$R$ is a region in the first octant of $\mathbb{R^3}$ that is bounded by $x^2+y^2+z^2=4$ and $z=\sqrt{x^2+y^2}$. Find the triple integral that would give the volume of $R$.
I know how to set this up in Cartesian form and Cylindrical form (thanks to a person that answered my previous question), and now I want the integral in spherical coordinates.
Recall that $x=\rho\sin\theta\cos\theta$, $y=\rho\sin\phi\sin\theta$, and $z=\rho\cos\phi$
Naturally, we have $0 \leq \phi \leq \pi$ and $0 \leq \theta \leq 2\pi$. For $\rho$, we take the radius $0 \leq \rho \leq 2$. Is this the correct set of boundaries for the integral?
Best Answer
First of all note that the question is ambiguous. While it is likely the question asks you to find the volume bound inside the cone and the sphere, it must use inequality signs to clearly state the region you have to find volume of. For the given surfaces, there are two regions bound in first octant - one which is inside the cone and one which is outside -
$(1)~$ Inside the cone: $x^2 + y^2 + z^2 \leq 4, z \geq \sqrt{x^2+y^2}$
$(2)~$ Outside the cone: $x^2 + y^2 + z^2 \leq 4, z \leq \sqrt{x^2+y^2}$
Using $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$, first find the equation of cone in spherical coordinates -
$z = \rho \cos\phi = \sqrt{x^2+y^2} = \rho \sin\phi$
$ \implies \tan \phi = 1 ~$ i.e. $\phi = \dfrac{\pi}{4}$
So for region inside the cone: $0 \leq \phi \leq \pi/4$
For region outside the cone in the first octant: $\pi/4 \leq \phi \leq \pi/2$
Also in first octant, $0 \leq \theta \leq \pi/2$
Bounds of $\rho$ is as you stated.
So the integral to find volume inside the cone is,
$ \displaystyle \int_0^{\pi/2} \int_0^{\pi/4} \int_0^2 \rho^2 \sin \phi ~ d\rho ~ d\phi ~ d\theta$
If it is volume outside the cone in first octant, then the integral will be
$ \displaystyle \int_0^{\pi/2} \int_{\pi/4}^{\pi/2} \int_0^2 \rho^2 \sin \phi ~ d\rho ~ d\phi ~ d\theta$