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Doss [256]
3 years ago
8

Use cylindrical coordinates. find the volume of the solid that is enclosed by the cone z = x2 + y2 and the sphere x2 + y2 + z2 =

72.
Mathematics
1 answer:
sashaice [31]3 years ago
7 0
Let R be the solid. Then the volume is

\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta

which follows from the facts that

\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta
(by computing the Jacobian)

and

z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8
(we take the positive solution, since it's clear that R lies above the x-y plane)
r^2+z^2=72\implies z=\pm\sqrt{72-r^2}
(again, taking the positive root for the same reason)
z=r^2\implies 8=r^2\implies r=\sqrt8

\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta
=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr
=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr
=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr
=\dfrac{32(27\sqrt2-35)\pi}3
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