prac3Bsol

prac3Bsol - Z 1 ydxdy = Z 4 ydy = [ y 2 / 2] 4 = 8. b) On C...

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18.02 Practice Exam 3 B – Solutions 1. a) 6 - ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¡ ¡ ¡ ¡ x = 1 y = x y = 2 x q q q (1,1) (1,2) b) Z 1 0 Z y y/ 2 dxdy + Z 2 1 Z 1 y/ 2 dxdy. (the frst integral corresponds to the bottom halF 0 y 1, the second integral to the top halF 1 y 2.) 2. a) δdA = r sin θ r 2 rdrdθ = sin θdrdθ . M = ZZ R δdA = Z π 0 Z 3 1 sin θ drdθ = Z π 0 2 sin θdθ = £ - 2 cos θ / π 0 = 4 . b) ¯ x = 1 M ZZ R xδdA = 1 4 Z π 0 Z 3 1 r cos θ sin θdrdθ 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 - 6 y 2 f = x 3 - 6 y 2 x + c ( y ) f y = - 12 xy + c 0 ( y ) = - 12 xy + 4 y . So c 0 ( y ) = 4 y , thus c ( y ) = 2 y 2 (+ 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 Z C F · d r = f (1 , 1) - f (1 , 0) = (1 - 6 + 2) - 1 = - 4 . 4. a) The parametrization oF the circle C is x = cos t , y = sin t , For 0 t < 2 π ; then dx = - sin tdt , dy = cos tdt and W = Z 2 π 0 (5 cos t + 3 sin t )( - sin t ) dt + (1 + cos(sin t )) cos tdt. b) Let R be the unit disc inside C ; I C F · d r = ZZ R ( N x - M y ) dA = ZZ R (0 - 3) dA = - 3 area( R ) = - 3 π. 5. a) (0 , 4) C 4 ² C 3 o (1 , 4) (0 , 0) C 1 / C 2 O (1 , 0) I C F · ˆ n ds = ZZ R div F dxdy = ZZ R ( y + cos x cos y - cos x cos y ) dxdy = ZZ R ydxdy = Z 4
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Unformatted text preview: Z 1 ydxdy = Z 4 ydy = [ y 2 / 2] 4 = 8. b) On C 4 , x = 0, so F =-sin y , whereas n =- . Hence F n = 0 . Therefore the Fux of F through C 4 equals 0. Thus Z C 1 + C 2 + C 3 F n ds = I C F n ds-Z C 4 F n ds = I C F n ds ; and the total Fux through C 1 + C 2 + C 3 is equal to the Fux through C . 6. Let u = 2 x-y and v = x + y-1. The Jacobian f f f f u x u y v x v y f f f f = f f f f 2-1 1 1 f f f f = 3. Hence dudv = 3 dxdy and dxdy = 1 3 dudv , so that V = ZZ (2 x-y ) 2 +( x + y-1) 2 &lt; 4 (4-(2 x-y ) 2-( x + y-1) 2 ) dxdy = ZZ u 2 + v 2 &lt; 4 (4-u 2-v 2 ) 1 3 dudv = Z 2 Z 2 (4-r 2 ) 1 3 rdrd = Z 2 2 3 r 2-1 12 r 4 2 d = Z 2 4 3 d = 8 3 ....
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This note was uploaded on 04/18/2008 for the course 18 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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prac3Bsol - Z 1 ydxdy = Z 4 ydy = [ y 2 / 2] 4 = 8. b) On C...

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