15_3_13 - 2 t = = a 2 2 ± sin t p 1 + sin 2 t + ln | sin t...

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13. The first octant part of the curve x 2 + y 2 = a 2 , z = x , can be parametrized r = a cos t i + a sin t j + a cos t k , ( 0 t π/ 2 ). We have ds = a 1 + sin 2 t dt , so Z x ds = a 2 Z π/ 2 0 cos t p 1 + sin 2 t dt Let sin t = tan θ cos t dt = sec 2 θ d θ = a 2 Z t = π/ 2 t = 0 sec 3 θ d θ = a 2 2 ± sec θ tan θ + ln | sec θ + tan θ | ² ³ ³ ³ ³ t = π/
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Unformatted text preview: 2 t = = a 2 2 ± sin t p 1 + sin 2 t + ln | sin t + p 1 + sin 2 t | ² ³ ³ ³ ³ π/ 2 = a 2 2 ± √ 2 + ln ( 1 + √ 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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