prelim3_spring2006solutions - Math 192 Prelim 3 Solutions,...

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Math 192 Prelim 3 Solutions, April 27, 2006 1. i) V = 2 Z 2 π 0 Z 2 1 Z 2 - r 2 0 dz r dr dθ ii) V = Z 2 π 0 Z π/ 4 0 Z 2 secφ sec φ ρ 2 sin φ dρ dφ dθ 2. ~ r ( t ) = ( t 2 - 1) ~ j + 2 t ~ k d~ r dt = 2 t ~ j + 2 ~ k → | d~ r dt | = 2 t 2 + 1. Mass = R C δds = R 1 0 (3 t/ 2)2 t 2 + 1 dt = ( t 2 + 1) 3 / 2 ] 1 0 = 2 2 - 1. 3. d~ r dt = cos t ~ i - sin t ~ j + 1 6 ~ k. R C ± ~ F · d~ r dt ² dt = R π 0 ( t cos t - sin t cos 2 t + 2 sin t ) dt = 4 / 3. 4. M = Axz + sin y, N = Bx cos y, P = x 2 . Therefore, ∂P ∂y = 0 = ∂N ∂z . Set ∂M ∂z = Ax = ∂P ∂x = 2 x A = 2. Set ∂N ∂x = B cos y = ∂M ∂y = cos y B = 1. 5. i) ∂f ∂x = M = yz + e x cos y f = xyz + e x cos y + g ( y, z ). Then ∂f ∂y = xz - e x sin y + ∂g ∂y = N = xz - e x sin y ∂g ∂y = 0 g = h ( z ) f = xyz + e x cos y + h ( z ) ∂f ∂z = xy + h 0 ( z ) = P = xy + z h 0 ( z ) = z h ( z ) = z 2 2 + C f = xyz + e x cos y + z 2 2 + C . ii) Work = f (1 , π, 2) - f (0 , 0 , 0) = (2 π - e + 2) - 1 = 2 π - e + 1. 6. The given surface is a level surface of g ( x, y, z ) = y 2 + z 2 → ∇ g = 2 y ~ j + z 2 ~ k . We project S onto the xy-plane and so ~ p = ~ k, |∇ g · ~ k | = | 2 z | = 2 z, since z 0 . Now, ( ~ F
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This note was uploaded on 06/01/2008 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).

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