15_3_2 - 2 Thus z ds = x = t cos t y = t sin t z = t(0 t 2...

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2. : x = t cos t , y = t sin t , z = t , ( 0 t 2 π) . We have ds = p ( cos t - t sin t ) 2 + ( sin t + t cos t ) 2 + 1 dt = p 2 + t 2 dt . Thus Z z ds = Z 2 π 0 t p 2 + t 2 dt Let u = 2 + t 2 du = 2 t dt = 1 2 Z 2 + 4 π 2 2 u 1 / 2 du = 1 3 u 3 / 2 ± ± ± ± 2
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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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