15_3_1 - 1. : r = a cos t sin ti + a sin2 tj + a cos tk, 0...

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Unformatted text preview: 1. : r = a cos t sin ti + a sin2 tj + a cos tk, 0 t /2. Since |r|2 = a 2 (cos2 t sin2 t + sin4 t + cos2 t) = a 2 for all t, must lie on the sphere of radius a centred at the origin. We have ds = a (cos2 t - sin2 t)2 + 4 sin2 t cos2 t + sin2 t dt = a cos2 2t + sin2 2t + sin2 t dt = a 1 + sin2 t dt. Thus z ds = /2 0 2 0 /4 a cos t a 1 + sin2 t dt 1 Let u = sin t du = cos t dt =a 1 + u 2 du sec3 d Let u = tan du = sec2 d /4 0 = a2 = 0 a2 sec tan + ln | sec + tan | 2 a2 = 2 + ln(1 + 2) . 2 ...
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