Writing triple integrals in spherical coordinates over nonspherical/nonconical regions

Question

Defining upper and lower limits of integration for $\rho$, $\theta$, and $\phi$ is relatively easy when writing a triple integral in spherical coordinates if the region of integration is defined by spheres and/or cones centered at the origin. However, this becomes more involved when the region is bounded by any other type of surface, such as a cylinder or a paraboloid.

The following is an example I came across recently, and I can't figure out how to represent the upper and lower bounds for $\rho$:

Here's what I've come up with so far:

$\iiint\limits_E\mathrm{d}V=\int_{\arcsin(\frac{1}{\sqrt{6}})}^{\pi}\int_0^{2\pi}\int_0^\sqrt{6} \rho^2 \sin(\phi)\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\phi + \int_{\pi}^{\arcsin(\frac{1}{\sqrt{6}})}\int_0^{2\pi}\int_0^{\textbf{???}} \rho^2 \sin(\phi)\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\phi$

How does one going about finding the equation for $\rho$ on the segment of the region of integration where it meets the sides of the cylinder?

Answer 1

$$x^2+y^2=1\to x^2+y^2+z^2=1+z^2\to\rho^2=1+\rho^2\cos^2\phi\to\rho^2=\frac{1}{\sin^2\phi}$$

Answer 2

Hint: see the right triangle with angle $\phi$.

Answer 3

FIRST EDIT: Accidentally took twice the area desired. Now revised.

SECOND EDIT Fixed two errors: (i) forgot Jacobian for $V_{2}$ and (ii) had to integrals flipped (with differential forms in right order)

So, here's another approach!

First thing you should note is this is azimuthally independent and we can immediately reduce this to a two-dimensional problem (maybe this helps?)

You can now look at twice the size of the region (symmetric about $z=0$) and break it up into 3 regions, two of which are the same (where $r = \sqrt{x^2+y^2+z^2} = \sqrt{6}$ over which our bounds will be invariant; that is the maximal distance in the radial direction is constant. For this problem, let's call this $V_{1}$ and for the region where this maximal distance is changing, let's call this $V_{2}$.

We get that one region has $V_{1} = \oint_{V_{1}} r^2 \sin(\theta)\; dr\; d\theta\; d\phi = \int_{0}^{2\pi}\int_{0}^{\arcsin(1/\sqrt{6})}\int_{0}^{\sqrt{6}} r^2 \sin(\theta)\; dr\; d\theta\; d\phi$ and the other $V_{2} = \int_{0}^{2\pi}\int_{\arcsin(1/\sqrt{6})}^{\pi-\arcsin(1/\sqrt{6})}\int_{0}^{\sqrt{1+6\cos^2(\theta)}}r^2 \sin(\theta)\; dr\; d\theta\; d\phi$.

We conclude by stating that $V = \frac{1}{2}\cdot (2V_{1}+V_{2})$.