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exam4sol - 18.02 Exam 4 Solutions Problem 1/2 1 1 0 0 r2 r...

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