Find the volume of the solid within the sphere $x^2+y^2+z^2=9$, outside the cone $z=\sqrt{x^2+y^2}$, and above the $xy$-plane.

Using Cylindrical coordinates,
$r^2+z^2=9$ and $z=r$. Intersection is $x^2+y^2=\frac{9}{2}$

\begin{align} V=\int_0^{2\pi}\int_{\frac{3}{\sqrt 2}}^{3}\int_{0}^{\sqrt{9-r^2}}r\:dz\:dr\:d\theta+\int_0^{2\pi}\int_0^{\frac{3}{\sqrt2}}\int_0^r r\:dz\:dr\:d\theta \end{align} Again Using Spherical coordinates,
I can't figure out $\phi$ and $\rho$. I just need those limit anyone can skip evaluation. And is my first approach correct$?$ Any help will be appreciated.
Thanks in advances.

  • $\begingroup$ Note that $\sqrt{9/2}$ is not $3/2$ $\endgroup$ – Empy2 Dec 21 '19 at 17:18
  • $\begingroup$ @Empy2 it's a typo. thanks for mentioning $\endgroup$ – Dr.Antidode Dec 21 '19 at 17:19

Using the following substitutions for spherical coordinates: $$z = \rho \cos(\phi)$$ $$x = \rho \sin(\phi)\cos(\theta)$$ $$y = \rho \sin(\phi)\sin(\theta)$$ \begin{align} \rho \cos(\phi)&=\sqrt{\rho^2 \sin^2(\phi)\cos^2(\theta)+\rho^2 \sin^2(\phi)\sin^2(\theta)}\\ &=\sqrt{\rho^2\sin^2(\phi)}\\ \phi&=\frac{\pi}{4} \end{align} Above the $xy$-plane hence $\phi\leq \frac{\pi}{2}\implies \frac{\pi}{4}\leq\phi\leq\frac{\pi}{2}$ $$\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_0^{3}\rho^2 \sin(\phi) d\rho d\phi d\theta$$You can think the volume is whole upper hemisphere except the cone that's why $0\leq \theta\leq 2\pi$ and $0\leq \rho\leq 3$

  • $\begingroup$ Your last comment give me intuition of $\rho$ and $\phi$ @emonHR thanks. Did I am correct in Cylindrical coordinate$?$ $\endgroup$ – Dr.Antidode Dec 21 '19 at 17:56

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