quiz3 - 1 [6 marks, 3 marks each part] Evaluate (a) C zds...

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1 [6 marks, 3 marks each part] Evaluate (a) R C zds where C is parametrized by g ( t ) = (cos t, sin t,t ) , 0 t 2 π. g 0 ( t ) = ( - sin t, cos t, 1) Z C zds = 2 π Z 0 t p ( - sin t ) 2 + (cos t ) 2 + 1 2 dt = 2 2 π Z 0 t 1 2 dt = = 2 2 3 t 3 / 2 2 π 0 = 2 2 3 (2 π ) 3 / 2 = 8 3 π 3 / 2 (b) R C ydx + zdy + xdz, where C is the line segment from (1 , 1 , 1) to (1 , 2 , 3) . Parametrize C by g ( t ) = (1 , 1 , 1) + t (0 , 1 , 2) = (1 , 1 + t, 1 + 2 t ) , 0 t 1 g 0 ( t ) = (0 , 1 , 2) R C ydx + zdy + xdz = 1 R 0 [(1+ t ) · 0+(1+2 t ) · 1+1 · 2] dt = 1 R 0 (2 t +3) dt = = t 2 + 3 t | 1 0 = 4 2 [6 marks] Evaluate R C + ( xy + arctan e x ) dx - ( x 2 + cos 4 y ) dy, where C is the triangle with vertices (0 , 0) , (1 , 0) , and (1 , 1) oriented clockwise. To use Green's Theorem, we have to change the orientation of C to positive with respect to the interior of the triangle, i.e to the counterclockwise ( C + ). This changes the sign of the integral, so we have
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quiz3 - 1 [6 marks, 3 marks each part] Evaluate (a) C zds...

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