Hi, I need to find the volume of the solid that lies above the cone with equation (in spherical coordinates) [tex] \phi = \frac{\Pi}{3} [/tex] and inside the torus with equation [tex] \rho = 4\sin\phi [/tex]. I thought that the bounds are: [tex] 0\leq\rho\leq4\sin\phi[/tex], [tex]\frac{\Pi}{3}\leq\phi\leq\frac{\Pi}{2}[/tex], and [tex]0\leq\theta\leq2\Pi[/tex] but when I evaluated the integral (using Mathematica) of [tex]\rho^2\sin\phi[/tex] (the Jacobian) using these bounds I got the wrong answer. Any help would be greatly appreciated.(adsbygoogle = window.adsbygoogle || []).push({});

Thanks.

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# Homework Help: Triple Integral

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