# Be the curve consisting of the line segment from 0 1

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be the curve consisting of the line segment from (0 , - 1) to (0 , 1) followed by the semi-circle from (0 , 1) through (1 , 0) to (0 , - 1) . Compute R C ~ F · d~ r .
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using polar coordinates, we have Z C ~ F · d~ r = - ZZ D ( x 2 - ( x + x 2 )) dA = - Z π/ 2 - π/ 2 Z 1 0 ( - r cos θ ) r dr dθ = Z π/ 2 - π/ 2 cos θ ! Z 1 0 r 2 dr = [sin θ ] π/ 2 - π/ 2 1 3 r 3 1 0 = 2 · 1 3 = 2 3 . 11. Let S be the part of the surface z = 1 2 x 2 that lies in the first octant and also satisfies 1 2 z 2 and y 1 . If f ( x, y, z ) = yz x , compute RR S f dS . The graph of z = 1 2 x 2 satisfies 1 2 z 2 and x 0 (recall that S lies in the first octant) exactly when 1 x 2. Hence we can parametrize S as the graph of z = 1 2 x 2 over { 1 x 2 , 0 y 1 } in the xy -plane. This gives ZZ S f dS = Z 2 1 Z 1 0 y ( x 2 / 2 ) x p x 2 + 0 2 + 1 dy dx = Z 1 0 y dy Z 2 1 1 2 x p 1 + x 2 dx = 1 2 y 2 1 0 1 6 ( 1 + x 2 ) 3 / 2 2 1 = 1 2 5 3 / 2 6 - 2 3 / 2 6 = 5 5 - 2 2 12 . 12. Let ~ F ( x, y, z ) = yz~ ı + ( y cos y ) ~ + 3 x ~ k . Also, let C be the curve of intersection of the cylinder x 2 + y 2 = 4 and the plane z - 2 y = 0 , with counter-clockwise orientation when viewed from above. Compute R C ~ F · d~ r .
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If we let g ( x, y ) = 2 y , as usual, then the integrand that appears in the integral over D for RR S curl ~ F ( x, y, z ) · d ~ S is - P ∂g ∂x - Q ∂g ∂y + R = - 0 · 0 - ( y - 3) · 2 + ( - 2 y ) = 6 - 4 y. Thus, using Stokes’ theorem and writing the resulting integral over D in polar coordinates, we have Z C ~ F · d~ r = ZZ S curl ~ F ( x, y, z ) · d ~ S = ZZ D (6 - 4 y ) dA = 6 · Area( D ) - 4 Z 2 π 0 Z 2 0 ( r sin θ ) r dr dθ = 6 · 4 π - 4 Z 2 π 0 sin θ dθ Z 2 0 r 2 dr = 24 π - 4 [ - cos θ ] 2 π 0 1 3 r 3 2 0 = 24 π - 4( - 1 + 1)(8 / 3) = 24 π. 13. Consider the solid tetrahedron bounded by the planes x = 0 , y = 0 , z = 0 , and 2 x + y + 2 z = 2 . Let S be the boundary of this solid tetrahedron, with orientation given by the outward unit normal. Also, let ~ F ( x, y, x ) = e yz ~ ı - (3 xy + 1) ~ + ( x + y ) ~ k . Compute RR S ~ F · d ~ S .
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Thus we have ZZ S ~ F · d ~ S =