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13 Nov 2019

Solve in spherical coordinates for ∫ ∫ ∫ z dV bounded above by the sphere x^2+y^2+z^2=2 and below by the cone z = (x^2+y^2)^1/2 (in spherical coordinates this is phi = pi/4)

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Irving Heathcote
Irving HeathcoteLv2
5 Oct 2019

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