Note in the last integration above the only non zero

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Unformatted text preview: tion above, the only non-zero contribution comes from the integral of 1/16 as it’s easy to see that 2π 2π cos 2 = cos 4 0 2π 2π cos 2 cos 4 = 0 E =0 0 0 − Evaluate 2− 2 1 cos(4 + 2) + cos(4 − 2) 2 (cos 2 + sin 2 =0 :) ). 2 + 2 =R2 S We apply Green’s Theorem here by noting that 2 2 Q = −2 − − sin 2 + 2 2 2 P = −2 − − cos 2 − 2 2 2 Q − P = 4 − − cos 2 − 2− 2 − 2− 2 cos 2 sin 2 By Green’s Theorem, the integral equals 4 − 2− 2 cos 2 =0 2 + 2 ≤R2 because the integral is anti-symmetric (odd) in in . 3 E and the area of integration is symmetric :) Questions on logical thinking Evaluate the integral − C 2+ 2 , where (a) C is a simple closed piecewise smooth curve not containing the origin in its inside...
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This document was uploaded on 02/10/2014.

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