hw7sol - Math 241 Solutions 7 6.1#7. Note that F(x, y ) =...

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Math 241 Solutions 7 6.1#7. Note that F ( x,y ) = ( y + 2 ,x ) , x ( t ) = (sin t, - cos t ) , 0 t π/ 2, so x 0 ( t ) = (cos t, sin t ) and hence Z x F · d s = Z π/ 2 0 F ( x ( t )) · x 0 ( t ) dt = Z π/ 2 0 ( - cos t + 2)(cos t ) + (sin t )(sin t ) dt = 2 . 6.2#1. Note that F ( x,y ) = ( - x 2 y,xy 2 ). Parametrizing ∂D obtains (2 cos t, 2 sin t ) , 0 t 2 π , so dx = - 2 sin tdt and dy = 2 cos tdt . Hence I ∂D Mdx + Ndy = I ∂D ( - x 2 y ) dx + xy 2 dy = Z 2 π 0 ( - 8 cos 2 t sin t )( - 2 sin tdt ) + (8 cos t sin 2 t )(2 cos tdt ) = Z 2 π 0 32 cos 2 t sin 2 tdt = 8 π. 6.2#8. Let D be region bounded by the ellipse x 2 + 4 y 2 = 4. Note that Area( D ) = 2 π . By Green’s Theorem, we have Z ∂D F · d s = Z Z D ∂N ∂x - ∂M ∂y dA = Z Z D (1 - 4) dA = - 3Area( D ) = - 6 π. 6.3#1. (a) Let the parametric function be x ( t ) = ( t,t,t ) , 0 t 1, so dx = dy = dz = dt and thus Z C z 2 dx + 2 ydy + xzdz = Z 1 0 t 2 dt + 2 tdt + t 2 dt = 5 / 3 . (b) Note
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This note was uploaded on 09/02/2010 for the course MATH 241 taught by Professor Any during the Spring '08 term at University of Illinois at Urbana–Champaign.

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hw7sol - Math 241 Solutions 7 6.1#7. Note that F(x, y ) =...

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