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prac4Bsol

# prac4Bsol - 18.02 Practice Exam 4B Solutions Problem 1/2 0...

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18.02 Practice Exam 4B – Solutions Problem 1. ¡ ¡ ¡ ¡ Z π/ 2 0 Z 1 0 Z 1 0 r 2 r dz dr dθ . Problem 2. ¡ ¡ a) sphere: ρ = 2 a cos φ . b) plane: ρ = a sec φ . c) Z 2 π 0 Z π/ 4 0 Z 2 a cos φ a sec φ ρ 2 sin φ dρ dφ dθ . Problem 3. a) ∂y (2 xy + z 3 ) = 2 x = ∂x ( x 2 + 2 yz ); ∂z (2 xy + z 3 ) = 3 z 2 = ∂x ( y 2 + 3 xz 2 - 1); ∂z ( x 2 + 2 yz ) = 2 y = ∂y ( y 2 + 3 xz 2 - 1); so ~ F is conservative. ¡ ¡ ¡ ¡ p ( x 1 , y 1 , z 1 ) C 1 C 2 C 3 b) Method 1: f ( x, y, z ) = R C 1 + C 2 + C 3 ~ F · d~ r ; R C 1 ~ F · d~ r = R x 1 0 (2 xy + z 3 ) dx = R x 1 0 0 dx = 0 ( y = 0, z = 0) R C 2 ~ F · d~ r = R y 1 0 ( x 2 + 2 yz ) dy = R y 1 0 x 2 1 dy = x 2 1 y 1 ( x = x 1 , z = 0) R C 3 ~ F · d~ r = R z 1 0 ( y 2 + 3 xz 2 - 1) dz = R z 1 0 ( y 2 1 + 3 x 1 z 2 - 1) dz = y 2 1 z 1 + x 1 z 3 1 - z 1 ( x = x 1 , y = y 1 ) So f ( x, y, z ) = x 2 y + y 2 z + xz 3 - z + c . Method 2: ∂f ∂x = 2 xy + z 3 , so f ( x, y, z ) = x 2 y + xz 3 + g ( y, z ). ∂f ∂y = x 2 + ∂g ∂y = x 2 + 2 yz , so ∂g ∂y = 2 yz . Therefore g ( y, z ) = y 2 z + h ( z ), and f ( x, y, z ) = x 2 y + xz 3 + y 2 z + h ( z ). ∂f ∂z = 3 xz 2 + y 2 + h 0 ( z ) = y 2 + 3 xz 2 - 1, so h 0 ( z ) = - 1.
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