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Unformatted text preview: Mathematics 250 Fall 2008 Selected Solutions Assignment # 9 MVM, 8.3.2 : Calculate the following line integrals. a) R C xy 3 dx , where C is the unit circle, C = x y : x 2 + y 2 = 1 , oriented counterclockwise. b) R C zdx + xdy + ydz , where C is the line segment from 1 2 to 1 1 3 c) R C y 2 dx + zdy 3 xydz , where C is the line segment from 1 1 to 2 3 1 . d) R C ydx , where C is the intersection of the unit sphere with the plane x + y + z = 0, oriented counterclockwise as viewed from above. Response: a) We can parametrize C by γ ( t ) = cos t sin t , for 0 ≤ θ ≤ 2 π . This parametrization is consistent with the specified orientation. Then γ * ( xy 3 dx ) = γ * ( x ) γ * ( y ) 3 γ * ( dx ) = (cos t )(sin 3 t )( sin tdt ) = sin 4 t cos tdt. Thus, we have Z C xy 3 dx = Z 2 π sin 4 t cos tdt = ( sin 5 t 5  2 π = 0 0 = 0 . Indeed, this result could have been predicted by symmetry: the curve C is symmetric under reflection in the yaxis, and the contribution of an angular interval in the right half plane is cancelled by the contribution of its reflection in the yaxis. b) We can parametrize C by γ ( t ) = 1 2 + t 1 1 3  1 2 = t 1 2 t 2 + t , for 0 ≤ t ≤ 1. Then Z C ydx = Z 1 (1 2 t ) dt = ( t t 2 )  1 = 1 1 (0 0) = 0 . c) We can parametrize C by γ ( t ) = 1 1 + t 2 3 1  1 1 = 1 + t 3 t 1 2 t , for 0 ≤ t ≤ 1. Then we have Z C y 2 dx + zdy 3 xydz = Z 1 ( (3 t ) 2 dt + (1 2 t )(3 dt ) 3(1 + t )(3 t )( 2 dt ) ) = Z 1 (27 t 2 + 12 t + 3) dt = (9 t 3 + 6 t 2 + 3 t )  1 = (9 + 6 + 3) (0 + 0 + 0) = 18 . d) To parametrize this curve, we follow the suggestion in the book, and construct an orthonormal basis for the plane in question. Let P be the plane defined by x + y + z = 0. One vector in P is the vector 1 f 1 = 1 √ 2 1 1 , which spans the intersection of P with the ( x, y )plane. We want a second vector in P orthogonal to f 1 . A little thought will produce the vector 1 √ 6 1 1 22 ....
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 Fall '06
 RogerHowe
 Integrals, Unit Circle

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