Spring 2006 Final Review (Solutions)

Spring 2006 Final Review (Solutions) - Math 53 - Solutions...

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Math 53 - Solutions to Final Exam Review GSI: Santiago Canez Disclaimer: As always, there may be some typos in this write up. If you think you have found a mistake, please let me know! Happy studying. 1. Compute the line integral of F ( x,y ) = (2 x 2 3 y 2 ) i + (2 x +3 y 2 ) j over the triangle with vertices ( 2 , 0), (2 , 0), and (0 , 2) oriented counterclockwise. Solution. Let C denote the triangle. By Green’s Theorem, we have ° C F · d r = °° D curl F · k dA where D is the region enclosed by the triangle. The left side of D is given by y x =2 and the right side is given by y + x = 2. Since curl F · k =2+6 y , we have °° D curl F · k dA = ° 2 0 ° y +2 y 2 (2 + 6 y ) dxdy = ° 2 0 (2 x +6 yx ) ± ± ± y +2 y 2 dy = ° 2 0 ( 12 y 2 + 20 y + 8) dy = 24 . Hence ² C F · d r = 24. 2. (a) DeFne the curl and divergence of a vector Feld. (b) Determine whether or not the vector Feld F = x 2 y i + zye x j 2 xyz k is the curl of another vector Feld. Solution. (a) Let F = P i + Q j + R k . Then curl F =( R y Q z ) i +( P z R x ) j +( Q x P y ) k and div F = P x + Q y + R z . (b) Since div curl G = 0 for any vector Feld G , if F was the curl of some other vector Feld, its divergence would have to be zero. But div F =2 xy + ze x 2 xy = ze x ° =0so F is not the curl of another vector Feld. Date : May 11, 2006 1
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3. Let S be the sphere of radius ρ . (a) Compute the surface area of S . (b) Compute the surface integral °° S z dS . Solution. (a) Parameterize the sphere by r ( φ, θ )= ρ sin φ cos θ i + ρ sin φ sin θ j + ρ cos φ k , 0 φ π, 0 θ 2 π. Then r φ = ρ cos φ cos θ i + ρ cos φ sin θ j ρ sin φ k r θ = ρ sin φ sin θ i + ρ sin φ cos θ j . We can then compute | r φ × r θ | = ρ 2 sin φ. Thus the surface area of S is (with D the region where φ and θ lie) ±± D | r φ × r θ | dA = ± 2 π 0 ± π 0 ρ 2 sin φ dφ dθ =4 πρ 2 . (b) We use the same parameterization as above. We have for f ( x,y,z )= z ±± S f ( x,y,z ) dS = ±± D f ( r ( u,v )) | r u × r v | dA = ±± D ρ cos φ ( ρ 2 sin φ ) dA = ρ 3 ± 2 π 0 ± π 0 sin φ cos φdφdθ =0 . Note that this makes sense since the function f ( x,y,z )= z is symmetric about the xy -plane, so the integral over the bottom half of the sphere should be the negative of the integral over the top half of the sphere and hence the integral over the whole sphere should be 0. 4. Let F ( x,y,z )= x i + y j + z k . Determine whether the Fux of F across S is negative, positive, or zero for each surface S below. (a) the cone z 2 = x 2 + y 2 oriented outward (b) the sphere x 2 + y 2 + z 2 = 1 oriented inward (c) the plane y = 2 oriented with normals having positive j component Solution. Here, F is just the “radial” vector ±eld, which always points radially away from the origin.
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Spring 2006 Final Review (Solutions) - Math 53 - Solutions...

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