I want to find the volume of the solid within the sphere $x^2+y^2+z^2=16$ above the xy-plane and outside the cone $z=3\sqrt{x^2+y^2}$
I'm not sure if I'm interpreting this correctly, but this is basically a sphere with a triangular cross section cut out and the base is flat, right?
My biggest issue is coming up with the bounds. if we're above the xy plane, then $z\geq 0$. The projection on the xy-plane is given by $x^2+y^2=16$, a circle with radius 4. The projection on the yz-plane is given by $z=3\sqrt{y^2}=3\vert y\vert$
Constructing a triple integral in spherical coordinates I need my bounds for $\rho, \phi,$ and $\theta$. The most I know is $0\leq \rho \leq 4$. I don't know what the other angle measures should be since I'm taking the solid outside the cone.
Best Answer
To find equation of the cone in spherical coordinates,
$z=3\sqrt{x^2+y^2} \implies \rho \cos\phi = 3 \rho \sin\phi$
i.e. $~\tan \phi = \dfrac13 ~$ where $\phi$ is the polar angle (angle made with z-axis).
As we have to find the volume outside of the cone, $\phi = \arctan(1/3)$ is the lower bound of $\phi$.
The upper bound of $\phi$ is $\dfrac{\pi}{2}$ as we are above xy plane.
If you see the projection of the region in xy-plane, it covers all $4$ quadrants so $0 \leq \theta \leq 2\pi$.
So the integral should be,
$ \displaystyle \int_0^{2\pi} \int_{\arctan(1/3)}^{\pi/2} \int_0^4 \rho^2 \sin \phi ~ d\rho ~ d\phi ~ d\theta$