251-f06-finalsolutions

# 251-f06-finalsolutions - i| = r we have ZZ S 2 f x y z dS =...

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MATH 251.504 Solutions for the Final Examination Fall 2006 1. (a) true, (b) true, (c) true, (d) false, (e) false, (f) true, (g) true, (h) true. 2. (a) curl F = 0 , y cos( yz ) , - z cos( yz ) and div F = - sin y + e z . (b) curl G = - 1 , - 1 , - 2 y , and so curl F = - 2 , 0 , 0 and div F = 0. 3. We have F = f where f ( x, y, z ) = x 3 yz 2 + sin x + y , and so C F · d r = f ( π, - 2 , 0) - f (0 , 1 , 3) = - 3 by the Fundamental Theorem for Line Integrals. 4. We have div F = 6 z ( x 2 + y 2 + z 2 ) 2 and so by the Divergence Theorem the given surface integral is equal to E 6 z ( x 2 + y 2 + z 2 ) 2 dV = 1 0 π 2 0 2 π 0 (6 ρ 5 cos ϕ ) ρ 2 sin ϕ dθ dϕ dρ = 6 1 0 ρ 7 π 2 0 1 2 sin 2 ϕ dϕ 2 π 0 = 3 4 π. 5. By Green’s Theorem the given line integral is equal to 3 0 3 - x - 3+ x x dy dx = 3 0 (2 x 2 - 6 x ) dx = - 9 . 6. The cylindrical part S 1 of the surface can be parametrized by r ( θ, z ) = cos θ, sin θ, z where 0 θ 2 π and - 3 z 3, and since | r θ × r z | = | cos θ, sin θ, 0 | = 1 we have S 1 f ( x, y, z ) dS = 2 π 0 3 - 3 (2 + z 2 ) dz dθ = 60 π. The base S 2 can be parametrized by r ( r, θ ) = r cos θ, r sin θ, - 3 where 0 r 1 and 0 θ 2 π , and since

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Unformatted text preview: i| = r we have ZZ S 2 f ( x, y, z ) dS = Z 2 π Z 1 (2 r 2 + 9) r dr dθ = 10 π. Thus RR S f ( x, y, z ) dS = RR S 1 f ( x, y, z ) dS + RR S 2 f ( x, y, z ) dS = 70 π . 1 7. The area is equal to ZZ D q 1 + ( ∂z ∂x ) 2 + ( ∂z ∂y ) 2 dA = 2 Z 1 Z x √ 1 + x 2 dy, dx = 2 3 (2 3 / 2-1) . 8. The line integral around each of the four smooth parts of the boundary of S is zero, and so the given surface integral is equal to zero. 9. Using the parametrization r ( t ) = h cos t, sin t i for 0 ≤ t ≤ 2 π (which gives the opposite orientation) we have r ( t ) = h-sin t, cos t i and hence Z C F · d r =-Z 2 π h sin t,-cos t i · h-sin t, cos t i dt = Z 2 π dt = 2 π. 2...
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