CHAPTE SIXTEEN 6. SOLUTIONS 5 Solutions for Section 6. Eercises. Mark the values of the function on the plane, as shown in Figure 6., so that ou can guess respectivel at the smallest and largest values the function takes on each small rectangle. Lower sum f i, i) Upper sum + 6 + + 7 7.).).. f i, i) 7 + + 6 + 8.).).6.. 5 6 8.. 5 7..... Figure 6.. In the subrectangle in the top left in Figure 6., it appears that f, ) has a maimum value of about 9. In the subrectangle in the top middle, f, ) has a maimum value of. Continuing in this wa, and multipling b and, we have Overestimate 9 + + + 7 + 8 + + 5 + 7 + 8))5) 8. Similarl, we find Thus, we epect that Underestimate 7 + 7 + 8 + + 5 + 7 + + + 6))5). f, )da 8.. a) If we take the partition of consisting of just itself, we get Lower bound for integral min f A ) ). Similarl, we get Upper bound for integral ma f A ) ) 6.
6 Chapter Siteen /SOLUTIONS b) The estimates asked for are just the upper and lower sums. We partition into subrectangles a,b) of width and height, where a, b) is the lower-left corner of a,b). The subrectangles are then,),,),,), and,)., ), ), ), ) Then we find the lower sum Lower sum Similarl, the upper sum is a,b) Figure 6. A a,b) min a,b) f Min of f on a,b) ) a,b) Min of f on a,b) ) a,b) f, ) + f, ) + f, ) + f, )) + + + ) 8. Upper sum Ma of f on a,b) ) a,b) f, ) + f, ) + f, ) + f, )) + + + ) + 6 6.6. The upper sum is an overestimate and the lower sum is an underestimate, so we can get a better estimate b averaging them to get 6 + 8 7... a) We first find an over- and underestimate of the integral, using four subrectangles. On the first subrectangle, ), the function f, ) appears to have a maimum of and a minimum of 79. Continuing in this wa, and using the fact that and, we have and Overestimate + 9 + 85 + 79))) 8, Underestimate 79 + 68 + 6 + 55))) 56. A better estimate of the integral is the average of the overestimate and the underestimate: Better estimate 8 + 56 7. b) The average value of f, ) on this region is the value of the integral divided b the area of the region. Since the area of is 6)8) 8, we approimate Average value f, )da 7 77.5. Area 8 We see in the table that the values of f, ) on this region var between 55 and, so an average value of 77.5 is reasonable. 5. Partition into subrectangles with the lines,.5,,.5, and and the lines,,,, and. Then we have 6 subrectangles, each of which we denote a,b), where a, b) is the location of the lower-left corner of the subrectangle.
6. SOLUTIONS 7 We want to find a lower bound and an upper bound for the volume above each subrectangle. The lower bound for the volume of a,b) is.5min of f on a,b) ) because the area of a,b) is.5.5. The function f, ) + increases with both and over the whole region, as shown in Figure 6.. Thus, Min of f on a,b) fa, b) + ab, because the minimum on each subrectangle is at the corner closest to the origin. a, b) a, b) Figure 6. Figure 6. Similarl, Ma of f on a,b) fa +.5, b + ) + a +.5)b + ). So we have Lower sum.5 + ab).5 + ab) a,b) a,b) 6 +.5 ab Since a,.5,,.5 and b,,,, epanding this sum gives Lower sum 6 +.5 + + + +.5 +.5 +.5 +.5 + + + + +.5 +.5 +.5 +.5 ) a,b) Similarl, we can compute the upper sum: 5. Upper sum.5 + a +.5)b + )).5 + a +.5)b + )) a,b) a,b) 6 +.5 a +.5)b + ). a,b) 6. Since f, ) is measured in micrograms per square meter, and we are integrating over an area measured in square meters, the units of the integral are micrograms. The integral represents the total quantit of pollution, in micrograms, in the region. 7. The eact value of the integral is /. 8. The value of the integral is around.. Problems 9. The function being integrated is f, ), which is positive everwhere. Thus, its integral over an region is positive.
8 Chapter Siteen /SOLUTIONS. The function being integrated is f, ), which is positive everwhere. Thus, its integral over an region is positive.. The function being integrated is f, ) 5. Since > in, f is positive in and thus the integral is positive.. The function being integrated is f, ) 5, which is an odd function in. Since B is smmetric with respect to, the contributions to the integral cancel out, as f, ) f, ). Thus, the integral is ero.. The region D is smmetric both with respect to and aes. The function being integrated is f, ) 5, which is an odd function in. Since D is smmetric with respect to, the contributions to the integral cancel out. Thus, the integral of the function over the region D is ero.. The function being integrated, f, ) + 5, is an odd function in while D is smmetric with respect to. Then, b smmetr, the positive and negative contributions of f will cancel out and thus its integral is ero. 5. In a region such as B in which <, the quantit + 5 is less than ero. Thus, its integral is negative. 6. The region is smmetric with respect to and the integrand is an odd function in, so the integral over is ero. 7. The function being integrated, f, ) is alwas negative in the region B since in that region < < and <. Thus, the integral is negative. 8. The function being integrated, f, ), is an odd function in while D is smmetric with respect to. B smmetr, the integral is ero. 9. The region D is smmetric both with respect to and aes. The function being integrated is odd with respect to in the region D. Thus, its integral is ero.. Since D is a disk of radius, in the region D, we have <. Thus, π/ < < π/. Thus, cos is alwas positive in the region D and thus its integral is positive.. The function f, ) e is positive for an value of. Thus, its integral is alwas positive for an region, such as D, with nonero area.. The region D is smmetric both with respect to and aes. Looking at the contributions to the integral of the function f, ) e, we can see that an contribution made b the point, ), where >, is greater than the corresponding contribution made b, ), since e > > e for >. Thus, the integral of f in the region D is positive.. The region D is smmetric both with respect to and aes. The function f, ) is odd with respect to, and thus the contributions to the integral from, ) and, ) cancel. Thus the integral is ero in the region D.. The function f, ) is odd with respect to, and thus the integral is ero in region B, which is smmetric with respect to. 5. We use four subrectangles to find an overestimate and underestimate of the integral: A better estimate of the integral is the average of the two: Overestimate 5 + 9 + 9 + 5))) 56, Underestimate 5 + + + ))). f, )da 56 + 9. The units of the integral are milligrams, and the integral represents the total number of mg of mosquito larvae in this 8 meter b 6 meter section of swamp. 6. The question is asking which graph has more volume under it, and from inspection, it appears that it would be the graph for the mosquitos. 7. Let s break up the room into 5 sections, each of which is meter b meter and has area A. We shall begin our sum as an upper estimate starting with the lower left corner of the room and continue across the bottom and moving upward using the highest temperature, T i, in each case. So the upper iemann sum becomes 5 Ti A T A + T A + T A + + T5 A i AT + T + T + + T 5) ) +9 + 8 + 7 + 7+ 9 +8 + 7 + 7 + 6+ 7 +7 + 6 + 6 + 6+ 6 +6 + 5 + 5 + 5+ 5 + + + + ) )659) 659.
In the same wa, the lower iemann sum is formed b taking the lowest temperature, t i, in each case: 5 ti A t A + t A + t A + + t5 A i At + t + t + + t 5) ) 7 +7 + 6 + 6 + 5+ 6 +6 + 5 + 5 + 5+ 5 + + + + + + + + + + + + + + ) )6) 6. So, averaging the upper and lower sums we get: 6.5. To compute the average temperature, we divide b the area of the room, giving Average temperature 6.5 5)5) 5. C. Alternativel we can use the temperature at the central point of each section A. Then the sum becomes 5 T i i A A 5 T i i ) 9 +8 + 7 + 6.5 + 6+ 7 +7 + 6 + 6 + 5.5+ 6 +5.5 + 5 + 5 + 5+ 5 + + + + + + + +.5 + ) )6) 6. Then we get Average temperature 5 T i i A 6 Area 5)5) 5. C. 6. SOLUTIONS 9 8. The total area of the square is.5).5).5. See Figure 6.5. On a disk of radius.5 the function has a value of or more, giving a total contribution to the integral of at least ) π.5 ).. On less than half of the rest of the square the function has a value between and, giving a contribution to the integral of between /.5) ).5 and. Since the positive contribution to the integral is therefore greater in magnitude than the negative contribution, f da is positive...5..5.5...5.5..5. Figure 6.5
Chapter Siteen /SOLUTIONS 9. The total number of tornados, per ear, in a certain region, is the integral of the frequenc of tornados in that region. In order to approimate it, we first subdivide the states into smaller regions of miles, as shown in the figure in the tet. We will find the upper and lower bounds for the frequenc of tornados, and then take the average of the two. We do this b finding the highest frequenc of tornados in each subdivision, and then add up all the frequencies, and then do the same for the low frequencies. a) For the high frequenc in Teas, going left to right, top to bottom, we get: [9 + 8) + 9 + 8 + 9 + 9 + 8) + 7 + 7 + 9 + 9 + 8) + + + + 6 + 8 + 7 + 7 + 6) + + + + 6 + 7 + 7) + + + 5) + + )]. This equals 8 tornados. For the low frequenc, we get: [7 + 7) + 7 + 7 + 7 + 7 + 5) + + 5 + 6 + 7 + 5) + + + + + + 6 + 7 + 6) + + + + + 5 + 5) + + + ) + + )]. This equals tornados. The average of the two equals 8 + )/ 8 tornados. b) For the high frequenc in Florida, going left to right, top to bottom, we get: [5+5+5+9) +9+9) +7+7) +5)]. This equals 6 tornados. For the low frequenc, we get: [5 + 5 + 5 + 5) + 7 + 7) + 7 + 5) + 5)]. This equals 5 tornados. The average of the two is equal to 6 + 5)/ 56 tornados. c) For the high frequenc in Ariona, going left to right, top to bottom, we get: [++) +++) +++) + + )]. This equals tornados. For the low frequenc, we get: [ + + ) + + + ) + + + ) + + )]. This equals tornados. The average of the two is equal to + )/ tornados.. Let be the region 6, 8. Then Volume w, ) da Lower estimate: ++8+++8++++6+6+++++++++++++) 58. Upper estimate: 8++6+7+7+6++8+++++++6+7+7+6++++++) 8. The average of the two estimates is 7 cubic feet.. a) The graph of f looks like the graph of g in the -plane slid parallel to itself forward and backward in the direction, since the value of does not affect the value of f. b) The solid bounded b the graph of f is a clinder, hence f da Volume under surface d c) b a g) d Length in direction ) ) Cross-sectional area in -plane. a) On the -plane,, so the equation of the edge of the base is + 5, a circle of radius 5. b) The area of the base is π 5) 5π meters. c) The cross-section at has equation + 5, a circle of radius 5. d) The area of the cross-section is π 5) 5π meters. e) At height, the cross-section is the circle + 5.. Let This is a circle of radius 5, so f) The approimate volume of the slice is A) meters. g) We have and let Volume of pile A) π 5 ) 5 )π meters. 5 5 )π d f, ) g, ) 5 + a + + a + ) π 5 5π meters.
6. SOLUTIONS where < a < a. For all points, ) we have f, ) g, ) >, so in iemann sum approimations for f da and g da using the same subdivision of for both integrals, we have f i, j) g i, j). It follows that f da g da, and so increasing the value of a decreases the value of the integral. Take a iemann sum approimation to fda f i, i) A. Then, using the fact that a + b a + b repeatedl, we have: fda f i, i) A i,j i,j f i, i) A. Now f i, j) A f i, j A since A is non-negative, so fda f i, i) A f i, j) A. i,j But the last epression on the right is a iemann sum approimation to the integral f da, so we have fda f i, j) A f i, j ) A f da. Thus, i,j fda i,j i,j f da. +a + da. Solutions for Section 6. Eercises. We evaluate the inside integral first: + ) d + ) +. Therefore, we have. We evaluate the inside integral first: Therefore, we have + ) dd. We evaluate the inside integral first: Therefore, we have + ) d + ) dd + ) d + 6 ) 5. + ) + 9. + 9) d + 9) 6. 6) d ). 6) dd ) d 6 ) 5.
Chapter Siteen /SOLUTIONS. We evaluate the inside integral first: ) ) d. Therefore, we have 5. f d d or f d d ) dd ) d ). 6. This region lies between and and between the lines and, and so the iterated integral is Alternativel, we could have set up the integral as follows: / f, ) dd. f, ) dd. 7. The line connecting, ) and, ) is or So the integral becomes )/ f d d + or )/ f d d 8. The line on the left through points, ) and, 6)) is the line ; the line on the right through points, 6) and 5, )) is the line + 5. See Figure 6.6. One wa to set up this iterated integral is: 6 5 )/ / f, ) dd. The other option for setting up this integral requires two separate integrals, as follows: f, ) dd + 5 +5 f, ) dd. + 5 Figure 6.6 9. Two of the sides of the triangle have equations 5) and 5. So the integral is f d d )
. The line connecting, ) and, ) is. So the integral is e + dd e e d ) f d d )/ e e ) d e )e )e. See Figure 6.7. 6. SOLUTIONS Figure 6.7 Figure 6.8.. e dd e d 5 sin d d e d e e ). See Figure 6.8. 5 5 sin d sin d sin cos ) 5 sin 5 5 cos 5) sin cos ).68. See Figure 6.9. 5 Figure 6.9 Figure 6.. dd d
Chapter Siteen /SOLUTIONS See Figure 6.. 6 9 ) d ) 7 7 / / [ ) 7 7 / 7 ) ] 656.8 5. The region of integration ranges from to and from to, as shown in Figure 6.. To evaluate the integral, we evaluate the inside integral first: ) + ) d + ) + ) + 8. Therefore, we have + ) dd ) d ) 9.5. 6 + 9 Figure 6. Figure 6. 6. See Figure 6.. dd 9 d 9 9 ) d 9) d 9 ) + 8 7. + da + d d + ) d + ) ) d
5 [ + ) 5 5 ] 5 5 5 ) )) 5 9 ).876 6. SOLUTIONS 5 8. In the other order, the integral is + d d. First we keep fied and calculate the inside integral with respect to : + d + )/ [ + ) / /]. Then the outside integral becomes 9. [ + ) / /] d [ 5 + )5/ ] 5 5/ [ 5/ 5/].876 5 Note that the answer is the same as the one we got in Eercise 7. 5 + ) sin da π/ 5 + ) 5 + ) d 5 + ) 5 + ) sin d d ) π/ cos d + ) 9. The region of integration,, is shown in Figure 6.. Integrating first over, as shown in the diagram, we obtain ) da Now integrating with respect to gives d da d + 8 ) d. ) d
6 Chapter Siteen /SOLUTIONS + + Figure 6. Figure 6.. It would be easier to integrate first in the direction from to +, because integrating first in the direction would involve two separate integrals. + ) da + + [ + ) d d + + 9 ) d d [ ] + + 6 + 9 d [ 8 + ) + 9 + )] d + ) 9 + 6 ] See Figure 6.. ) 9 + 6 6. The region is bounded b,,, and. Thus Volume 6 ) dd. To evaluate this integral, we evaluate the inside integral first: 6 ) d ) ) ). Therefore, we have The volume of this object is.. 6 ) dd. To find the average value, we evaluate the integral ) ) d 5 5.. 6 + ) dd, and then divide b the area of the base region. To evaluate this integral, we evaluate the inside integral first: 6 + ) d + ) 6 6 + 7.
Therefore, we have 6 + )dd 6 + 7)d + 7) 7. 6. SOLUTIONS 7 The value of the integral is 7. The area of the base region is 6 8. To find the average value of the function, we divide the value of the integral b the area of the base region: Average value Area 6 + ) dd 7 5. 8 The average value is 5. This is reasonable, since the smallest value of f, ) on this region is, and the largest value is + 6.. To find the average value, we first find the value of the integral We evaluate the inside integral first: Therefore, we have ) dd. ) ) d 9. ) dd ) 9 9) d 7. The value of the integral is 7. To find the average value, we divide the value of the integral b the area of the region: Average value Area ) dd 7 6. The average value of f, ) on this rectangle is 6. This is reasonable since the smallest value of on this region is and the largest value is 6. 5. a) See Figure 6.5. /, /) + b) If we integrate with respect to first, we have f, ) da Figure 6.5 / f, ) d d. If we integrate with respect to first, the integral must be split into two parts, so c) If f, ), f, ) da da / / / f, ) d d + d d / / ) d / ) 8. / f, ) d d. d d
8 Chapter Siteen /SOLUTIONS Alternativel, da / / / d d + d + / / d d + ) d / ) + / / + 8 + 8. d d 6. a) )/ or + Figure 6.6 b) Problems + g, ) d d. 7. As given, the region of integration is as shown in Figure 6.7. eversing the limits gives ) e dd e d e d e e. Figure 6.7 Figure 6.8
6. SOLUTIONS 9 8. The function sin ) has no elementar antiderivative, so we tr integrating with respect to first. The region of integration is shown in Figure 6.8. Changing the order of integration, we get sin ) d d sin ) d d sin ) cos ) cos sin ) d d + cos ).. 9. As given, the region of integration is as shown in Figure 6.9. eversing the limits gives + dd + ) d + d 9 + ) 9 ). 9 9 Figure 6.9 Figure 6.. As given, the region of integration is as shown in Figure 6.. eversing the limits gives 9 sin ) dd 9 9 cos ) sin ) sin ) d 9 cos 8) ).56. d
Chapter Siteen /SOLUTIONS. The region of the integration is shown in Figure 6.. To make the integration easier, we want to change the order of the integration and get e e e ln d d e e ln d d ln ln ln d e d e ). e, ) + 8 8 + 8 e Figure 6. Figure 6.. Order reversed: 8 8 )/ 8)/ f, ) d d. See Figure 6... a) We divide the base region into four subrectangles as shown in Figure 6.. The height of the object at each point, ) is given b f, ), we label each corner of the subrectangles with the value of the function at that point. See Figure 6..) Since Volume Height Length Width, and and, we have and We average these to obtain Overestimate + + 6 + ))), Underestimate + 6 + + ))) 6. f 6 Volume f + 6 f 8. f f 6 f f f f b) We have f, ), so Volume 6 dd Figure 6. ) 6 d 8 d 9. The volume of this object is. Notice that is between the over- and underestimates, and 6, found in part a).
6. SOLUTIONS. a) The contour f, ) lies in the -plane and has equation so e ), ) ln/) ) + ln.69. This is the equation of a circle centered at, ) in the -plane. Other contours are of the form e ) c ) lnc/). Thus, all the contours are circles centered at the point, ). b) The cross-section has equation f, ) e. If, the base region in the -plane etends from to. See Figure 6., which shows the circular region below W in the -plane. So c) Slicing parallel to the -ais, we get Area e d. Volume e ) d d., ), ) Figure 6.: egion beneath W in the -plane 5. The intersection of the graph of f, ) 5 and -plane is a circle + 5. The given solid is shown in Figure 6.5. Thus the volume of the solid is V f, ) da 5 5 5 5 ) d d. 5
Chapter Siteen /SOLUTIONS f, ) 5 f, ) 5 6 Figure 6.5 Figure 6.6 6. The intersection of the graph of f, ) 5 and the plane 6 is a circle, +. The given solid is shown in Figure 6.6. Thus, the volume of the solid is V f, ) 6) da 9 9 ) d d. 9 7. The solid is shown in Figure 6.7, and the base of the integral is the triangle as shown in Figure 6.8. backside + + + Figure 6.7 Figure 6.8 Thus, the volume of the solid is V da ) da )/ ) d d.
6. SOLUTIONS 8. Volume d d d d 9. The region of integration is shown in Figure 6.9. Thus Volume + ) d d + ) d d. 9 9 Figure 6.9 Figure 6.. The region of integration is shown in Figure 6.. Thus, Volume 9 9 + ) d d / + ) d 9 ) + d ) 9 5 5/ + 9 7.5.. The plane + + cuts the -plane in the line +, so the region of integration is the triangle shown in Figure 6.. We want to find the volume under the graph of. Thus, + ) Volume ) d d + d + ) + ) + ) ) d 8 + 8) d + 8) 6.
Chapter Siteen /SOLUTIONS )/ or + Figure 6.. Let be the triangle with vertices, ),, ) and, ). Note that + + ) + ) + + > for, >, so + + is above + on. We want to find Volume We need to epress this in terms of double integrals. + + ) + )) da + + ) da., ) +.5 O Figure 6. To do this, divide into two regions with the line to make regions for and for. See Figure 6.. We want to find + + ) da + + ) da + + + ) da. Note that the line connecting, ) and, ) is, and the line connecting, ) and, ) is +.5. So + + ) da The line between, ) and, ) is, so + + ) da +.5 +.5 + + ) d d. + + ) d d.
We can now compute the double integral for : and the double integral for : +.5 +.5 + + ) d d + + ) d d ) +.5 + + d 8 + ) d 7 8 + ) d 9 8, + / + ) 9 8 + 9 + +.5 ) d 8 + 9 + ) 9 8. d 6. SOLUTIONS 5 So, Volume 9 8 + 9 8 8 8 6.. We want to calculate the volume of the tetrahedron shown in Figure 6.. /c /a a + b /b Figure 6. We first find the region in the -plane where the graph of a + b + c is above the -plane. When we have a + b. So the region over which we want to integrate is bounded b, and a + b. Integrating with respect to first, we have Volume /a a)/b d d /a a)/b b a c d d /a /a 6abc. c b c a c ) a)/b d bc a + a ) d
6 Chapter Siteen /SOLUTIONS. The region bounded b the -ais and the graph of is shown in Figure 6.. The area of this region is A )d ) 6. Figure 6. So the average distance to the -ais for points in the region is Average distance da area) da d ) + d ) d 6 + 6. Therefore the average distance is /6 /. /6 5. Assume the length of the two legs of the right triangle are a and b, respectivel. See Figure 6.5. The line through a, ) and, b) is given b +. So the area of this triangle is b a A ab. b a Figure 6.5 Thus the average distance from the points in the triangle to the -ais one of the legs) is Average distance A ab ab ab a b a +b a b a b 6 d d b a + b ) d a + b ) a ) a.
6. SOLUTIONS 7 Similarl, the average distance from the points in the triangle to the -ais the other leg) is Average distance A b a b +a b 6. a) We have Average value of f f da Area of Square Square ab ) ab b ab 6. d d a b + a a + b + c ) dd ) d ) a + b + c a + b + 8 ) c d a + b + 8 ) c 6 a + b + 6 ) c a + b + c The average value will be if and onl if /)a + b + /)c. b) Since /)a + b + /)c, we must have b /)a /)c. An function f, ) a + /)a /)c) + c where a and c are an real numbers is a correct solution. For eample, a, c leads to the function f, ) + /) +, and a, c leads to the function f, ) +, both of which have average value on the given square. See Figures 6.6 and 6.7. d Figure 6.6: f, ) + + Figure 6.7: f, ) + 7. a) We have Average value of f f da Area of ectangle ectangle 6 6 6a + 9b) 6 + b) dd a 6 a + 9 ) b d 6 a + 9 ) b a + b ) d a + b.
8 Chapter Siteen /SOLUTIONS The average value will be if and onl if a + /)b. This equation can also be epressed as a + b, which shows that f, ) a + b has average value of on the rectangle, if and onl if f, ). b) Since a + b, we must have b /) /)a. An function f, ) a + /) /)a) where a is an real number is a correct solution. For eample, a leads to the function f, ) + 8/), and a leads to the function f, ) +6/), both of which have average value on the given rectangle. See Figure 6.8 and 6.9. Figure 6.8: f, ) + 8 Figure 6.9: f, ) + 6 8. a) One solution would be to arrange that the minimum values of f on the square occur at the corners, so that the corner values give an underestimate of the average. See Figure 6.. 9 96 98... Figure 6. Figure 6. b) One solution would be to arrange that the maimum values of f on the square occur at the corners, so that the corner values give an overestimate of the average. See Figure 6.. 9. The force, F, acting on A, a small piece of area, is given b F p A, where p is the pressure at that point. Thus, if is the rectangle, the total force is given b F p da. We choose coordinates with the origin at one corner of the plate. See Figure 6.. b a, b) a Figure 6.
6. SOLUTIONS 9 Suppose p is proportional to the square of the distance from the corner represented b the origin. Then we have Thus, we want to compute F p k + ), for some positive constant k. k + )da. ewriting as an iterated integral, we have b a b k + ) da k b k + ) dd k a + a ) d k k a b + ab ). a) + d ) + a b a Solutions for Section 6. Eercises. W f dv + 5 ) d d d + 5 ) d d + 5 5 ) d d + 5 5 ) + 5 ) 6 5) d d. f dv W a + b + c) d d d a + b + c) d d a + b + c) d a + b + c
Chapter Siteen /SOLUTIONS. f dv a b c W a b c e d d d e e e d d d a b c e e e ) d d a b e e e c + ) d d a e c ) a b e e ) d e b ) e c ) e d e a ) e b ) e c ). W f dv π π π π π π π π π π π sin d π cos ) )) 8 sin cos + ) d d d π sin sin + ) d d sin [sin + π) sin ] d d sin sin ) d d π sin cos ) d 5. The region is the half clinder in Figure 6.. 6. The region is the half clinder in Figure 6.. 7. The region is the quarter sphere in Figure 6.5. Figure 6. Figure 6. Figure 6.5
6. SOLUTIONS 8. The region is the half clinder in Figure 6.6. 9. The region is the clinder in Figure 6.7.. The region is the hemisphere in Figure 6.8. Figure 6.6 Figure 6.7 Figure 6.8. The region is the hemisphere in Figure 6.9.. The region is the quarter sphere in Figure 6.5.. The region is the quarter sphere in Figure 6.5. Figure 6.9 Figure 6.5 Figure 6.5 Problems. A slice through W for a fied value of is a semi-circle the boundar of which is r, so the inner integral is r f,, ) d. Lining up these stacks parallel to -ais gives a slice from r to r giving r r r f,, ) d d. Finall, there is a slice for each between r and r, so the integral we want is r r r r r f,, ) d d d.
Chapter Siteen /SOLUTIONS 5. A slice through W for a fied value of is a semi-circle the boundar of which is r, so the inner integral is r f,, ) d. r Lining up these stacks parallel to -ais gives a slice from to r giving r r f,, ) d d. r Finall, there is a slice for each between and r, so the integral we want is r r r f,, ) d d d. r 6. A slice through W for a fied value of is a semi-circle the boundar of which is, so the inner integral is f,, ) d. Lining up these stacks parallel to -ais gives a slice from to giving f,, ) d d. Finall, there is a slice for each between and, so the integral we want is f,, ) d d d. 7. A slice through W for a fied value of is a semi-circle the boundar of which is r, so the inner integral is r f,, ) d. Lining up these stacks parallel to -ais gives a slice from r to r giving r r r f,, ) d d. Finall, there is a slice for each between and, so the integral we want is r 8. The required volume, V, is given b r V r 5 f,, ) d d d. + ddd + )) dd [ ] d ) d
6. SOLUTIONS 9. The pramid is shown in Figure 6.5. The planes, and, and + + intersect the plane 6 in the lines,, + on the 6 plane as shown in Figure 6.5. backside + +,, 6), 6, 6) 6 bottom 5,, 6) Figure 6.5, 6, 6) 6 plane + 6,, 6) 5,, 6) Figure 6.5 These three lines intersect at the points,, 6), 5,, 6), and, 6, 6). Let be the triangle in the planes 6 with the above three points as vertices. Then, the volume of the solid is V 6 )/ 6 )/ 6 6 6 6 ) 9 d d d ) d d 6 )/ 7 + 8) d d. Figure 6.5 shows a slice through the region for a fied. Figure 6.5
Chapter Siteen /SOLUTIONS The required volume, V, is given b V. d d d d d d d d d. Figure 6.55 shows a slice through the region for a fied value of. Figure 6.55 We break the region into small cubes of volume V. A stack of cubes verticall above the point, ) in the -plane gives the strip shown in Figure 6.55 and so the inner integral is The plane and the surface meet when, giving ), so or. Lining up the stacks parallel to the -ais gives a slice from to. Thus, the limits on the middle integral are d d d. Finall, there is a slice for each between and, so the integral we want is The required volume, V, is given b d d d. V d d d
6. SOLUTIONS 5 6 ) d d 6 d 6. d d. The required volume, V, is given b V 5 5 + 5 5 5 5 5. d d d + ) d d + 5 d 5 ) + 5 ) ) d. The required volume, V, is given b 5 V d d d 5 d d 5 5 5. d d. a) The vectors u i j and v i k lie in the required plane so p u v i + j + k is perpendicular to this plane. Let,, ) be a point in the plane, then ) i + j + k is perpendicular to p, so ) i + j + k ) i + j + j ) and so ) + +. Therefore, the equation of the required plane is + +. b) The required volume, V, is given b V d d d
6 Chapter Siteen /SOLUTIONS 6. ) d d d ) ) ) d ) d 5. a) The equation of the surface of the whole clinder along the -ais is +. The part we want is. See Figure 6.56. Figure 6.56 b) The integral is f,, ) dv D f,, ) ddd. 6. The region of integration is shown in Figure 6.57, and the mass of the given solid is given b 6 + + 6 or + 6 + or + Figure 6.57 mass δ dv
6. SOLUTIONS 7 + +6 + + + + ) + ) ddd +6 dd + ) + 6) dd 5 + 6 + 6) dd 5 ) + + 6 + d 7 8 ) + + d 7 5 8 ) 9 + + 7 5 8 9 + + 5. 7. The pramid is shown in Figure 6.58. The planes, and, and + + intersect the plane 6 in the lines,, + as shown in Figure 6.59. backside + +,, 6), 6, 6) 6 bottom 5,, 6) Figure 6.58, 6, 6) 6 plane + 6,, 6) 5,, 6) Figure 6.59 These three lines the edges of the pramid) intersect the plane 6 at the points,, 6), 5,, 6), and, 6, 6). Let be the triangle in the plane 6 with these three points as vertices. Then, the mass of the solid is Mass 6 )/ 6 6 )/ 6 )/ 6 6. 6 ) δ,, ) d d d d d d ) d d )/ 9 7 + 8) d d
8 Chapter Siteen /SOLUTIONS 8. From the problem, we know that,, ) is in the cube which is bounded b the three coordinate planes,,, and the planes,,. We can regard the value + + as the densit of the cube. The average value of + + is given b V average value + + ) dv volumev ) + + ) ddd 8 ) + + ) dd 8 8 + + ) dd 8 8 + + ) d 8 6 + 6 + ) d 8 + ) 8 6 + ). 8 9. Positive. The function + is positive, so its integral over the solid W is positive.. Positive. If,, ) is an point inside the solid W then + <. Thus the integrand + >, and so its integral over the solid W is positive.. Zero. The value of is positive above the first and fourth quadrants in the -plane, and negative and of equal absolute value) above the second and third quadrants. The integral of over the entire solid cone is ero because the integrals over the two halves of the cone cancel.. Zero. The value of is positive on the half of the cone above the second and third quadrants and negative of equal absolute value) on the half of the cone above the third and fourth quadrants. The integral of over the entire solid cone is ero because the integrals over the four quadrants cancel.. Positive. Since is positive on W, its integral is positive.. Zero. You can see this in several was. One wa is to observe that is positive on part of the cone above the first and third quadrants where and are of the same sign) and negative of equal absolute value) on the part of the cone above the second and fourth quadrants where and have opposite signs). These add up to ero in the integral of over all of W. Another wa to see that the integral is ero is to write the triple integral as an iterated integral, sa integrating first with respect to. For fied and, the -integral is over an interval smmetric about. The integral of over such an interval is ero. If an of the inner integrals in an iterated integral is ero, then the triple integral is ero. 5. Zero. Write the triple integral as an iterated integral, sa integrating first with respect to. For fied and, the -integral is over an interval smmetric about. The integral of over such an interval is ero. If an of the inner integrals in an iterated integral is ero, then the triple integral is ero. 6. Negative. If,, ) is an point inside the cone then <. Hence the function is negative on W and so is its integral. 7. Positive. The function e is a positive function everwhere so its integral over W is positive. 8. Positive. The function + is positive, so its integral over the solid W is positive. 9. Positive. If,, ) is an point inside the solid W then + <. Thus + >, and so its integral over the solid W is positive.. Positive. The value of is positive on the half-cone, so its integral is positive.. Zero. is positive on the half of the half-cone above the first quadrant in the -plane and negative of equal absolute value) on the half of the half-cone above the fourth quadrant. The integral of over W is ero because the integrals over
each half add up to ero.. Positive. Since is positive on W, its integral is positive. 6. SOLUTIONS 9. Zero. You can see this in several was. One wa is to observe that is positive on part of the cone above the first quadrant where and are of the same sign) and negative of equal absolute value) on the part of the cone above the fourth quadrant where and have opposite signs). These add up to ero in the integral of over all of W. Another wa to see that the integral is ero is to write the triple integral as an iterated integral, sa integrating first with respect to. For fied and, the -integral is over an interval smmetric about. The integral of over such an interval is ero. If an of the inner integrals in an iterated integral is ero, then the triple integral is ero.. Zero. Write the triple integral as an iterated integral, sa integrating first with respect to. For fied and, the -integral is over an interval smmetric about. The integral of over such an interval is ero. If an of the inner integrals in an iterated integral is ero, then the triple integral is ero. 5. Negative. If,, ) is an point inside the cone then <. Hence the function is negative on W and so is its integral. 6. Positive. The function e is a positive function everwhere so its integral over W is positive. 7. Orient the region as shown in Figure 6.6 and use Cartesian coordinates with origin at the center of the sphere. The equation of the sphere is + + 5, and we want the volume between the planes and 5. The plane cuts the sphere in the circle + + 5, or + 6. Volume 6 5 6 ddd. 5 Sphere is + + 5 Circle is + 6 + 6 Figure 6.6 Figure 6.6 8. The intersection of two clinders + and + is shown in Figure 6.6. This region is bounded b four surfaces:,,, and So the volume of the given solid is 9. Set up aes as in Figure 6.6. V d d d
Chapter Siteen /SOLUTIONS Figure 6.6 The slanting plane has equation and the line where it intersects the -plane has equation. The mass of the bottom part is δ times its volume, V Bottom, where V Bottom d d d ) d d ) d ) 6 ) m. d d ) d Since the cube has volume 8 m, the upper part has volume V Top 8 / / m. Thus Mass V Bottomδ + V Topδ δ + δ gm. 5. The mass m is given b m dv W ++ + + ) d d + / + ) d + /) d gm. Then the -coordinate of the center of mass is given b W dv ++ + + ) d d + / + ) d + /) d / cm. d d d d d d An essentiall identical calculation since the region is smmetric in and ) gives ȳ / cm.
Finall, we compute : W So, ȳ, ) /, /, 5/). 5. The mass m is given b m dv dv W )/ Then the coordinates of the center of mass are given b and and 6 ȳ 6 6 W W W ++ + + ) / d d + + ) /6 d d d d + ) + ) ) d 5/ cm. )/ )/ d d )/ ) d ) ) d ) ) ) d ) ) /6 gm. dv 6 dv 6 dv 6 )/ )/ )/ )/ )/ )/ 5. The volume V of the solid is 6. We need to compute m + dv m 6 6 W m 6 m m d d d d d d / cm. d d d /8 cm. d d d / cm. + d d d + ) d d + /) d + 8/) d 5m/ 6. SOLUTIONS
Chapter Siteen /SOLUTIONS 5. The volume of the solid is 8abc, so we need to evaluate m + ) dv m 8abc 8abc W 5. B the definition, we have that a + b m V m V m V c + m V m 8abc m bc m c c b a c b c b c c c c + ) d d d a a + ) d d b / + ) b b d b / + ) d c mb + c )/ + ) dv + m V W + + ) dv W + ) dv + m V W ) dv W + ) dv W ) dv W Since is alwas positive, the integral W ) dv will be positive, thus a + b > c. Solutions for Section 6. Eercises... π π/ / π/ π/ f rdr dθ f rdr dθ f rdr dθ π/. f rdr dθ π/ 5. See Figure 6.6. θ π/ r r θ π/ Figure 6.6 Figure 6.6 6. See Figure 6.6.
6. SOLUTIONS 7. See Figure 6.65. 8. See Figure 6.66. r θ π/ r θ π/ r θ π/6 r r θ π/ Figure 6.65 Figure 6.66 Figure 6.67 9. See Figure 6.67.. See Figure 6.68. θ π/ θ π/ r / sin θ or r sin θ or r / cos θ or r cos θ or Figure 6.68 Figure 6.69. See Figure 6.69.. B using polar coordinates, we get sin + )da π π sinr )r dr dθ cosr ) dθ π cos cos ) dθ cos ) π π cos ). The region is pictured in Figure 6.7. Figure 6.7
Chapter Siteen /SOLUTIONS Using polar coordinates, we get )da π/ r cos θ sin θ)rdr dθ π/ cos θ sin θ) r dθ 5 5 π/ π/ 5 cos θ sin θ) dθ cos θ dθ sin θ. The presence of the term + suggests that we should convert the integral into polar coordinates. Since + r, the integral becomes + dd π π/ r drdθ. π r dθ π 9 dθ 8π. 5. a) / Figure 6.7 b) f, ) d d. c) For polar coordinates, on the line /, tan θ / /, so θ tan /). On the -ais, θ π/. The quantit r goes from to the line, or r sin θ, giving r / sin θ and f, ) fr cos θ, r sin θ). Thus the integral is π/ tan /) / sin θ fr cos θ, r sin θ)r dr dθ. Problems 6. B the given limits, and, the region of integration is in Figure 6.7. In polar coordinates, we have π/ π/ r cos θ r dr dθ π/ π/ π/ π/ ) cos θ r cos θ dθ dθ π/ sin θ ) π/
6. SOLUTIONS 5 6 6 6 Figure 6.7 Figure 6.7 7. From the given limits, the region of integration is in Figure 6.7. In polar coordinates, π/ θ π/. Also, 6 r cos θ. Hence, r 6/ cos θ. The integral becomes 6 d d π/ 6/cos θ π/ π/ π/ tan θ r π/ π/ 6/cos θ r dr dθ ) dθ π/ π/ )) 6. 6 cos θ dθ Notice that we can check this answer because the integral gives the area of the shaded triangular region which is 6 6) 6. 8. From the given limits, the region of integration is in Figure 6.7. π/ Figure 6.7 So, in polar coordinates, we have, π/ π/ r cos θ sin θ)r dr dθ π/ ) cos θ sin θ r dθ sinθ) dθ π/ cosθ) ).
6 Chapter Siteen /SOLUTIONS 9. The graph of f, ) 5 is an upside down bowl, and the region whose volume we want is contained between the bowl above) and the -plane below). We must first find the region in the -plane where f, ) is positive. To do that, we set f, ) and get + 5. The disk + 5 is the region over which we integrate. Volume 5 ) da π 65 65π π 5 5 r ) 5 r dθ π dθ 5 r ) rdr dθ. First, let s find where the two surfaces intersect. 8 + So at the intersection. See Figure 6.75. 8 + + + Figure 6.75 The volume of the ice cream cone has two parts. The first part which is the volume of the cone) is the volume of the solid bounded b the plane and the cone +. Hence, this volume is given b + ) da, where is the disk of radius centered at the origin, in the -plane. Using polar coordinates, we have: ) π + da r) r dr dθ [ π ) ] r r dθ π 8π/ dθ
6. SOLUTIONS 7 The second part is the volume of the region above the plane but inside the sphere + + 8, which is given b. a) 8 ) da where is the same disk as before. Now 8 ) da π π π π 8 r )rdr dθ r 8 r dr dθ ) 8 r ) / / 8 / ) dθ π8 6 ) 8π π 6 8) 8π 8π 5) Thus, the total volume is the sum of the two volumes, which is π )/. Total Population π/ π/ π dθ π δr, θ) rdr dθ. π dθ r dr dθ r dθ b) We know that δr, θ) decreases as r increases, so that eliminates iii). We also know that δr, θ) decreases as the -coordinate decreases, but r cos θ. With a fied r, is proportional to cos θ. So as the -coordinate decreases, cos θ decreases and i) δr, θ) r) + cos θ) best describes this situation. c) π/ π/ r) + cos θ) rdr dθ + cos θ)r r ) dθ π/ Thus, the population is around 9,.. a) The volume, V, is given b V Converting to polar coordinates gives π a V e r r dr dθ θ 9 9 π/ π/ [ + a e b) As a, the value of e a, so the volume tends to π.. The densit function is given b π π/ θ + sin θ + cos θ) dθ ] π/ π/ 8π ) 9 + ) da. a e r) π e a) π e a ). ρr) r where r is the distance from the center of the disk. So the mass of the disk in grams is ρr) da π 5 r)rdr dθ
8 Chapter Siteen /SOLUTIONS π π [ 5r r ] 5 dθ 5 dθ 5π grams). A rough graph of the base of the spring is in Figure 6.76, where the coil is roughl of width. inches. The volume is equal to the product of the base area and the height. To calculate the area we use polar coordinates, taking the following integral: Area π.6+.θ.5+.θ π π Therefore, the volume.66..7 in. rdrdθ.6.θ).5.θ) dθ..5 +.8θ)dθ.5 π + π.8θ ).66 Figure 6.76 5. The charge densit is δ k/r, where k is a constant. Total charge δ da Disk π k r r dθ dr k π Thus the total charge is proportional to with constant of proportionalit kπ. dθ dr k π dr kπ. 6. a) We must first decide where to put the origin. We locate the origin at the center of one disk and locate the center of the second disk at the point, ). See Figure 6.77. Other choices of origin are possible.) / + ) + r r cos θ π/ / Figure 6.77 Figure 6.78 B smmetr, the points of intersection of the circles are half-wa between the centers, at /. The -values at these points are given b ) ± ± ±.
6.5 SOLUTIONS 9 We integrate in the -direction first, so that it is not necessar to set up two integrals. The right-side of the circle + is given b. The left side of the circle ) + is given b Thus the area of overlap is given b Area. / / dd. b) In polar coordinates, the circle centered at the origin has equation r. See Figure 6.78. The other circle, ) +, can be written as + + +, so its equation in polar coordinates is r r cos θ, and, since r, r cos θ. At the top point of intersection of the two circles, /, /, so tan θ, giving θ π/. Figure 6.78 shows that if we integrate with respect to r first, we have to write the integral as the sum of two integrals. Thus, we integrate with respect to θ first. To do this, we rewrite r r cos θ as θ arccos. ) This gives the top half of the circle; the bottom half is given b r θ arccos. ) Thus the area is given b Area arccosr/) r dθdr. arccosr/) Solutions for Section 6.5 Eercises. a) A vertical plane perpendicular to the -ais:. b) A clinder: r. c) A sphere: ρ. d) A cone: φ π/. e) A horiontal plane: 5. f) A vertical half-plane: θ π/.. W f dv π/ π/ π/ π/ r + ) rdr dθ d 6 + 8 ) dθ d π 6 + 8 ) d 6π + 8 π π
5 Chapter Siteen /SOLUTIONS.. 5. W f dv π π π W π W sin r )) rdr dθ d cos r ) dθ d cos cos ) dθ d cos ) d πcos ) π cos ) f dv f dv 5 π π π/ 5 π π 5 π π 7 7 7 6 7 6 5 ρ ρ sin φ dφ dθ dρ ρ sin φ dφ dθ dρ π/ ρ dθ dρ ρ dρ 5π π π/ π π/ π π π π/ π π sin φ)ρ sin φ dρ dφ dθ ρ sin φ dρ dφ dθ sin φ dφ dθ cos φ dφ dθ φ π/ sin φ) dθ π ) dθ 7 6 π π 7ππ ) ) 6. Using Cartesian coordinates, we get: 7. Using clindrical coordinates, we get: 8. Using clindrical coordinates, we get: 5 π f d d d f rdr dθ d π/ f rdr dθ d
6.5 SOLUTIONS 5 9. Using spherical coordinates, we get: π π f ρ sin φ dρ dφ dθ. Using spherical coordinates, we get: π π/6 f ρ sin φ dρ dφ dθ. We use Cartesian coordinates, oriented as shown in Figure 6.79. The slanted top has equation m, where m is the slope in the -direction, so m /5. Then if f is an arbitrar funtion, the triple integral is Other answers are possible. 5 /5 5,, ) f ddd. 5,, ) 5 Figure 6.79 5,, ). We choose clindrical coordinates oriented as in Figure 6.8. The cone has equation r. Since we have a half cone scooped out of a half clinder, θ varies between and π. Thus, if f is an arbitrar function, the integral is Other answers are possible. π r fr ddrdθ. r π/ Figure 6.8. Figure 6.8 is one eighth of a sphere of radius, below the -plane and under the first quadrant.
5 Chapter Siteen /SOLUTIONS. a) The region of integration is the region between the cone r, the -plane and the clinder r. In spherical coordinates, r becomes ρ sin φ, so ρ / sin φ. The cone is φ π/ and the -plane is φ π/. See Figure 6.8. Thus, the integral becomes π π/ / sin φ π/ ρ sin φ dρdφdθ. r π r Figure 6.8: egion of integration is between the cone and the -plane b) The original integral is easier to evaluate, so π r r ddrdθ π r r drdθ π r drdθ π r 8π. Problems 5. In spherical coordinates, the spherical cap is part of the surface ρ. If α is the angle at the verte of the cone, we have tanα/) /, so α/ π/. Since the cone is below the -plane, the angle φ ranges from π/ to π. Thus, the integral is given b π π π/ fρ, φ, θ)ρ sin φ dρ dφ dθ. 6. In clindrical coordinates, the spherical cap has equation r. If α is the angle at the verte of the cone, we have tanα/) /, so α/ π/. The cone has equation r. Thus, the integral is π r r gr, θ, )r d dr dθ. 7. In rectangular coordinates, the spherical cap has equation. If α is the angle at the verte of the cone, we have tanα/) /, so α/ π/. The cone has equation +. Thus, the integral is + h,, ) d d d. 8. Orient the cone as shown in Figure 6.8 and use clindrical coordinates with the origin at the verte of the cone. Since the angle at the verte of the cone is a right angle, the angles AOB and COB are both π/. Thus, OB 5 cos π/ 5/. The curved surface of the cone has equation r, so Volume π 5/ 5/ r π 5/ 5/ r r r d dr dθ dr dθ π 5/ r 5 r ) dr dθ
θ π π 5 ) r r 5 ) C 5/ π 5 π 6 6.8 cm. B ) 5 5 5 A 6.5 SOLUTIONS 5 π/ π/ 5 O 5/ r Figure 6.8 9. a) In Cartesian coordinates, the bottom half of the sphere + + is given b. Thus W dv d d d. b) In clindrical coordinates, the sphere is r + and the bottom half is given b r. Thus W dv c) In spherical coordinates, the sphere is ρ. Thus, W dv π/ π/ π. a) Since the cone has a right angle at its verte, it has equation π/ r d dr dθ. r +. The sphere has equation + +, so the top half is given b ρ sin φ dρ dφ dθ. The cone and the sphere intersect in the circle. See Figure 6.8. Thus W dv +,. / / /) /) + d d d. Circle is π + + Figure 6.8
5 Chapter Siteen /SOLUTIONS b) In clindrical coordinates, the cone has equation r and the sphere has equation r. Thus W dv π / r r r d dr dθ. c) In spherical coordinates, the cone has equation φ π/ and the sphere is ρ. Thus W dv π π/. a) Since the cone has a right angle at its verte, it has equation +. ρ sin φ dρ dφ dθ. Figure 6.85 shows the plane with equation /. The plane and the cone intersect in the circle + /. Thus, /) W dv / / /) + d d d. π Figure 6.85 b) In clindrical coordinates the cone has equation r, so W dv π / r r d dr dθ. c) In spherical coordinates, the cone has equation φ π/ and the plane / has equation ρ cos φ /. Thus W dv π π/ / cos φ) ρ sin φ dρ dφ dθ.. The region is a solid clinder of height, radius with base on the -plane and ais on the -ais. Use clindrical coordinates: + ) / d d d π π π r dr dθ d r r dθ d dθ d π.
6.5 SOLUTIONS 55. The region of integration is half of a ball centered at the origin, radius, on the side. Since the integral is smmetric, we can integrate over the quarter unit ball, ) and multipl the result b. Use spherical coordinates: π/ π π/ π π/ d d d + + ) / ρ ρ sin φ dρ dφ dθ ρ sin φ dφ dθ π cos φ) dθ ) ) π π.. Use clindrical coordinates: when r +, then + + + so ±. The region W is shown in Figure 6.86. + ) dv W π π π π r r r drdθd dθd + ) d + 5 5 ) π 8π 5. ) ) dθd Clinder + Sphere + + Figure 6.86 5. a) The angle φ takes on values in the range φ π. Thus, sin φ is nonnegative everwhere in W, and so its integral is positive. b) The function φ is smmetric across the plane, such that for an point,, ) in W, with, the point,, ) has a cos φ value with the same magnitude but opposite sign of the cos φ value for,, ). Furthermore, if, then,, ) has a cos φ value of. Thus, with cos φ positive on the top half of the sphere and negative on the bottom half, the integral will cancel out and be equal to ero. 6. a) The integral is negative. In W, we have < <. Thus, is negative throughout W and thus its integral is negative. b) On the top half of the sphere, is nonnegative, but can be both positive and negative. Thus, since W is smmetric with respect to the plane, the contribution of a point,, ) will be canceled out b its reflection,, ). Thus, the integral is ero.
56 Chapter Siteen /SOLUTIONS 7. The region whose volume we want is shown in Figure 6.87: θ π 6 θ π 5 Figure 6.87 Using clindrical coordinates, the volume is given b the integral: V π/ 5 π/6 π/ 5 5 r π/6 π/ π/6 r dr dθ d 5 dθ d dθ d π π ) d 6 5 π 6 5π 6. 8. a) In clindrical coordinates, the cone is r and the sphere is r +. The surfaces intersect where +. So and r. r 9. a) Volume π b) In spherical coordinates, the cone is φ π/ and the sphere is ρ. b) π π π ρ sin φ dρdφdθ. r r ddrdθ r Volume π r π π/ r r ddrdθ. r r ddrdθ. ρ sin φ dρdφdθ.. Orient the region as shown in Figure 6.88 and use clindrical coordinates with the origin at the center of the sphere. The equation of the sphere is + + 5, or r + 5. If, then r + 5, so r 6 and r. π 5 r π 5 r Volume r ddrdθ r drdθ
π π r 5 r r) drdθ θ ) ) 5π 5 7 π 5.5 cm. Sphere 5 r + 5 5 r Figure 6.88 5 r ) / r 6.5 SOLUTIONS 57 ). Orient the region as shown in Figure 6.89 and use spherical coordinates with the origin at the center of the sphere. The equation of the sphere is + + 5, or ρ 5. The plane is the plane ρ cos φ, so ρ / cos φ. In Figure 6.89, angle AOB is given b The volume is given b V π arccos/5) 5 π arccos/5) arccos/5) cos φ, so φ arccos/5). 5 / cos φ 5 9 cos φ ρ sin φ dρdφdθ θ ) sin φ dφ arccos/5) 5 π sin φ dφ π 5 ) arccos/5) cos φ 9 π 5 ) )) 5 9 /5) π 5 ) 9 )) 6 5 9 ) 5 π 8 5π 5.5 cm. π arccos/5) ) 9 cos φ sin φ dφ cos φ ) arccos/5) ) sin φ ρ p5 p/ cos φ dφ A φ O 5 Sphere ρ 5 B Figure 6.89
58 Chapter Siteen /SOLUTIONS. The densit function can be rewritten as δρ, φ, θ) ρ. So the mass is W δp ) dv π π/ π π/ 8 π ρ ρ sin φ dρ dφ dθ 8 sin φ dφ dθ + ) dθ 8 π + ) 8 π + ). Using spherical coordinates: M π π π π 7 [ π π ρ)ρ sin φ dρ dθ dφ ρ ρ ] sin φ dθ dφ sin φ dθ dφ 7 π π cos φ) 7 π [ ) + ] 7π.. a) We use spherical coordinates. Since δ 9 where ρ 6 and δ where ρ 7, the densit increases at gm/cm for each cm increase in radius. Thus, since densit is a linear function of radius, the slope of the linear function is. Its equation is δ ρ 7) so δ ρ. b) Thus, Mass π π 7 6 ρ )ρ sin φ dρ dφ dθ. c) Evaluating the integral, we have π) ) Mass π cos φ ρ ρ 7 π 5.5) 7π gm 56.99 gm. 6 5. The distance from a point,, ) to the origin is given b + +. Thus we want to evaluate + + dv Vol) where is the region bounded b the hemisphere 8 and the cone +. See Figure 6.9. We will use spherical coordinates.
6.5 SOLUTIONS 59 + Figure 6.9 In spherical coordinates, the quantit ρ goes from to 8, and θ goes from to π, and φ goes from to π/ because the angle of the cone is π/). Thus we have π π/ 8 + + dv ρρ sin φ) dρdφdθ π π/ π π/ π π sin φ ρ 8 6 sin φ dφdθ π/ 6 cos φ) dθ ) 6 dθ ) π dφdθ From Problem of Section 6. we know that Vol) π )/, therefore + + dv Average distance Vol) ) π [ )π/]. 6. a) First we must choose a coordinate sstem, since none is given. We pick the -plane to be the fied plane and the -ais to be the line perpendicular to the plane. Then the distance from a point to the plane is, so the densit at a point is given b Densit ρ k. Using clindrical coordinates for the integral, we find a r Mass π a k r ddrdθ. a r
6 Chapter Siteen /SOLUTIONS b) B smmetr, we can evaluate this integral over the top half of the sphere, where. Then π a a r π a a r Mass kr ddrdθ k r drdθ π a ) k ra r r a ) drdθ kπ a r ) a πk a πka. 7. a) We use the aes shown in Figure 6.9. Then the sphere is given b r + 5, so Volume π 5 5 r r ddrdθ. 5 r b) Evaluating gives Volume π 5 r 5 r dr π 5 r π 5 r 5 r dr ) 5 r ) / 5 π )/ 6 6π 9.5 mm. 5 r + 5 + + 5 5 a b Figure 6.9 Figure 6.9 8. We must first decide on coordinates. We pick spherical coordinates with the common center of the two spheres as the origin. We imagine the half-melon with the flat side horiontal and the positive -ais going through the curved surface. See Figure 6.9. The volume is given b the integral Evaluating gives Volume π π/ Volume sin φ ρ pb pa π π/ b dφdθ π cos φ) a ρ sin φ dρdφdθ. π/ b a ) π b a ). To check our answer, notice that the volume is the difference between the volumes of two half spheres of radius a and b. These half spheres have volumes πb / and πa /, respectivel.