CHAPTER EIGHT. Area of region. (2x)dx = x 2. Area = Area of region ( x 2 +6x) x. ( x 2 +6x)dx = x3 3 +3x2. Area of region. dy = 3y y2.

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1 CHAPTER EIGHT 8. SOLUTIONS 7 Solutions for Section 8. Eercises. (a) The strip stretches between = and =, so (b) We have Area of region Area = (). ()d = = 9.. (a) The strip stretches between = and = +6, so Area of region ( +6). (b) We have Area = 6. (a) The strip stretches from = / to =, so (b) We have Area = ( +6)d = + Area of region 6 4. (a) The equation = +6 can be solved for as 6+ = ( ). 6 = 6. ( ) ( ) 6 d = = 9. 4 = 6± 6 4 = ± 9. The left end of the strip is given b = 9, and the right end is given b = + 9. Thus, Area of region ((+ 9 ) ( 9 )) = 9. (b) We have Area = 9 5. Each strip is a rectangle of length and width, so 9 d = (9 )/ / Area of strip =, so 5 Area of region = d = Check: This area can also be computed using Length Width = 5 = = 5. = 6.

2 7 Chapter Eight /SOLUTIONS 6. Using similar triangles, the height,, of the strip is given b Thus, so = 6 so =. Area of strip =, Area of region = 6 d = 4 Check: This area can also be computed using the formula Base Height = 6 = B similar triangles, if w is the length of the strip at height h, we have w = 5 h ( so w = h ). 5 5 Thus, ( Area of strip w h = h ) h. 5 ( Area of region = 5 ( h ) dh = 5 6 = 9. h h ) 5 = 5. Check: This area can also be computed using the formula Base Height = 5 = Suppose the length of the strip shown is w. Then the Pthagorean theorem gives ( w ) h + = so w = h. Thus Area of strip w h = h h, Area of region = h dh. Using VI- in the Table of Integrals, we have ( ( h Area = h h + arcsin = 9(arcsin arcsin( )) = 9π. )) Check: This area can also be computed using the formula πr = 9π. 9. The strip has width, so the variable of integration is. The length of the strip is. Since + = and the region is in the first quadrant, solving for gives =. Thus The region stretches from = to =, so Area of strip = d. Area of region = Evaluating using VI- from the Table of Integrals, we have Area = ( +arcsin ( )) d. Check: This area can also be computed using the formula 4 πr = 4 π( ) = 5 π. = 5(arcsin arcsin) = 5 π.

3 8. SOLUTIONS 7. The strip has width, so the variable of integration is. The length of the strip is for. For positive, we have =. Thus, Area of strip =. Since the region etends from = to = 4, Area of region = 4 4 d = = 6. Check: The area of the region can be computed b Base Height = 8 4 = 6.. The width of the strip is, so the variable of integration is. Since the graphs are = and =, the length of the strip is, and Area of strip ( ). The curves cross at the points (,) and (,), so Area of region = ( )d =. The width of the strip is, so the variable of integration is. The line has equation = 6. The length of the strip is6 ( 4) =. (Since 4 is negative where the graph is below the-ais, subtracting 4 there adds the length below the -ais.) Thus Both graphs cross the -ais where =, so Area of region = Area of strip ( ).. Each slice is a circular disk with radius r = cm. Summing over all disks, we have Taking a limit as, we get Evaluating gives = 6. ( )d = Volume of disk = πr = 4π cm. Total volume 4π cm. Total volume = lim 4π = Total volume = 4π 9 9 = 6π cm. 4πd cm. = 4. Check: The volume of the clinder can also be calculated using the formula V = πr h = π 9 = 6π cm. 4. Each slice is a circular disk. Since the radius of the cone is cm and the length is 6 cm, the radius is one-third of the distance from the verte. Thus, the radius atisr = / cm. See Figure 8.. Summing over all disks, we have Taking a limit as, we get Volume of slice πr = π 9 cm. Total volume π 9 cm. Total volume = lim π 9 = 6 π 9 d cm.

4 74 Chapter Eight /SOLUTIONS Evaluating, we get Total volume = π 9 6 = π 9 6 = 8π cm. Check: The volume of the cone can also be calculated using the formula V = πr h = π 6 = 8π cm. / cm 6 cm Figure Each slice is a circular disk. From Figure 8., we see that the radius at height isr = cm. Thus 5 Volume of disk πr = π( 5 ) = 4 5 π cm. Summing over all disks, we have Taking the limit as, we get Total volume 4π 5 cm. Evaluating gives 4π Total volume = lim 5 = Total volume = 4π = π cm. 4π 5 d cm. Check: The volume of the cone can also be calculated using the formula V = πr h = π 5 = π cm. 5 cm 5 Figure Each slice is a rectangular slab of length m and width that decreases with height. See Figure 8.. At height, the length is given b the Pthagorean Theorem + = 7. Solving gives = 7 m. Thus the width of the slab is = 7 and Volume of slab = Length Width Height = 7 = 7 m.

5 8. SOLUTIONS 75 Summing over all slabs, we have Taking a limit as, we get Total volume 7 m. Total volume = lim 7 = To evaluate, we use the table of integrals or the fact that 7, so Total volume = d m. 7 d represents the area of a quarter circle of radius 7 d = 4 π7 = 45π m. Check: the volume of a half clinder can also be calculated using the formula V = πr h = π7 = 45π m. 7 Figure Each slice is a circular disk. See Figure 8.4. The radius of the sphere is 5 mm, and the radius r at height is given b the Pthagorean Theorem +r = 5. Solving gives r = 5 mm. Thus, Summing over all disks, we have Taking the limit as, we get Evaluating gives Volume of disk πr = π(5 ) mm. Total volume π(5 ) mm. Total volume = lim π(5 ) = Total volume = π ( 5 ) 5 5 π(5 )d mm. = 5 π mm. Check: The volume of a hemisphere can be calculated using the formula V = πr = π5 = 5 π mm. r 5 Figure 8.4

6 76 Chapter Eight /SOLUTIONS 8. Each slice is a square; the side length decreases as we go up the pramid. See Figure 8.5. Since the base of the pramid is equal to its vertical height, the slice at distance from the base, or ( ) from the top, has side ( ). Thus Summing over all slices, we get Evaluating, we find Volume of slice ( ) m. Total volume ( ) m. Total volume = lim ( ) = Total volume = (4 4 + )d = ( 4 + ( ) d m. ) = 8 m. Check: The volume of the pramid can also be calculated using the formula V = b h = = 8 m. ( ) ( ) m m Figure 8.5 Problems 9. Triangle of base and height and. See Figure 8.6. (Either or can be the base. A non-right triangle is also possible.) Figure 8.6. Semicircle of radius r = 9. See Figure Figure 8.7

7 8. SOLUTIONS 77. Quarter circle of radius r = 5. See Figure h h h Figure 8.8 h. Triangle of base and height 7 and 5. See Figure 8.9. (Either 7 or 5 can be the base. A non-right triangle is also possible.) 5 h h 7 Figure 8.9. There are at least two possible answers. Since = when, one possibilit is that the integral gives the area between the parabola = and the line = as shown in Figure 8.. Alternativel, since when, the integral gives the area between the curves = and =, as shown in Figure 8.. Figure 8. Figure (a) The area is shown in Figure 8.(a) with one vertical slice shown. Each slice is bounded on the top b = and on the bottom b =, so the height of each slice is. The width of each slice is, so the area is represented b A = ( )d = = ( ) = 4.5.

8 78 Chapter Eight /SOLUTIONS (a) (b) 9 9 = = / = = Figure 8. (b) The area is shown in Figure 8.(b) with one horizontal slice shown. Each slice is bounded on the right b =, which is = distance out from the -ais, and is bounded on the left b =, which is = / distance out from the -ais. The length of each slice is /. The width of each slice is, and the area etends from = to = 9 so the area is represented b 9 ( A = )d = / 9 6 = 9/ 6 9 ( ) = 4.5. Notice that the two was of computing the area give the same answer, as we epect. 5. (a) The area is shown in Figure 8.(a) with one vertical slice shown. Each slice is bounded on the top b = and on the bottom b =, so the height of each slice is ( ). The width of each slice is and the area ranges from = to = 4, so the area is represented b A = 4 (( ) )d = 4 d = 4 = 4 4 ( ) = 4. (a) = (b) = 8 8 = 4 = / 4 Figure 8. (b) The area is shown in Figure 8.(b) with two horizontal slices shown. Because some of the slices are bounded on the right b = while others are bounded on the right b =, we use two separate integrals to calculate this area using horizontal slices. The slices between = and = 8 are bounded on the right b =, which is = / distance from the -ais, and are bounded on the left b the -ais, which is =. This part of the area is given b 8 Bottom part of the area = d = 8 4 = 4 8 = 6. The slices between = 8 and = are bounded on the right b =, which is = distance from the -ais, and are bounded on the left b the -ais, which is =. This part of the area is given b Top part of the area = ( )d = = ( 8 ) 8 =

9 8. SOLUTIONS 79 The total area is given b Total area = 8 d + ( )d = 6+8 = 4. Notice that the two was of computing the area give the same answer, as we epect. 6. (a) The area is shown in Figure 8.4(a) with two vertical slices shown. Because some of the slices are bounded on the top b = while others are bounded on the top b = 6, we use two separate integrals to calculate this area using vertical slices. The slices between = and = are bounded on the top b = and are bounded on the bottom b the -ais, which is =. This part of the area is given b Left part of the area = 8 d = = =.667. The slices between = and = 6 are bounded on the top b = 6 and are bounded on the bottom b the -ais, which is =. This part of the area is given b The total area is given b Right part of the area = Total area = d+ 6 (6 )d = 6 6 = (6 ) = 8. 6 (6 )d = =.667. (a) 4 = (b) 4 = = 6 = Figure 8.4 (b) The area is shown in Figure 8.4(b) with one horizontal slice shown. The slices are all bounded on the right b = 6, which is = 6 distance from the -ais, and are bounded on the left b =, which is = distance from the -ais. The area ranges from = to = 4 so the area is given b A = 4 ((6 ) )d = 6 / 4 = / =.667. Notice that the two was of computing the area give the same answer, as we epect. 7. (a) The area is shown in Figure 8.5(a) with two vertical slices shown. Because some of the slices are bounded on the top b = while others are bounded on the top b = 6, we need to use two separate integrals to calculate this area using vertical slices. The slices between = and = are bounded on the top b = and are bounded on the bottom b the -ais, which is =. This part of the area is given b Left part of the area = d = = = 9.

10 7 Chapter Eight /SOLUTIONS The slices between = and = 5 are bounded on the top b = 6 and are bounded on the bottom b the -ais, which is =. This part of the area is given b 5 5 Right part of the area = 6d = 6 = =. The total area is given b Total area = d+ 5 6d = 9+ =. (a) 6 = 6 = (b) 6 = 6 = / = 5 = Figure 8.5 (b) The area is shown in Figure 8.5(b) with one horizontal slice shown. The slices are all bounded on the right b = 5 and are bounded on the left b =, which is = / distance from the -ais. The area ranges from = to = 6 so the area is given b A = 6 (5 (/))d = = =. Notice that the two was of computing the area give the same answer, as we epect. 8. Hemisphere with radius. See Figure 8.6. h 44 h h Figure Cone with height and radius / = 4. See Figure Figure 8.7

11 8. SOLUTIONS 7. Cone with height 6 and radius. See Figure / Figure 8.8. Hemisphere with radius. See Figure 8.9. ( ) ( ) Figure 8.9. Radius = r We slice up the sphere in planes perpendicular to the -ais. Each slice is a circle, with radius = r ; that s the radius because + = r when z =. Then the volume is V π( ) = π(r ). Therefore, as tends to zero, we get V = =r = ( = π(r )d = r =r = π(r )d ) r πr π = 4πr.. We slice the cone horizontall into clindrical disks with radius r and thickness h. See Figure 8.. The volume of each disk isπr h. We use the similar triangles in Figure 8. to writer as a function of h: r h = so r = 4 h. The volume of the disk at height h is π( 4 h) h. To find the total volume, we integrate this quantit from h = to h =. ( ) V = π 4 h dh = π h 6 = 6π =.97 m.

12 7 Chapter Eight /SOLUTIONS 6 m m h h m r h Figure 8. Figure Radius = r(h ) h This cone is what ou get when ou rotate the line = r(h )/h about the ais. So slicing perpendicular to the ais ields =h h ( ) (h )r V = π d = π d h = h = π r (h h + )d h [ = πr h h + h ] h = πr h. 5. (a) A vertical slice has a triangular shape and thickness. See Figure 8.. Volume of slice = Area of triangle = Base Height = = cm. Thus, Total volume = lim = 4 d = 4 = cm. cm cm cm cm Figure 8. (b) A horizontal slice has a rectangular shape and thickness h. See Figure 8.. Using similar triangles, we see that w = h, so w = ( h) = h.

13 8. SOLUTIONS 7 Thus So, Volume of slice 4w h = 4 ( ) h h = (8 8 ) h h. Total volume = lim (8 8 ) h h h = (8 8 ) ( ) h dh = 8h 4h = cm. h h cm 4 cm cm h Figure 8. w 4 cm cm h w 6. We slice the water into horizontal slices, each of which is a rectangle. See Figure 8.4. Volume of slice 5w h km. To findw in terms of h, we use the similar triangles in Figure 8.5: So Summing over all slices and letting h gives Evaluating the integral gives w = h. so w = 5h. Volume of slice 5 5h h = 5h h km. Total volume = lim 5h h = h Total volume = 5 h.. = 45 km. 5hdh km. w 5 km km. km km w h h Figure 8.4 Figure 8.5

14 74 Chapter Eight /SOLUTIONS 7. To calculate the volume of material, we slice the dam horizontall. See Figure 8.6. The slices are rectangular, so Volume of slice 4w h m. Since w is a linear function of h, and w = 6 when h =, and w = when h = 5, this function has slope = ( 6)/5 =. Thus w = 6 hmeters, so Volume of slice 4(6 h) h m. Summing over all slices and taking the limit as h gives Evaluating the integral gives Total volume = lim 4(6 h) h = h Total volume = 4 m ( 6h h 5 4(6 h)dh m. ) 5 = m. m w w 6 m 4 m 5 m 6 m h h Figure 8.6 Figure 8.7 Strengthen Your Understanding 8. The horizontal slice at height goes from = to a point on the line =. At this point, = /. The correct integral for the area is 8 /d. 9. Slice the sphere + = into slices of thickness perpendicular to the-ais. The tpical slice is approimatel a clinder of radius = and heighth =. The volume of the slice is approimatelπ h = π( ). The volume of the sphere is thus π( ) d. 4. It would be hard to use vertical slices to find the area of the shaded region in Figure 8.8 because the vertical slices on the right portion of the figure are in two pieces. Figure 8.8

15 8. SOLUTIONS One possible answer is the region between the positive -ais, the positive -ais and the line =. For horizontal slices, the width of a slice at height is =, and for vertical slices the height of a slice at positionis =. 4. True. Since = ± 9 represent the top and bottom halves of the sphere, slicing disks perpendicular to the -ais gives Volume of slice π = π(9 ) Volume = π(9 )d. 4. False. Evaluating does not give the volume of a cone πr h/: h ( ) h ( ) π(r )d = π r = π rh h. Alternativel, ou can show b slicing that the integral representing this volume is h πr ( /h) d. 44. False. Using the table of integrals (VI-8 and VI-) or a trigonometric substitution gives r π r d = π ( ( r +r arcsin r)) r = πr (arcsin arcsin) = π r 4. The volume of a hemisphere is πr /. Alternativel, ou can show b slicing that the integral representing this volume is r π(r )d. 45. True. Horizontal slicing gives rectangular slabs of length l, thickness, and widthw = r. So the volume of one slab is l r, and the integral is r r l r d. Solutions for Section 8. Eercises. (a) The volume of a disk is given b so (b) We have. (a) The volume of a disk is given b so (b) We have Volume = 6 V π() = 4π, Volume = Volume = 4π 4π d. = 6π. V π( +6), Volume = 6 π( 4 +6 )d = π π( +6) d. ( ) 6 = 96π. 5

16 76 Chapter Eight /SOLUTIONS. (a) The strip stretches from = / to =. The volume of a disk with a hole in it is ( ( ) V π = ) π 4 (6 ), so (b) We have Volume = Volume = π 4 4. (a) The equation = +6 can be solved for as 6+ = 6 ( π 4 (6 )d. 6 = 6± 6 4 ) 6 = 6π. = ± 9. The left end of the strip is given b = 9, and the right end is given b = + 9. Thus, the volume of a disk with a hole is V π((+ 9 ) ( 9 ) ) = π(( ) ( )) = π 9, so (b) We have 5. The volume is given b 6. The volume is given b V = V = Volume = π 9 Volume = π(9 )/ / π d = π d = 9 d. 9 π 4 d = π 5 5 = 6π. π(+) 4 d = π(+)5 5 = π 5. = π The volume is given b V = 8. The volume is given b 9. The volume is given b π(4 ) d = π V = V = π( +) d = π π d = ( )d = π π(e ) d = ( ) ( ) (+)d = π + = π. πe d = π e = π (e e ). = 56π 5.

17 . The volume is given b Integration b parts gives. The volume is given b V =. The volume is given b V = π V = π/ ( ) π d = π + π d = π/ V = π (cossin+) π/ ( cosh) d = π πcos d. = π 4. d (+) = π(+) ( = π ) = π. coshd = π sinh = π sinh.. Since the graph of = is below the graph of = for, the volume is given b ( ) V = π d π( ) d = π ( 4 )d = π SOLUTIONS 77 = π Since the graph of = e is above the graph of = e for, the volume is given b ( ) V = π(e ) d π(e ) d = π(e 6 e e 6 ( ) )d = π 6 e e 6 = π 6 e Since f () =, we evaluate the integral numericall or using the table to get Arc length = + d = ln( 5+) 6. Since f () = sin, we evaluate the integral numericall to get Arc length = +sin d = = Since f () = /(+), we evaluate the integral numericall to get ( ) Arc length = + d = Note that this function is actuall / in disguise. So L = = 4 9 w= w= = 8 7 w where we set w = + 9 4, sod = 4 9 dw. +[ ] d = w dw = = d = 8 ( ( ) ).56, 7 9. This is a one-quarter of the circumference of a circle of radius. That circumference is π = 4π, so the length is 4π 4 = π.

18 78 Chapter Eight /SOLUTIONS. Since f () = sinh, the arc length is given b. The length is L = +sinh d = cosh d = coshd = sinh = sinh. ( (t)) +( (t)) +(z (t)) dt = ( ) dt = 4. This is the length of a straight line from the point (8,5,) to(,9,).. We have D = = = e. ( et sin(e t )) +(e t cos(e t )) dt e t dt = e t dt This is the length of the arc of a unit circle from the point (cos,sin) to (cose,sine) in other words between the angles θ = andθ = e. The length of this arc is (e ).. We have D = π ( sint) +(5cos5t) dt. We cannot find this integral smbolicall, but numerical methods show D Since d/dt = cos tsint, d/dt = sin tcost, we have π Arc length = 9cos 4 tsin t+9sin 4 tcos tdt = = π π costsint cos t+sin tdt costsint dt. We calculate the integral over the interval t π/, where both cost andsint are positive, and multipl b 4: π/ π/ Arc length = costsintdt = 6sin t = 6. Problems 5. The two functions intersect at (,) and (8,). We slice the volume with planes perpendicular to the line = 9. This divides the solid into thin washers with volume Volume of slice = πr out πr in. The outer radius is the horizontal distance from the line = 9 to the curve =, sor out = 9. Similarl, the inner radius is the horizontal distance from the line = 9 to the curve = 4, so r in = 9 4. Integrating from = to = we have V = [π(9 ) π(9 4) ]d. 6. The two functions intersect at (,) and (8,). We slice the volume with planes perpendicular to the line =. This divides the solid into thin washers with Volume of slice = πr out πr in. The outer radius is the vertical distance from the line = to the curve =, 4 sorout =. Similarl, the inner 4 radius is the vertical distance from the line = to the curve =, so r in =. Integrating from = to = 8 we have 8 V = [π( 4 ) π( ] ) d.

19 8. SOLUTIONS Note that the lines = 9 and = intersect at (9,). We slice the volume with planes that are perpendicular to the line =. This divides the solid into thin washers with Volume of slice = πr out πr in. Note that the inner radius is the vertical distance from the line = to the -ais, so r in =. Similarl, the outer radius is the vertical distance from the line = to the line =, so rout = +. Integrating from = to = 9 we have 9 V = [π(+ ] ) π d. 8. Note that the lines = 9 and = intersect at (9,). We slice the volume with planes that are perpendicular to the line =. This divides the solid into thin washers with Volume of slice = πr out πr in. Note that the inner radius is the horizontal distance from the line = to the line =, sor in = +. Similarl, the outer radius is the horizontal distance from the line = to the line = 9, so r out = + 9. Integrating from = to = we have V = [π(+9) π(+) ]d. 9. The two functions intersect at(, ) and(5, 5). We slice the volume with planes perpendicular to the -ais. This divides the solid into thin washers with volume Volume of slice = π((r out) (r in) ). The outer radius is the vertical distance from the -ais to the curve = 5, so r out = 5. Similarl, the inner radius is the vertical distance from the -ais to the curve =, sor in =. Integrating from = to = 5 we have V = 5 π((5) ( ) )d.. The two functions intersect at(, ) and(5, 5). We slice the volume with planes perpendicular to the -ais. This divides the solid into thin washers with volume Volume of slice = π((r out) (r in) ). The outer radius is the horizontal distance from the-ais to the curve =, sor out =. Similarl, the inner radius is the horizontal distance from the -ais to the curve = /5, sor in = /5. Integrating from = to = 5 we have V = 5 π(( ) (/5) )d.. The two functions intersect at (, ) and (5, 5). We slice the volume with planes perpendicular to the horizontal line = 4. This divides the solid into thin washers with volume Volume of slice = π((r out) (r in) ). The outer radius is the vertical distance from the line = 4 to the curve = 5, sor out = 4+5. Similarl, the inner radius is the vertical distance from the line = 4 to the curve =, so r in = 4 +. Integrating from = to = 5 we have V = 5 π((4+5) (4+ ) )d.. The two functions intersect at (,) and (5,5). We slice the volume with planes perpendicular to the vertical line =. This divides the solid into thin washers with volume Volume of slice = π((r out) (r in) ). The outer radius is the horizontal distance from the line = to the curve =, so r out = +. Similarl, the inner radius is the horizontal distance from the line = to the curve = /5, so r in = +/5. Integrating from = to = 5 we have V = 5 π((+ ) (+/5) )d.

20 7 Chapter Eight /SOLUTIONS. One arch of the sine curve lies between = and = π. Sinced(sin)/d = cos, evaluating the integral numericall gives π Arc length = +cos d = The curves cross at (,); see Figure 8.9. The straight side has length =.44; as this is the hpotenuse of a right triangle with sides,,. The curved side has Arc length = +() d = ln( 5+) =.479. Thus the perimeter is =.89. = (,) = Figure (a) Slicing the region perpendicular to the -ais gives disks of radius. See Figure 8.. Volume of slice π = π( ). Thus, Total volume = lim π( ) = ( )) 8 = π ( 9 = = 6π. π( )d = π ( ) Figure 8. (b) The arc length, L, of the curve = f() is given b L = b a +(f ()) d. In this problem is an implicit function of. Solving forgives = as the equation of the top half of the hperbola. Differentiating gives Thus Arc length = d d = ( ) / () =. ( + d = ) + d = d =.48.

21 6. Radius= b a a z = b ( a ). V = a π d = π a ( = πb a a a b ( a ) d a ) d = πb [ a ] a = πb ( a a a ) = πb ( a a ) 8. SOLUTIONS 7 = 4 πab. 7. Radius= + ( = ) We slice the region perpendicular to the ais. The Riemann sum we get is π( +). So the volume V is the integral V = π( +) d = π ( 6 + +)d ( ) 7 = π = (6/7)π z We slice the region perpendicular to the ais. The Riemann sum we get is π( ) = π( ). So the volume V is the integral Radius= ( = ) V = = π = π π( ) d ( + 4 )d ( ) = (8/5)π z We take slices perpendicular to the ais. The Riemann sum for approimating the volume is πsin. The volume is the integral corresponding to that sum, namel Radius= sin V = π πsin d = π [ sincos+ ] π = π Slice the object into disks horizontall, as in Figure 8.. A tpical disk has thickness and radius =. Thus Volume of slice π = π. Volume of solid = lim π = πd = π = π.

22 7 Chapter Eight /SOLUTIONS = Figure Slice the object into rings verticall, as is Figure 8.. A tpical ring has thickness and outer radius = and inner radius =. Volume of slice π π = π( 4 ). Volume of solid = lim π( 4 ) = π( 4 )d = π ( 5 5 ) = 4 5 π. = Outer radius Inner radius Figure 8.: Cross-section of solid 4. The region is clindrical with a hole around the ais of rotation, =. Slice it into rings verticall, as in Figure 8.. A tpical ring has thickness and outer radius + = and inner radius + = +. Thus Volume of slice π π( +) = π(5 4 4 ). Volume of solid = ( π(5 4 4 ) = π ) = 5π 5. = Ais of rotation: = 4 Inner radius 5 Outer radius Figure 8.: Cross-section of solid

23 8. SOLUTIONS 7 4. Slicing perpendicularl to the -ais gives squares whose thickness is and whose side is =. See Figure 8.4. Thus Volume of square slice ( ) = ( + 4 ). Volume of solid = ( + 4 )d = = 8 5. = Base of square (standing on paper) Figure 8.4: Base of solid 44. Slicing perpendicularl to the-ais gives semicircles whose thickness is and whose diameter is =. See Figure 8.5. Thus ( ) Volume of semicircular slice π = π 4 ( + 4 ). Volume of solid = ( ) π 4 ( + 4 )d = π = π = π 5. = Diameter of semicircle (standing on paper) Figure 8.5: Base of solid 45. An equilateral triangle of side s has height s/ and Area = s s = 4 s. Slicing perpendicularl to the -ais gives equilateral triangles whose thickness is and whose side is =. See Figure 8.6. Thus Volume of triangular slice 4 ( ) = 4.

24 74 Chapter Eight /SOLUTIONS Volume of solid = 4 d = 4 = 8. = Base of triangle (standing on paper) Figure 8.6: Base of solid 46. r = e z This is the volume of revolution gotten from the rotating the curve = e. Take slices perpendicular to the -ais. The will be circles with radius e, so V = = = = πe π d = π = π(e ) e d We slice the volume with planes perpendicular to the line =. This divides the curve into thin washers, as in Eample on page 44 of the tet, whose volumes are r out = e r in = ( = ) πr outd πr ind = π(+) d π d. So the integral we get from adding all these washers up is 48. r in = 7 e r out = 7 z = V = [π(+) π ]d = = π [(+e ) 9]d = π [e +6e ]d = π[ e +6e ] = π[(e /+6e) (/+6)] 4.4. This problem can be done b slicing the volume into washers with planes perpendicular to the ais of rotation, = 7, just like in Eample. This time the outside radius of a washer is 7, and the inside radius is7 e. Therefore, the volume V is = V = [π7 π(7 e ) ]d = π (4e e )d ( = 7) = = π [4e ] [ e = π 4e ( e 4 )]

25 8. SOLUTIONS We now slice perpendicular to the -ais. As stated in the problem, the cross-sections obtained thereb will be squares, with base length e. The volume of one square slice is (e ) d. (See Figure 8.7.) Adding the volumes of the slices ields Volume = = = d = e d = e = e =.95. z e e Figure z We slice perpendicular to the -ais. As stated in the problem, the cross-sections obtained thereb will be semicircles, with radius e. The volume of one semicircular slice is π( ) e d. (Look at the picture.) Adding up the volumes of the slices ields r = e Volume = = = = πe 6 π d = π 8 = π(e ) 6 e d If we revolve the region g() around the-ais, the volume of the resulting solid is given b Taking the hint, we can write our integral in this form: π π(4 4cos )d = π Therefore, g() = sin or g() = sin. 4π ( cos ) d = π 4πsin d = π π π(sin) d. }{{} (g()) π(g()) d. 5. At time t = 6, the particle has traveled 6 = 8 cm. Suppose it is then at the point = b. Then the arc length from the origin to this point is 8 cm. Since ( ) d d / = /, we have Arc length = b b +( / ) d = +d = 8

26 76 Chapter Eight /SOLUTIONS (+)/ b = 8 (+b)/ = 8 (+b) / = 8 b = 8 / = 8.. When = 8., = We want to approimate A(h)dh, where h is height, and A(h) represents the cross-sectional area of the trunk at height h. Since A = πr (circular cross-sections), and c = πr, where c is the circumference, we have A = πr = π[c/(π)] = c /(4π). We make a table of A(h) based on this: Table 8. height (feet) Area (square feet) We now form left & right sums using the chart: LEFT(6) = = RIGHT(6) = = 7. So TRAP(6) = RIGHT(6)+LEFT(6) = = 67. cubic feet. 54. If = e /, then = ln. (Note that since <, ln.) A tpical slice has thickness and radius. See Figure 8.8. So Volume of slice = π = πln. Thus, Total volume = π Since ln is not defined at =, this is an improper integral: Total Volume = π lnd = π lim a = π lim(ln ) a a lnd. a lnd = π lim( alna+a). a B looking at the graph of ln on a calculator, we see that lim alna =. Thus, a Total volume = π( ) = π. Figure 8.8

27 8. SOLUTIONS (a) We can begin b slicing the pie into horizontal slabs of thickness h located at height h. To find the radius of each slice, we note that radius increases linearl with height. Since r = 4.5 when h = and r =.5 when h =, we should have r =.5 +h/. Then the volume of each slab will be πr h = π(.5 +h/) h. To find the total volume of the pie, we integrate this from h = toh = : V = π (.5+ h ) dh [ ] h = π 7 + 7h h 4 [ = π 7 + 7( ) + 49() ] 5 in. 6 4 (b) We use.5 in as a rough estimate of the radius of an apple. This gives us a volume of (4/)π(.5) in. Since 5/ 5, we would need about 5 apples to make a pie. 56. (a) The volume can be computed b several methods, not all of them requiring integration. We will slice horizontall, forming rectangular slabs of length cm, height, widthwand integrate. See Figure 8.9. w Figure 8.9 cm 5 cm Figure 8.4 d 6 6 so Thus From the right triangle, we see d = tan6 = d =. w = 5+d = 5+. The volume of the slab is V w = ( 5+ ), so the total volume is given b Volume = lim V = lim h ( ) = 5+ d = ( ( 5+ ) 5 + ) h ( ) = 5h+ h cm. (b) The maimum value of h is h = 5sin6 = 5 / cm 4. cm. (c) The maimum volume of water that the gutter can hold is given b substituting h = 5 / into the volume: ( ( Maimum volume = 5 5 ) ) 5 / + = 5 4 ( + ) = cm. (d) Because the gutter is narrower at the bottom than the top, if it is filled with half the maimum possible volume of water, the gutter will be filled to a depth of more than half of 4. cm.

28 78 Chapter Eight /SOLUTIONS (e) We want to solve for the value of h such that ( Volume = 5h+ h ) = 875 = Vma 5h+ h = Solving gives h =.5 and h =.8. Since onl positive values of h are meaningful, h =.5 cm. H = a We divide the interior of the boat into flat slabs of thickness and width = /a. (See above.) We have Volume of slab L = L a. We are interested in the total volume of the region H, so If L andh are in meters, Total volume = lim L = L a H ( a) (/) = H (/) d = 4LH(/). a Buoanc force = 4,LH(/) a ( ) (/)d L a newtons. 58. We can find the volume of the tree b slicing it into a series of thin horizontal clinders of height dh and circumference C. The volume of each clindrical disk will then be ( ) C V = πr dh = π dh = C dh π 4π. Summing all such clinders, we have the total volume of the tree as Total volume = 4π C dh. We can estimate this volume using a trapezoidal approimation to the integral with h = : LEFT estimate = 4π [( )] = 4π (566). RIGHT estimate = 4π [( )] = 4π (45). TRAP = (449) 59 cubic inches. 4π

29 8. SOLUTIONS (a) The volume, V, contained in the bowl when the surface has height h is However, since = 4, we have = so that V = V = h h π d. π d = πh/. Differentiating gives dv/dh = πh / = π h. We are given that dv/dt = 6 h, where the negative sign reflects the fact that V is decreasing. Using the chain rule we have dh dt = dh dv dv dt = dv/dh dv dt = π h ( 6 h) = 6 π. Thus, dh/dt = 6/π, a constant. (b) Since dh/dt = 6/π we know that h = 6t/π +C. However, when t =, h =, therefore h = 6t/π. The bowl is empt when h =, that is when t = π/6 units. 6. The problem appears complicated, because we are now working in three dimensions. However, if we take one dimension at a time, we will see that the solution is not too difficult. For eample, let s just work at a constant depth, sa. We appl the trapezoid rule to find the approimate area along the length of the boat. For eample, b the trapezoid rule the approimate area at depthfrom the front of the boat to feet toward the back is (+8) = 5. Overall, at depthwe have that the area for each length span is as follows: Table 8. length span: depth We can fill in the whole chart the same wa: Table 8. length span: depth Now, to find the volume, we just appl the trapezoid rule to the depths and areas. For eample, according to the trapezoid rule the approimate volume as the depth goes from toand the length goes from to is (5+5) = 75. Again, we fill in a chart: Table 8.4 length span: depth span Adding all this up, we find the volume is approimatel 595 cubic feet. You might wonder what would have happened if we had done our trapezoids along the depth ais first instead of along the length ais. If ou tr this, ou d find that ou come up with the same answers in the volume chart! For the trapezoid rule, it does not matter which ais ou choose first.

30 74 Chapter Eight /SOLUTIONS 6. (a) The equation of a circle of radiusr around the origin is + = r. This means that = r, so(d/d) =, and d/d = /. Since the circle is smmetric about both aes, its arc length is 4 times the arc length in the first quadrant, namel r ( ) d r ( 4 + d = 4 + d. d ) (b) Evaluating this integral ields ( 4 r + This is the epected answer. ) d = 4 = 4r r r r + r d = 4 r r d r d = 4r(arcsin(/r)) r = πr. 6. As can be seen in Figure 8.4, the region has three straight sides and one curved one. The lengths of the straight sides are,, ande. The curved side is given b the equation = f() = e. We can find its length b the formula +f () d = +(e ) d = +e d. Evaluating the integral numericall gives.5. The total length, therefore, is about + + e f() = e Figure The graph of f is a downward-opening parabola with -intercepts at =,4, and f () = ( ( 4)) = 4. The arc length of the graph of f from = to = 4 is given b 4 Arc length = +(f ()) d 4 = +(4 ) d. because f () = 4 This integral can be simplified as We would like to write the integrand in the form This suggests: ( f () ) = / 4 6+7d. + = +(f ()). f () = /4 alternativelf () = /4 f() = 4 5 5/4 +C =.8 5/4 lettingc = for convenience.

31 8. SOLUTIONS 74 Thus, 4 + d = 4 + ( (.8 5/4 ) ) d, which is the arc length of the curve =.8 5/4 from = to = 4. Other possible solutions are an of the curves = ±.8 5/4 +C. 65. The graph of f is concave down where f < : This means that f () < where f() = e. f () = e f () = e +4 e ( = e ). < <. Thus, the graph is concave down for / < < /. The arc length of this portion of the graph of f is given b / Arc length = +(f ()) d 66. The graph of f is concave down where f () < : / / = / / = / + ( e ) d +4 e d. f() = f () = f () = 48+6 = ( )( ). Hence f () < for < <. The arc length of this portion of the graph of f is given b Arc length = +(f ()) d = +( ) d. 67. The arc length of the catenar between = b and = b is meters. Since we have Thus Arc length = = b b b b d d (cosh) = sinh = (e e ), + 4 (e e ) d = b b (e +e ) d = b b 4+e +e d (e +e )d = (e e ) e b e b =. b b = e b e b. Solving numericall gives b =. meters. Thus, the ends of the chain are (.) = 4.64 meters apart.

32 74 Chapter Eight /SOLUTIONS 68. Here are man functions which work. An linear function = m+b works. This follows because d d b ) b d = a + ( d d The function = works : d d = ( / ), and + ( d ) d = + d 4 = [ 4 = a = m is constant for such functions. So +m d = (b a) +m. ( ) d = 4 ( + ) d = 8 4 ] +C d 6 ( + ) d One more function that works is = ln(cos); we have d = sin/cos. Hence d + ( ) d d. = + d where the last integral comes from IV- of the integral tables. ( ) sin d = + sin cos cos d sin +cos = d = cos cos d = cos d = ln sin+ +C, sin 69. (a) If f() = g (t) dt, then, b the Fundamental Theorem of Calculus, f () = g (). So the arc length of f from tois +(f (t)) dt = +( g (t) ) dt = = +g (t) dt g (t)dt = g() g() = g(). (b) If g is the arc length of an function f, then b the Fundamental Theorem of Calculus, g () = +f (). So ifg () <,g cannot be the arc length of a function. (c) We find a functionf whose arc length from tois g() =. Using part (a), we see that f() = (g (t)) dt = dt =. This is the equation of a line. Does it make sense to ou that the arc length of a line segment depends linearl on its right endpoint? 7. (a) For n =, we have In the first quadrant, the equation is the line + =. + =. B smmetr, the graph in the other quadrants gives the square in Figure 8.4. Forn =, the equation is of a circle of radius, centered at the origin: For n = 4, the equation is + = =. The graph is similar to a circle, but bulging out more. See Figure 8.4.

33 8. SOLUTIONS 74 (b) For n =, the arc length is the perimeter of the square. Each side is the hpotenuse of a right triangle of sides,,. Thus Arc length = 4 = For n =, the arc length is the perimeter of the circle of radius. Thus For n = 4, we find the arc length using the formula Arc length = π = π = 6.8. L = b a +(f ()) d. We find the arc length of the top half of the curve, given b = ( 4 ) /4, and double it. Since Arc length = d d = 4 ( 4 ) /4 ( 4 ) = ( ) + d = ( 4 ) /4 ( 4 ) /4, + 6 d. ( 4 ) / The integral is improper because the integral is not defined at = ±. Using numerical methods, we find Arc length = + 6 d = 7.8. ( 4 ) / = + = + = Figure 8.4 Strengthen Your Understanding 7. The volume is 5( π() π() ) d, since the slices are disks-with-holes. 7. The arc length is given b π/4 +(f ()) d with f() = sin. Thus, the correct formula for the arc length is π/4 +cos d. 7. The curve begins at the point (,) and ends at the point (,). The distance between these points is + >. Since the arc length of an curve is greater than or equal to the length of the straight line connecting the points, the arc length of = 5 between = and = must be greater than. 74. The triangular region with vertices (, ),(, ) and (, ) gives the eact same solid shape whether rotated about the - or -ais. 75. One eample is the region bounded b = and the-ais for. When the region is rotated about the-ais, we get a cone of radius and height. When it is rotated about the -ais, we get a cone of radius and height. Since Volume around -ais = π = 4π Volume around -ais = π = π, the volume is greater around the -ais than around the -ais.

34 744 Chapter Eight /SOLUTIONS 76. The circle of radius 5 centered at(5,) goes through both the points(,) and(,). The upper semicircle and the lower semicircle are different curves between these two points that have the same arc length, namel half the circumference of the circle. 77. The distance from(,) to(,) is, so an curve other than a straight line between the points has an arc length greater than. One possible eample isf() =. 78. False. The volume also depends on how far awa the region is from the ais of revolution. For eample, let R be the rectangle 8, and let S be the rectangle,. Then rectangle R has area greater than rectangle S. However, when ou revolve R about the -ais ou get a clinder, ling on its side, of radius and length 8, which has volume 8π. When ou revolve S about the -ais, ou get a clinder of radius and length, which has volume π. Thus the second volume is larger, even though the region revolved has smaller area. 79. False. Suppose that the graph of f starts at the point (,) and then goes down to (,) and from there on goes along the -ais. For eample, if f() = ( ) on the interval [,] and f() = on the interval [,], then f is differentiable on the interval [,]. The arc length of the graph of f on the interval [,] is at least, while the arc length on the interval [,] is9. 8. True. Since f is concave up, f is an increasing function, so f () f () = /4 on the interval [,4]. Thus +(f ()) +9/6 = 5/4. Then we have: Arc length = 4 +(f ()) d 4 5 d = False. Since f is concave down, this means that f () is decreasing, so f () f () = /4 on the interval [,4]. However, it could be that f () becomes negative so that (f ()) becomes large, making the integral for the arc length large also. For eample, f() = (/4) is concave down and f () = /4, but f() = and f(4) =, so the graph of f on the interval [,4] has arc length at least. Solutions for Section 8. Eercises. Withr = andθ = π/, we find = rcosθ = cos(π/) = / and = rsinθ = sin(π/) = /. The rectangular coordinates are( /, /).. With r = and θ = π/4, we find = rcosθ = cos( π/4) = ( /) = 6/ and = rsinθ = sin( π/4) = ( /) = 6/. The rectangular coordinates are( 6/, 6/).. With r = and θ = π/6, we find = rcosθ = cos( π/6) = / = and = rsinθ = sin( π/6) = ( /) =. The rectangular coordinates are(, ). 4. Withr = andθ = 5π/6, we find = rcosθ = cos(5π/6) = ( /) = and = rsinθ = sin(5π/6) = (/) =. The rectangular coordinates are(,). 5. With = and =, find r from r = + = + =. Find θ from tanθ = / = / =. Thus, θ = tan () = π/4. Since(,) is in the first quadrant this is a correct θ. The polar coordinates are (,π/4). 6. With = and =, find r = + = ( ) + =. Find θ from tanθ = / = /( ) =. Thus, θ = tan () =. Since (,) is on the -ais between the second and third quadrant, θ = π. The polar coordinates are (,π). 7. With = 6 and =, find r = ( 6) +( ) = 8 =. Find θ from tanθ = / = / 6 = /. Thus,θ = tan ( / ) = π/6. Since( 6, ) is in the fourth quadrant, this is the correctθ. The polar coordinates are(, π/6). 8. With = and =, find r = ( ) + = 4 =. Find θ from tanθ = / = /( ). Thus, θ = tan ( / ) = π/6. Since(,) is in the second quadrant,θ = π/6+π = 5π/6. The polar coordinates are (,5π/6).

35 8. SOLUTIONS (a) Table 8.5 contains values of r = sin θ, both eact and rounded to one decimal. Table 8.5 θ π/ π/ π/ π 4π/ π/ 5π/ π 7π/ 5π/ 8π/ r / / + / + / / / r (b) See Figure 8.4. Figure 8.4 Figure 8.44 (c) The circle has equation r = /. The cardioid is r = sinθ. Solving these two simultaneousl gives or / = sinθ, sinθ = /. Thus,θ = π/6 or5π/6. This gives the points(,) = ((/)cosπ/6,(/)sinπ/6) = ( /4,/4) and(,) = ((/)cos5π/6,(/)sin5π/6) = ( /4,/4) as the location of intersection. (d) The curve r = sinθ, pictured in Figure 8.44, has two regions instead of the one region that r = sinθ has. This is because sinθ will be twice for ever π ccle in θ, as opposed to once for ever π ccle in θ for sinθ.. There will be n loops. See Figures Figure 8.45: n = Figure 8.46: n = Figure 8.47: n = Figure 8.48: n = 4. The graph will begin to draw over itself for anθ π so the graph will look the same in all three cases. See Figure Figure 8.49

36 746 Chapter Eight /SOLUTIONS. The curve will be a smaller loop inside a larger loop with an intersection point at the origin. Larger n values increase the size of the loops. See Figures Figure 8.5:n = 4 Figure 8.5: n = 4 5 Figure 8.5: n = 4. See Figures 8.5 and The first curve will be similar to the second curve, ecept the cardioid (heart) will be rotated clockwise b9 (π/ radians). This makes sense because of the identitsinθ = cos(θ π/). Figure 8.5:r = cosθ Figure 8.54: r = sinθ 4. Let θ π and/6 r /. 5. A loop starts and ends at the origin, that is, when r =. This happens first when θ = π/4 and net when θ = 5π/4. This can also be seen b using a trace mode on a calculator. Thus restricting θ so that π/4 θ 5π/4 will graph the upper loop onl. See Figure To show onl the other loop use θ π/4 and 5π/4 θ π. See Figure Figure 8.55: π/4 θ 5π/4 Figure 8.56: θ π/4 and 5π/4 θ π 6. (a) Let θ π/4 and r. (b) Break the region into two pieces: one with / and, the other with / and. 7. The region is given b 8 r 8 andπ/4 θ π/.

37 8. The region is given b r and π/6 θ π/6. 8. SOLUTIONS The circular arc has equation r =, for θ π/. the vertical line = has polar equation rcosθ =, or r = /cosθ. So the region is described b θ π/ and r /cosθ.. Epressing and in terms of θ, we have = cosθ and = sinθ. The slope is given b At θ = π/4, we have d d = cosθ sinθ = cosθ sinθ. d d θ=π/4 = / / =.. Epressing and in terms of θ, we have = e θ cosθ and = e θ sinθ. The slope is given b d d = eθ sinθ +e θ cosθ sinθ +cosθ = e θ cosθ e θ sinθ cosθ sinθ. At θ = π/, we have sinθ = andcosθ =, so d = + d =. θ=π/. Epressing and in terms of θ, we have = ( cosθ)cosθ and = ( cosθ)sinθ. The slope is given b At θ = π/, we have cosθ = and sinθ =, so that d d d ( cosθ)cosθ sinθsinθ = d ( cosθ)sinθ sinθcosθ. θ=π/ =.. The curve is given parametricall b Thus, calculating d/dθ and d/dθ, gives = e θ cosθ and = e θ sinθ. Arc length = = = π π/ π (eθ cosθ e θ sinθ) +(e θ sinθ +e θ cosθ) dθ e θ (cosθ sinθ) +(sinθ +cosθ) dθ π/ π e θ dθ π/ = (e π e π/ ).

38 748 Chapter Eight /SOLUTIONS 4. The curve is given parametricall b Thus, calculating d/dθ and d/dθ, gives Arc length = = = π π π = θ cosθ and = θ sinθ. (θcosθ θ sinθ) +(θsinθ +θ cosθ) dθ 4θ +θ 4 dθ θ 4+θ dθ = (4+4π ) / 4/. Problems 5. The formula for area is Therefore, since π/ A = β A = sin (θ)dθ = (sinθ) dθ, we have r = sinθ. The integral represents the shaded area inside one petal of the three-petaled rose curve, r = sinθ, in Figure α r dθ π/ Figure 8.57: Graph of r = sinθ 6. The spiral is shown in Figure Area = π θ dθ = 6 θ π = 8π 6. π/ π π Figure 8.58: Spiral r = θ

39 8. SOLUTIONS The region between the spirals is shaded in Figure Area = π ((θ) θ )dθ = π θ dθ = π θ = 4π. π 4π Figure 8.59: Region between the inner spiral, r = θ, and the outer spiral,r = θ 8. The cardioid is shown in Figure 8.6. The following integral can be evaluated using a calculator or b parts or using the table of integrals. π π Area = (+cosθ) dθ = ( +cosθ +cos θ ) dθ = (θ +sinθ + cosθsinθ + ) π θ = (π +++π) = π. Figure 8.6: Cardioid r = +cosθ 9. (a) See Figure 8.6. In polar coordinates, the line = isrcosθ =, so its equation is The circle of radius centered at the origin has equation (b) The line and circle intersect where r = cosθ. r =. cosθ = cosθ = θ = π, π. Thus, Area = π/ π/ ( ( ) ) dθ. cosθ

40 75 Chapter Eight /SOLUTIONS (c) Evaluating gives Area = π/ π/ ( 4 ) dθ = π/ cos θ (4θ tanθ) = 4π π/. θ = π/ = θ = π/ Figure 8.6. Using the formula Area= (/) β α (f(θ)) dθ gives Area = π a dθ = a θ π = πa, which is the formula for the area of a disk of radius a.. Using the formula Arc length = β (d/dθ) +(d/dθ) dθ, where = acosθ, = asinθ and a > gives α Arc length = π ( asinθ) +(acosθ) dθ = π which is the formula for the circumference of a circle of radius a. adθ = aθ π = πa,. See Figure 8.6. Notice that the curves intersect at(,), where θ =,π, and at(,), where θ = π, so Area = π π ( (+sinθ) )dθ = π π ( sinθ sin θ)dθ. Using a calculator, integration b parts, or formula IV-7 in the integral table, we have Area = (cosθ + sinθcosθ ) π θ = ( + ) π = π 4. π r = +sinθ r = Figure 8.6

41 8. SOLUTIONS 75. The two curves intersect where sinθ = sinθ = θ = π 6, 5π 6. See Figure 8.6. We find the area of the right half and multipl that answer bto get the entire area. The integrals can be computed numericall with a calculator or, as we show, using integration b parts or formula IV-7 in the integral tables. Area of right half = π/6 ( ( ( sinθ) ) dθ ) π/ π/6 ( sinθ +sin θ ) dθ 4 = π/ = π/6 ( θ) 4 sinθ +sin dθ π/ = ( 4 θ +cosθ sinθcosθ + ) π/6 θ ( = 5π ). 8 π/ Thus, Total area = 5π r = / π/6 r = sinθ Figure Figure 8.64 shows the curves which touch at (,) and the origin. However, the circle lies entirel inside the cardioid, so we find the area b subtracting the area of the circle from that of the cardioid. To find the areas, we take the integrals. The cardioid, r = + cosθ, starts at (,) when θ = and traces the top half, reaching the origin when θ = π. Thus Area of cardioid = π (+cosθ) dθ. The circle starts at (,) when θ = and traces the top half, reaching the origin whenθ = π/. Thus Area of circle = π/ (cosθ) dθ.

42 75 Chapter Eight /SOLUTIONS The area, A, we want is therefore Area = = π π (+cosθ) dθ (+cosθ +cos θ)dθ π/ π/ (cosθ) dθ 4cos θdθ = (θ +sinθ + ) π (sinθcosθ +θ) 4 π/ (sinθcosθ +θ) = π π = π. Alternativel, we could compute the area of the cardioid and subtract the area of the circle of radius from it. The integrals can be computed numericall using a calculator, or, as we show, using integration b parts or formula IV-8 from the integral tables. r = +cosθ r = cosθ Figure (a) The graph of r = cosθ is a circle of radius centered at (,); the graph of r = sinθ is a circle of radius centered at (,). See Figure (b) The Cartesian coordinates of the points of intersection are at(,) and(,). The origin corresponds to θ = π/ on r = cosθ and to θ = on r = sinθ. The point (,) has polar coordinates r =,θ = π/4. We find the area below the line θ = π/4 and above r = sin θ and double it: Area = π/4 (sinθ) dθ = 4 π/4 sin θdθ. Using a calculator, integration b parts or formula IV-7 from the integral tables, ( Area = 4 sinθcosθ + θ ) π/4 = +π 4 = π. r = sinθ θ = π/4 r = cosθ Figure 8.65

43 8. SOLUTIONS The area is A = a r dθ = a ( θ θ dθ = ) a = a 6 = a = 6 a = (a) See Figure (b) The curves intersect whenr = 4cosθ = cosθ =. In the first quadrant: θ = π so θ = π 6. Using smmetr, the area in the first quadrant can be multiplied b 4 to find the area of the total bounded region. ( Area = 4 ) π/6 (4cosθ )dθ ( 4sinθ = = 4sin π π = 4 π ) π/6 θ = π =.7..5 r = r = 4cosθ.5 Figure The slope of the tangent line at θ = π/ is d/d = /5. Since = sin(θ)cosθ and = sin(θ)sinθ, when θ = π/, we have = /4 and = 9/4. Thus, the equation of the tangent line is 9 ( ) 4 = 5 4 = =

44 754 Chapter Eight /SOLUTIONS 9. We first find the points with horizontal and vertical tangents in the first quadrant and then use smmetr to obtain the points in other quadrants. The slope of the tangent line is The curve has a horizontal tangent where d 6cos(θ)sinθ +sin(θ)cosθ = d 6cos(θ)cosθ sin(θ)sinθ. 6cos(θ)sinθ +sin(θ)cosθ =. Solving this equation numericall for < θ < π/, we have θ =.955; in addition θ = is a solution. Thus, there are horizontal tangents where =.6 and =.9 and where =, =. Thus, the five points with horizontal tangents are (.6,.9); (.6,.9); (.6,.9); (.6,.9); (, ). The curve has vertical tangents where 6cos(θ)cosθ sin(θ)sinθ =. Solving this equation numericall for < θ < π/, we have θ =.655; in addition θ = π/ is a solution. Thus, there are vertical tangents where =.9, =.6, and where =, =. Thus, there are five points with vertical tangents: (.9,.6); (.9,.6); (.9,.6); (.9,.6); (, ). 4. We can epress and in terms of θ as a parameter. Sincer = θ, we have = rcosθ = θcosθ and = rsinθ = θsinθ. Calculating the slope using the parametric formula, d d = d/dθ d/dθ, we have d sinθ +θcosθ = d cosθ θsinθ. Horizontal tangents occur where d/d =, so Solving this equation numericall gives Vertical tangents occur where d/d is undefined, so sinθ +θcosθ = θ = tanθ. θ =,.9,4.9. Solving this equation numericall gives cosθ θsinθ = θ = tanθ = cotθ. θ =.86,.46.