# 15_3_4 - m = 1 2 Z π 2 p 2-cos 2 2 t sin 2 t dt Let v =...

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4. The wire of Example 3 lies in the first octant on the surfaces z = x 2 and z = 2 - x 2 - 2 y 2 , and, therefore, also on the surface x 2 = 2 - x 2 - 2 y 2 , or x 2 + y 2 = 1, a circular cylinder. Since it goes from ( 1 , 0 , 1 ) to ( 0 , 1 , 0 ) it can be parametrized r = cos t i + sin t j + cos 2 k , ( 0 t π/ 2 ) v = - sin t i + cos t j - 2 cos t sin t k v = p 1 + sin 2 ( 2 t ) = p 2 - cos 2 ( 2 t ). Since the wire has density δ = xy = sin t cos t = 1 2 sin ( 2 t ) , its mass is
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Unformatted text preview: m = 1 2 Z π/ 2 p 2-cos 2 ( 2 t ) sin ( 2 t ) dt Let v = cos ( 2 t ) d v = -2 sin ( 2 t ) dt = 1 4 Z 1-1 p 2-v 2 d v = 1 2 Z 1 p 2-v 2 d v, which is the same integral obtained in Example 3, and has value (π + 2 )/ 8....
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## This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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