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Unformatted text preview: Section 7.3
47. (a) Solve
d dc
[ 0.5, 2.5] by [ 0.5, 2.5]
2 2
2 307 c 4 c 2
2 2 0 0 2 c 0
2 c c 4 The intersection points are 1 1 , 1 , , 2 , and (1, 1). 4 4 1 1 , outer radius R , 4 y2 1 . The volume is 16 2 11 1 y . 48 16 1 1 x (a) A washer has inner radius r and area (R
2 1 This value of c gives a minimum for V because
a 2V dc 2 2 r ) 2 1 y4 1 dy 16 1 y4 1 3y 3 2 2 0.
2 2 2 Then the volume is 4 2 2 2 2 2 4 (b) A shell has radius x and height
1 1. The volume is
1 2 x 2
1 1/4 (b) Since the derivative with respect to c is not zero anywhere else besides c at c 0 or c
2 2 (x)
1/4 1 x 1 dx , x f (x) 0 1 x 2 2 3/2 x 3 11 . 48 , the maximum must occur
2 1. The volume for c
(3 2 8) 0 is 48. (a) For 0 For x x f (x) x x(sin x) x 2 4.935, sin x. sin x. So (c) and for c 1 it is 2.238. c 0 0, x f (x) sin x for 0 sin 0 . maximizes the volume. (b) Use cylindrical shells: a shell has radius x and height y. The volume is
0 2 xy dx, which from part (a) is 2 cos x
0 2 sin x dx
0 4 .
x 12 [0, 2] by [0, 6] 49. (a) A cross section has radius r A(x)
6 0 36 x 2 and area The volume gets large without limit. This makes sense, since the curve is sweeping out space in an everincreasing radius. 51. (a) Using d
C r2 (36x 2 144 (36x 2 x 4). The volume is
144 , and A 144 x 4) dx 12x 3 1 5 x 5 6 0 36 cm3. 5 d 2 2 C2 yields the 4 following areas (in square inches, rounded to the nearest tenth): 2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7, 10.7, 9.3, 6.4, 3.2. (b) 36 cm3 (8.5 g/cm3) 5 192.3 g 50. A cross section has radius r A(x) r2 (c sin x)2 c sin x and area (b) If C(y) is the circumference as a function of y, then the (c 2 2c sin x sin2 x). area of a cross section is
C(y)/ 2
2 The volume is A(y) (c2
0 2c sin x 2c cos x 2c 4 c
1 2
2 sin2 x) dx
1 x 2 1 sin 2x 4 C 2( y) , 4 1 4
6 6 0 and the volume is
0 C 2(y) dy. c2x c c
2 2c . (c) 1 4 6 A( y) dy
0 2 2 1 C 2(y) dy 4 0 1 6 0 [5.42 24 4 2(4.52 10.82 4.42 11.62 2 5.12 11.62 6.32 10.82 7.82 9.42 9.02) 6.32] 34.7 in.3 ...
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 Spring '08
 GERMAN

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