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Answer Guide 9 Math 408C: Unique Numbers 56975, 56980, and 56985 Tuesday, November 3, 2009 Homework problems Section 6.2 2. A cross section is a disc with radius 1 x 2 , so its area is A ( x ) = (1 x 2 ) 2 . Hence, V = Z 1 1 A ( x ) dx = Z 1 1 (1 x 2 ) 2 dx = 2 Z 1 0 (1 2 x 2 + x 4 ) dx = 16 15 : 6. A cross section has radius y y 2 . So A ( y ) = ( y y 2 ) 2 . Thus, V = Z 1 0 A ( y ) dy = Z 1 0 ( y 2 2 y 3 + y 4 ) dy = 30 : 14. V = Z 3 1 ( 1 x 2 ( 1) ± 2 [0 ( 1)] 2 ) dx = 134 81 : 20. V = Z 1 0 A ( y ) dy = Z 1 0 h 1 2 ( 3 p y ) 2 i dy = 2 5 : 34. V = Z & 0 ² [sin x ( 2)] 2 [0 ( 2)] 2 ³ dx = Z & 0 ´ (sin x + 2) 2 4 µ dx : 36. V = Z 2 & 0 ² (4 cos x ) 2 [4 (2 cos x )] 2 ³ dx = Z 2 & 0 ² (4 cos x ) 2 (2 + cos x ) 2 ³ dx : 1

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Section 7.1 14. Given h ( x ) = 1 + cos x for 0 x , we observe that h 0 ( x ) = ± sin x < 0 for 0 . So h is strictly decreasing and hence is one-to-one on [0 ] . 22.
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