Evaluate curlgrad s consider we have i curlgrad

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Unformatted text preview: s. Evaluate curl(grad ). S Consider ( ). We have i ∂ curl(grad ) = j ∂ k ∂ − =( )i − ( − − )j + ( )k = 0 :) E Let v = P i + Q j + Rk be an arbitrary vector field whose component functions have continuous partial derivatives. Evaluate div(curl v). S It is (R − Q ) + (P − R ) + (Q − P ) = R E 2 Evaluate +P , where C is given by −R +Q = ( − sin ), −P =0 :) = (1 − cos ) for C 0 ≤ ≤ 2π . Here, S −Q > 0 is some constant. First, 2 √√ 2 1 − cos 2 + = (1 − cos )2 + sin2 = = Thus, 2 = 3 √ 2π (1 − cos ) 2 C 5 2 = 3 √ 2 sin 0 π = 16 2 2 0 2π =8 2 sin 3 = 16 0 2 (1 − cos ) sin = 16 3 (1 − 2 2 + (1 − 22 ) −1 1 = 32 2 1 2 0 3 sin5 3 0 π 5 3 5 2 2π 4 ) = 32 0 2 3 1− 21 + 35 = 256 15 3 :) E (2 − ) Evaluate (1 − cos )...
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This document was uploaded on 02/10/2014.

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