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15_3_8 - = Z 2 π 1 sin 2 2 t dt = Z 2 π ± 1 1-cos 4 t 2...

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8. The curve of intersection of x 2 + z 2 = 1 and y = x 2 can be parametrized r = cos t i + cos 2 t j + sin t k , ( 0 t 2 π). Thus ds = p sin 2 t + 4 sin 2 t cos 2 t + cos 2 t dt = p 1 + sin 2 2 t dt . We have Z p 1 + 4 x 2 z 2 ds = Z 2 π 0 p 1 + 4 cos 2 t sin 2 t p 1 + sin
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Unformatted text preview: = Z 2 π ( 1 + sin 2 2 t ) dt = Z 2 π ± 1 + 1-cos 4 t 2 ² dt = 3 2 ( 2 π) = 3 π....
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