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Evaluate the triple integral of f(xyz)=z(x^2+y^2+z^2)^(−3/2) over the part of the ball x^2+y^2+z^2 less than or equal to64 defined by z greater than or equal to 4
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Converting the ball to cylindrical coordinates we get\[r^2+z^2\le64\]taken together with the fact that z is greater than or equal to 4 we get\[4\le z\le\sqrt{64-r^2}\]to find the region under the solid we need to look at the circle that composes the underside of the cap of the sphere, which is the circle at z=4\[r^2+4^2\le64\to r^2\le48\]\[0\le r\le4\sqrt3\]and because it is a circle we have\[0\le\theta\le2\pi\]so the integral becomes\[\int\int\int f(x,y,z)rdzdrd\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt3}\int_{4}^{\sqrt{64-r^2}}\frac{zr}{(r^2+z^2)^{3/2}}dzdrd\theta\]
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