Summer (Session 1) 2004 - Eggers' Class - Exam 2 (Version 1)

Summer(Session 1) - 0(Hint You may wish to compute ∇ F first ∇ F = ∂ ∂x y 2 = 0 By the Divergence Theorem RR S F d S = RRR V ∇ F dV

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Math 20E Midterm Exam 2 (version 1) Solution 1. Let F ( x, y, z ) = (2 y + 7 z, 2 x + 4 z, 7 x + 4 y ). (a) (4 points) Find a scalar field f for which f = F . ∂f ∂x = 2 y + 7 z implies f ( x, y, z ) = 2 xy + 7 xz + g ( y, z ). Then, ∂f ∂y = 2 x + ∂g ∂y = 2 x + 4 z implies f ( x, y, z ) = 2 xy + 7 xz + 4 yz + h ( z ). Finally, ∂f ∂z = 7 x + 4 y + h 0 ( z ) = 7 x + 4 y implies f ( x, y, z ) = 2 xy + 7 xz + 4 yz + C for some constant C . (b) (4 points) Compute the value of the line integral R C F · d R from (5 , 0 , 0) to (5 , 0 , 1) along the helix R ( t ) = (5 cos (2 πt ) , 5 sin (2 πt ) , t ). R C F · d R = R C f · d R = f (5 , 0 , 1) - f (5 , 0 , 0) = 35. 2. The function R : [ - π, π ] × [ - π 2 , π 2 ] R 3 ( u, v ) 7→ 3(cos v cos u, cos v sin u, sin v ) is a parameterization of the sphere of radius 3 centered at (0 , 0 , 0). (a) (4 points) Compute R ∂u × R ∂v , the normal vector to the parameterized sphere. R ∂u = 3( - cos v sin u, cos v cos u, 0) R ∂v = 3( - sin v cos u, - sin v sin u, cos v ) R ∂u × R ∂v = 9 ± ± ± ± ± ± i j k - cos v sin u cos v cos u 0 - sin v cos u - sin v sin u cos v ± ± ± ± ± ± = 9 cos v (cos v cos u, cos v sin u, sin v ) (b) (4 points) Let F ( x, y, z ) = ( y 2 , 0 , 0). Find the value of the surface integral RR S F · d S over the sphere of radius 3 centered at (0
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Unformatted text preview: , , 0). (Hint: You may wish to compute ∇· F first.) ∇· F = ∂ ∂x y 2 = 0. By the Divergence Theorem, RR S F · d S = RRR V ∇· F dV = 0. 3. Consider the iterated integral Z 3 y =0 Z √ 9-y 2 x =0 Z 3 z =0 z p x 2 + y 2 dz dx dy. (a) (4 points) Write the integral as an iterated integral in cylindrical coordinates. The region of integration is described in cylindrical coordinates by 0 ≤ ρ ≤ 3, 0 ≤ θ ≤ π 2 , 0 ≤ z ≤ 3. Thus, the iterated integral is written in cylindrical coordinates as Z π 2 θ =0 Z 3 ρ =0 Z 3 z =0 zρ · ρ dz dρ dθ. (b) (4 points) Compute the value of the integral. " Z π 2 dθ # ²Z 3 ρ 2 dρ ³²Z 3 z dz ³ = π 2 · 9 · 9 2 = 81 π 4 ....
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This note was uploaded on 04/30/2008 for the course MATH 20E taught by Professor Enright during the Summer '07 term at UCSD.

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