I am trying to find the volume within the sphere $x^2+y^2+z^2=4$, above the $xy-$plane, and below the cone $z=\sqrt{x^2+y^2}$.
Working in spherical coordinates, I came up with the following bounds: $0\leq p \leq 2$, $0\leq \theta\leq 2\pi$, $0\leq \phi\leq \frac{\pi}{2} $.
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_{0}^{2}p^2sin(\phi) \,\,dp\,d\theta\,d\phi$$
$$\int_{0}^{\frac{\pi}{2}}sin(\phi)\,\,d\phi\,\,\,\times \int_{0}^{2\pi}d\theta\,\,\,\times \int_{0}^{2} p^2\,\,dp$$
$$=\left.-cos\phi\right|_{0}^{\frac{\pi}{2}} \times \left.\theta\right|_{0}^{2\pi}\times \left.\frac{p^3}{3}\right|_{0}^{2}=\frac{16\pi}{3}$$
Have I done this correctly? I'm not sure if the $\phi$ bound is correct $-$ the maximum for the angle would be $\frac{\pi}{2}$ because the $xy-$plane is a constraint, but I don't know if the cone would prevent the angle from getting to zero (the $z-$axis). Any help would be appreciated.
Best Answer
The region is bound above by the cone and below by the xy-plane.
Equation of the cone is $z = \sqrt{x^2+y^2}$
$\implies \rho \cos\phi = \rho \sin\phi \implies \phi = \frac{\pi}{4} \ $ is the lower bound of $\phi$.
So integral should be,
$ \displaystyle \int_{\pi/4}^{\pi/2}\int_{0}^{2\pi}\int_{0}^{2} \rho^2 \sin\phi \,\,d\rho\,d\theta\,d\phi$