Consider the region shared by ρ=8cos(φ) and ρ=4. Find the volume of the region.

Question

Consider the region shared by ρ=8cos(φ) and ρ=4. Find the volume of the region.

I know it involves a triple integral, but do not understand how to set up the integral.

Answer 1

In order to appeal to geometric intuition, let's transform the equations back to cartesian coordinates and find out what is happening. Take $\rho = 8 \cos (\varphi)$ and multiply it by $\rho$, getting $\rho^2 = 8 \rho \cos(\varphi)$. If we consider the spherical coordinate change as given by

$$\begin{cases} x = \rho \cos \theta \sin \varphi, \\ y = \rho \sin \theta \sin \varphi, \\ z = \rho \cos \varphi, \end{cases}$$

we can write it as $x^2 +y^2 +z^2 = 8z$. This is simply a sphere with its center moved to the point $(0,0,4)$, since the equation can be written as $x^2 +y^2 +(z-4)^2 = 4^2$. As for the equation $\rho =4$ we have $x^2 +y^2 +z^2 = 4^2$.

Looks like we have to break this triple integral into two regions.

Region one: The volume of the top region is given by the following inequalities:

$$\begin{cases} 0 \leq \rho \leq 4, \\ 0 \leq \theta \leq 2 \pi, \\ 0 \leq \varphi \leq \frac{\pi}{6}. \end{cases} $$

Region two: The volume of the lower half is given by the following inequalities:

$$\begin{cases} 0 \leq \rho \leq 8 \cos \varphi, \\ 0 \leq \theta \leq 2 \pi, \\ \frac{\pi}{6} \leq \varphi \leq \frac{\pi}{2}. \end{cases}$$

Therefore the volume of the intersection of the spheres is given by

$$V = \int_0^{2 \pi} \int_0^{\pi/6} \int_0^4 \rho^2 \sin \varphi \, d \rho \, d \varphi \, d \theta + \int_0^{2 \pi} \int_{\pi/6}^{\pi/2} \int_0^{8 \cos \varphi} \rho^2 \sin \varphi \, d \rho \, d \varphi \, d \theta.$$

I managed to plot it in Mathematica to help visualization: