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AnswerKey9 - Answer Guide 9 Math 408C Unique Numbers 56975...

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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 < x < ° . So h is strictly decreasing and hence is one-to-one on [0 ; ° ] . 22. Let m = m 0 q 1 ° v 2 c 2 : Then 1 ° v 2 c 2 = ( m 0 m ) 2 and so v 2 = c 2 ´ 1 ° ( m 0 m ) 2 µ . We conclude that v = c r 1 ° m 2 0 m 2 ; which gives the velocity of the particle in terms of its mass.
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