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Unformatted text preview: 308 Section 7.3
2y and area
5 52. (a) A cross section has radius r r2 (b) V(h)
dV dt dh so dt 55. Solve ax
5 x2 0: This is true at x 0 and x a. For 2 y. The volume is
0 2 y dy A(h).
dh , dt y2 25 .
0 revolution about the xaxis, a cross section has radius r A(x) ax x 2 and area r2 (ax x 2)2 (a 2x 2 2ax 3 x 4). dV A(h) dh, so dh dV dh A(h) dh dt dV 1 A(h) dt 1 8 units3 3 3 = sec 8
y The volume is
a For h
dh so dt 4, the area is 2 (4) 8 ,
units3 . sec (a 2x 2 2ax 3 x 4) dx 0 1 2 3 a x 3 1 a 5. 30 1 4 ax 2 1 5 x 5 a 0 53. (a) For revolution about the yaxis, a cylindrical shell has
2 r x radius x and height ax
a x 2. The volume is 2
1 3 ax 3 1 4 x 4
a 0 2 (x)(ax
0 x 2) dx 1 4 a . 6 2 Setting the two volumes equal,
1 a5 30 1 4 1 a yields a 6 30 1 , so a 6 The remaining solid is that swept out by the shaded region in revolution. Use cylindrical shells: a shell has radius x and height 2 .r .
r2 22 5. r2 x 2. The volume is 56. The slant height s of a tiny horizontal slice can be
written as s x2 y2 1 ( g ( y))2 y. So the 2 (x)(2 r 2
2 2 (r 3 4 ( 8) 3
r x 2) dx surface area is approximated by the Riemann sum 2 x ) 2 3/2
r2 4 2 k 1 n g( yk) 1 (g ( y))2 y. The limit of that is the integral.
dx dy 1 32 . 3 57. g ( y)
2 , and
2
2 (b) The answer is independent of r. 54. Partition the appropriate interval in the axis of revolution and measure the radius r(x) of the shadow region at these points. Then use an approximation such as the trapezoidal
b 2
0 y 1 2 y 1 2 y dy
0 4y (4y 1 dy 1)3/2
2 0 6 13 3 13.614 rule to estimate the integral
a r 2(x) dx. 58. g ( y)
1 dx dy y3 3 y 2, and ( y 2)2 dy
2 3 1 (1 6 2
0 1 y 4)3/2 1 0 9 (2 2 1) 0.638. 59. g ( y)
3 dx dy 1 1/2 y , and 2 1 3/2 3 1 3/2 3 2
1 y1/2
3 1 1 1 1/2 2 y dy 2 1 dy. 4y 2
1 y1/2 Using NINT, this evaluates to 16.110 ...
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 Spring '08
 GERMAN

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