted 12 − y meters, so the total work needed to empty the tank is # 10 ! y "2 9800π (12 − y) dy = π(3.675 × 106 ) J ≈ 1.155 × 107 J. 2 0 Problem 6.5.21 21.Calculate Horizontal Figurerequired 11; water from a small hole at Evaluate the cylinder work (ininjoules) toexits pump all of the water outthe of top. the Hint: cylindrical tankthe of integral radius rby interpreting of it as the area of a circle. and lengthpart l shown below. Water exits here. r FIGURE 11 S OLUTION. Consider a horizontal “slice” of water of thickness dy at a distancepy above vertical solution Place the origin along the axis of the cylinder. At location y, the layer of water is a rectangular $ allow y to be negative.) The volume of the slice is 2` r$2 − y2 dy m3 , center of the tank. (We p volume of the layer is 2ℓ r 2 − y 2 "y, and slab of length ℓ, width 2 r 2 − y 2 and thickness "y. Thus, the $ is 2 · 9.8 · 1000` r2 − y2 dy N. The slice must be lifted a and so the force needed to lift the slice thedistance force needed to lift the layer is 19600ℓ r 2tank, − y 2 "y. Thetotal layerwork must done be lifted of r − y to get to the top of the so the is a distance r − y, so the total work needed to empty the tank is given by Zr # r p # rZ% # r Z r% p %p r 2 2 2 2 2 2 · 9.8 · 1000`r 2 − r y−2 (r y− (ry) −dy y) dy = 19600`r r 2 −r y− y − dr19600ℓ − 19600` y y y2 dr. 2 dy 2 dy. 19600ℓ = 19600ℓr r 2 −r y − −r −r −r −r −r −r The first integral on the right is equal to half of the area of a semicircle of radius r, and the Now, second integral is zero since the integrand # r % is odd (you can also use substitution to find this). Thus the work done is  y r 2 −y 2 du = 0 −r 1 19600`r πr2 = 9800`πr3 J. 6.5.21 2 because the integrand is an odd function and the integration interval is symmetric with respect to zero. Moreover, the other integral is one-half the area of a circle of radius r; thus, # r % 1 r 2 − y 2 dy = π r 2 . Problem 6.5.32 2 −r A 500-kg wrecking hangs from athe 12-meter Finally, the total work ball needed to empty tank is cable of density 15 kg/m attached to a crane. Calculate the work done if the crane lifts the ball&from ground level to 12 meters in the air by drawing in the cable. ' 1 2 19600ℓr π r − 19600ℓ(0) = 9800ℓπ r 3 J. S OLUTION. We will treat the cable2and the ball separately. Consider a segment of cable of length dy that must be lifted y feet. The work needed to lift the cable is W(y) = (15 dy)9.8y J, 22.soTrough in Figure 12;lift water the total work to the exits cableby is pouring over the sides. Z1 215 · 9.8y dy ≈ 10, 584 J. 0 Lifting the 500 kg ball to a height of 12 bmeters requires and additional c 500 · 12 · 9.8 h = 58, 800 J, so the total work done is 69,384 Joules. a FIGURE 12 6 6.5.32 0 9.8ρ(x) dx = 9.8 0 8x − 2x dx = 9.8 4x − x 3 & = 176.4 N & 0 another two meters requires an additional 352.8 J of work. The total work is therefore 661.5 J. Exercises 35–37: The gravitational force between two objects of mass m and M, separated by a distance r, has magnitude GMm/r 2 , where G = 6.67 × 10−11 m3 kg−1 s−1 . 35. Show that if two objects of mass M and m are separated by a distance r1 , then the work required to increase the separation to a distance r2 is equal to W = GMm(r1−1 − r2−1 ). solution The work required to increase the separation from a distance r1 to a distance r2 is & ! r2 GMm GMm &&r2 dr = − = GMm(r1−1 − r2−1 ). r &r1 r2 r1 Problem 6.5.35 The gravitational between two35objects of mass m and M, toseparated by a satellite distance r, has 36. Use force the result of Exercise to calculate the work required place a 2000-kg in an orbit magnitude GMm −11 3 −1 −1 6 m and where G = 6.67 ∗ 10 m kg s . 1200 km above the surface of the earth. Assume that the earth is a sphere of radius R = 6.37 × 10 e r2 mass Me = 5.98 × 1024 kg. Treat the satellite as a point mass. Show that if two objects of mass M and m are separated by a distance r1 , then the work required to solution The satellite will move from a distance r1 = Re to a distance Re + 1200000. Thus, from −1 r2 = −1 increase theExercise separation 35, to a distance r2 is equal to W = GMm(r1 − r2 ). S OLUTION. W = (6.67 10−11 )(5.98 × 1024 )(2000) From the×equation $ 1 1 − 6 6 6.37 Z × 10 6.37 × 10 + 1200000 b % ≈ 1.99 × 1010 J. W the = workF(x)dx 37. Use the result of Exercise 35 to compute required to move a 1500-kg satellite from an orbit a 1000 to an orbit 1500 km above the surface of the earth. we obtain solution The satellite will move from a distance r1 = Re + 1000000 to a distance r2 = Re + 1500000. from Exercise 35,  r2 Z r2 ZThus, r2 % $ 1 1 GMm −1 dr =−11 GMm dr =× GMm −1 W= = GMm(r1−1 24 1 − r2 ). W = (6.67 × 10 − )(5.98 × 10 )(1500) 2 2 6 6 r r 1000000 6.37 × 10 + 6.37 × 10 + 1500000 r1 r r1 r=r1 6.5.35 ≈ 5.16 × 109 J. 38. The pressure P and volume V of the gas in a cylinder of length 0.8 m and radius 0.2 m, with a movable piston, are related by P V 1.4 = k, where k is a constant (Figure 14). When the piston is fully extended, the gas pressure is 2000 kilopascals (kPa; 1 kilopascal is 103 newtons per square meter). k. The pressure(a)PCalculate and volume V of the gas in a cylinder of length 0.8 m and radius 0.2 m, with a movable (b) The force on the piston is P A, where A is the piston’s area. Calculate the force as a function of the length 1.4 piston, are xrelated by PV of the column of gas. = k, where k is a constant. When the piston is fully extended, the gas 3 pressure is 2000 kilopascals kPa isto10 newtons square meter). (c) Calculate the work(1 required compress the gas per column from 0.8 m to 0.5 m. Problem 6.5.38 0.2 x FIGURE 14 Gas in a cylinder with a piston. a) Calculate k. b) The force on the piston is PA, where A is the piston’s area. Calculate the force as a function of the length x of the column of gas. c) Calculate the work required to compress the gas column from 0.8 m to 0.5 m. S OLUTION. a) We have P = 2 × 106 Pa and V = π · 0.22 · 0.8 = 0.032π m3 , k = 2 × 106 · (0.032π)1.4 ≈ 80, 213.9. b) The are of the pistion is A = 0.04π, and the volume of the column of gas is V(x) = Ax. From P = k/V 1.4 we obtain F = PA = k · A = k(0.04π)−0....
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