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

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

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Math 20E Midterm Exam 2 (version 2) Solution 1. Let F ( x, y, z ) = (3 y + 6 z, 3 x + 5 z, 6 x + 5 y ). (a) (4 points) Find a scalar field f for which f = F . ∂f ∂x = 3 y + 6 z implies f ( x, y, z ) = 3 xy + 6 xz + g ( y, z ). Then, ∂f ∂y = 3 x + ∂g ∂y = 3 x + 5 z implies f ( x, y, z ) = 3 xy + 6 xz + 5 yz + h ( z ). Finally, ∂f ∂z = 6 x + 5 y + h 0 ( z ) = 6 x + 5 y implies f ( x, y, z ) = 3 xy + 6 xz + 5 yz + C for some constant C . (b) (4 points) Compute the value of the line integral R C F · d R from (3 , 0 , 0) to (3 , 0 , 1) along the helix R ( t ) = (3 cos (2 πt ) , 3 sin (2 πt ) , t ). R C F · d R = R C f · d R = f (3 , 0 , 1) - f (3 , 0 , 0) = 18. 2. The function R : [ - π, π ] × [ - π 2 , π 2 ] R 3 ( u, v ) 7→ 5(cos v cos u, cos v sin u, sin v ) is a parameterization of the sphere of radius 5 centered at (0 , 0 , 0). (a) (4 points) Compute R ∂u × R ∂v , the normal vector to the parameterized sphere. R ∂u = 5( - cos v sin u, cos v cos u, 0) R ∂v = 5( - sin v cos u, - sin v sin u, cos v ) R ∂u × R ∂v = 25 ± ± ± ± ± ± i j k - cos v sin u cos v cos u 0 - sin v cos u - sin v sin u cos v ± ± ± ± ± ± = 25 cos v (cos v cos u, cos v sin u, sin v ) (b) (4 points) Let F ( x, y, z ) = (0 , x 2 , 0). Find the value of the surface integral RR S F · d S over the
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Unformatted text preview: sphere of radius 5 centered at (0 , , 0). (Hint: You may wish to compute F rst.) F = y x 2 = 0. By the Divergence Theorem, RR S F d S = RRR V F dV = 0. 3. Consider the iterated integral Z 2 y =0 Z 4-y 2 x =0 Z 2 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 2, 0 2 , 0 z 2. Thus, the iterated integral is written in cylindrical coordinates as Z 2 =0 Z 2 =0 Z 2 z =0 z dz d d. (b) (4 points) Compute the value of the integral. " Z 2 d # Z 2 2 d Z 2 z dz = 2 8 3 2 = 8 3 ....
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