exam3sol

# exam3sol - 18.02 Exam 3 Solutions(1,2 y = 2x x=1 1 a(1,1 y...

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�� 18.02 Exam 3 Solutions (1,2) y = 2 x x = 1 (1,1) b) 1 y dxdy + 2 1 dxdy. 1. a) 0 y/ 2 1 y/ 2 �� y = x (the first integral corresponds to the bottom half 0 y 1, the second integral to the top half 1 y 2.) r sin θ 2. a) δdA = rdrdθ = sin θdrdθ . 2 r �� π 3 π π M = δdA = sin θ drdθ = 2 sin θdθ = 2 cos θ = 4 . 0 R 0 1 0 �� π 3 1 1 b) ¯ x = xδdA = r cos θ sin θdrdθ M R 4 0 1 The reason why one knows that ¯ x = 0 without computation is that the region and the density are symmetric with respect to the y -axis ( δ ( x, y ) = δ ( x, y )). 3. a) N x = 12 y = M y , hence F is conservative. b) f x = 3 x 2 2 6 y 2 f = x 3 6 y 2 x + c ( y ) f y = 12 xy + c ( y ) = 12 xy + 4 y . So c ( y ) = 4 y , thus c ( y ) = 2 y (+ constant). In conclusion f = x 3 6 xy 2 + 2 y 2 (+ constant) . c) The curve C starts at (1 , 0) and ends at (1 , 1), therefore F d r = f (1 , 1) f (1 , 0) = (1 6 + 2) 1 = 4 . · C 4. a) The parametrization of the circle C is x = cos t , y = sin t ,

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