15_2_22 - = r 2 cos r From the first equation r3(r = cos C 3 The second equation then gives C r3 sin = = r 1 sin 3 This equation can be solved for

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22. If F = r 2 cos θ ˆ r + α r β sin θ ˆ = φ( r , θ) , then we must have ∂φ r = r 2 cos θ, 1 r ∂φ ∂θ = α r β sin θ. From the first equation φ( r , θ) = r 3 3 cos θ + C (θ). The second equation then gives C 0 (θ) - r 3 3 sin θ = ∂φ ∂θ = α r β + 1 sin θ. This equation can be solved for a function C (θ) independent of
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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