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Solutions8

Solutions8 - 1 Mathematics 1c Solutions Homework Set 8 Due...

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Unformatted text preview: 1 Mathematics 1c: Solutions, Homework Set 8 Due: Tuesday, June 1 at 10am. 1. (10 Points) Section 8.1, Exercises 3c and 3d . Verify Green’s theorem for the disk D with center (0 , 0) and radius R and P ( x,y ) = xy = Q ( x,y ) and the same disk for P = 2 y,Q = x. . Solution. For 3c, let c ( t ) = ( R cos t,R sin t ) be the parameterization of ∂D . Then Z ∂D P dx + Qdy = Z 2 π ( R 2 cos t sin t,R 2 cos t sin t ) · (- R sin t,R cos t ) dt =- R 3 Z 2 π sin 2 t cos tdt + R 3 Z 2 π cos 2 t sin tdt = 0 + 0 = 0 . Also, ZZ D ∂Q ∂x- ∂P ∂y dxdy = ZZ D ( y- x ) dxdy = Z R Z 2 π ( r sin θ- r cos θ ) r dθ dr = Z R (0 + 0) r 2 dr = 0 . Hence, Green’s theorem for 3c is verified. For 3d, note that Green’s theorem Z ∂D Pdx + Qdy = ZZ D ∂Q ∂x- ∂P ∂y dxdy becomes Z ∂D 2 y dx + xdy = ZZ D (1- 2) dxdy =- ZZ D dxdy The right side is- πR 2 while the left side is, since x = R cos θ and y = R sin θ, Z 2 π (2 R sin θ )(- R sin θ ) dθ + ( R cos θ )( R cos θ ) dθ =- 2 R 2 Z 2 π sin 2 θdθ + R 2 Z 2 π cos 2 θdθ. Using the fact that sin 2 θ and cos 2 θ have averages 1 2 , namely 1 2 π Z 2 π sin 2 θdθ = 1 2 (this is one way of remembering the formula for the integrals of sin 2 θ and cos 2 θ on [0 , 2 π ] and [0 ,π...
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Solutions8 - 1 Mathematics 1c Solutions Homework Set 8 Due...

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