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Let Q be the region above by the plane $8z=4-x-y$ and below by the cone $64z^2=x^2+y^2$. How would I setup the triple integral to find volume of Q, using spherical coordinates? I just need help with the setup part, I will do the rest.

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1 Answer 1

up vote 2 down vote accepted

This is not going to be a nice integral, but if you insist :-), here's how to set it up:

With

$$ \begin{align} x&=r\sin\theta\cos\phi\;,\\ y&=r\sin\theta\sin\phi\;,\\ z&=r\cos\theta\;, \end{align} $$

the equation of the cone becomes

$$ 64r^2\cos^2\theta=r^2\sin^2\theta\,(\cos^2\phi+\sin^2\phi) $$

and thus

$$\theta=\arccos\frac1{\sqrt{65}}\;.$$

The equation of the plane becomes

$$ 8r\cos\theta=4-r\sin\theta\cos\phi-r\sin\theta\sin\phi\;, $$

and solving for $r$ yields

$$ r=\frac4{8\cos\theta+\sin\theta(\cos\phi+\sin\phi)}\;. $$

Thus the volume is given by the integral

$$ \int_0^{2\pi}\mathrm d\phi\int_0^{\arccos1/\sqrt{65}}\mathrm d\theta\sin\theta\int_0^{4/(8\cos\theta+\sin\theta(\cos\phi+\sin\phi))}\mathrm drr^2\;. $$

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