evaluate triple integral (x^2 + y^2 +z^2)^(-3/2) dV where D is the solid between two spheres of radius 2 and 6 centered at the origin
Added by Crystal C.
Step 1
Recall that in spherical coordinates, we have: $x = \rho \sin \phi \cos \theta$ $y = \rho \sin \phi \sin \theta$ $z = \rho \cos \phi$ and the Jacobian is $\rho^2 \sin \phi$. The given integral becomes: $\int_{0}^{2\pi} \int_{0}^{\pi} \int_{2}^{6} Show more…
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