I'm trying to find the volume between the spheres:
$x^2 + y^2 + z^2 = 9$
and:
$x^2 + y^2 + (z-2)^2 = 9$
I have calculated this, but have a strong feeling that little of what I did was actually correct. The main problem is with building the integral in spherical coordinates – I think I am missing something and did not take into account the intersection between the spheres, for example ($x^2 + y^2 = 8$).
So my line of thought was that $\rho$ can go from $0$ to the radius of the sphere $(3)$, $\phi$ goes from $0$ (top of the sphere) to $\pi/2$ (when $x, y, z = 0$) and $\theta$ from $0$ to $2\pi$.
The integral therefore looks like this:
$$\int_{0}^{3} \int_{0}^{\pi/2} \int_{0}^{2\pi} \rho\sin\phi \,d\theta \,d\phi \,d\rho$$
which results in $18\pi$.
Is this correct? If not what am I doing wrong?
Best Answer
Consider this figure which shows the two spheres and a plane at $z = 1$:
Find the volume of a "cap" of the yellow sphere (centered at the origin) above the blue plane. Multiply that volume by 2.0 due to symmetry, for the cap of the red sphere below the blue plane.
The differential volume of a thin slice disk of the yellow sphere is:
$\pi r^2$, where $r$ is measured perpendicular to the vertical ($z$) axis. Pythagoras tells us that $3^2 - z^2 = r^2$. (This incorporates your request to use polar coordinates.)
So your integral is
$\int\limits_{z=1}^3 \pi (3^2 - z^2) dz = {28 \pi \over 3}$.
But don't forget to multiply by 2.0 for the red cap beneath the blue plane:
$V = {56 \pi \over 3}$.