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**Unformatted text preview: **curve given by x = cos t , y = sin t from (1 , 0) to (0 , 1). Hint : Work = R C F d r Solution : The path is r ( t ) = h cos t, sin t i for 0 t 2 . Z C F d r = Z / 2 h-x 2 , xy i h-sin t, cos t i dt = Z / 2 h-cos 2 t, cos t sin t i h-sin t, cos t i dt = Z / 2 ( sin t cos 2 t + cos 2 t sin t ) dt = 2 Z / 2 sin t cos 2 t dt = 2 3 . 2. (3 points) Is the integral you evaluated above independent of path? How do you know? Solution : No, the integral is not independent of path because the vector eld given by F ( x, y ) is not the gradient of any function. We know this because P y 6 = Q x , where F ( x, y ) = h P, Q i = h-x 2 , xy i . Another way of saying this is that-x 2 dx + xy dy is not an exact dierential....

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