Volume Between Spheres – Spherical Coordinates

Question

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?

Answer 1

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:

$V = {56 \pi \over 3}$.

Answer 2

The figure illustrate the section of the two spheres in the plane $x-z$, that is the same as in the plane $y-z$.

Your volume is the double of the volume of a spherical cup that has radius $a= \overline{PA}$ and height $h=\overline {PD}$.

Intersecting the two sphere we can easily see that $P=(0,0,1)$, so, since $C=(0,0,2)$ we have:

$a=2\sqrt{2} \qquad h=2$

Now we can calculate the volume without integrals using the fact that the volume of a spherical cup is $$ V_{Cup}=\frac{\pi}{6}h(3a^2+h^2) $$ So we have: $$ V=2V_{Cup}=\frac{\pi}{3}h(3a^2+h^2)=\frac{56}{3}\pi $$

If you want to use an integral, given the symmetry around the $z$ axis, it is better to use cylindrical coordinates: $$ r^2=x^2+y^2 \qquad z \qquad \theta = \arccos \frac{x}{r} $$

In this case the limits of integration are: $$ 1\le z\le \sqrt{9-r^2} \quad 0 \le r \le 2\sqrt{2} \quad 0 \le \theta \le 2 \pi $$

and the volume is : $$ V=\int_0^{2\pi} \int_0^{2\sqrt{2}}\int_1^{\sqrt{9-r^2}}r\,dz\,dr\, d\theta $$

Answer 3

Thanks everyone for answering! I will mark someone else's response as solved, but since I wanted to do it with spherical coordinates, I figured it out. In fact, only the $\rho$ in the original integral was wrong: it needs to be calculated from $-1$ to $3$, so the solution is as follows:

$$\int_{-1}^{3} \int_{0}^{\pi/2} \int_{0}^{2\pi} \rho\sin\phi \,d\theta \,d\phi \,d\rho = 2\pi\int_{-1}^{3} \int_{0}^{\pi/2} \rho\sin\phi \,d\phi \,d\rho = -2\pi \int_{-1}^{3}(\rho^2cos\phi)\Big|_0^{\pi/2} \,d\rho =$$ $$= -2\pi \int_{-1}^{3}-\rho^2 \,d\rho = 2\pi \int_{-1}^{3} \rho^2 \,d\rho = \frac{2\pi}{3}(\rho^3)\Big|_{-1}^{3} = \frac{2\pi}{3}(27-(-1)) = \frac{56\pi}{3}.$$