Volume between two spheres

Question

Compute the volume enclosed by the two following spheres:

$$\left\{\begin{aligned}x^2+y^2+z^2=4\\ x^2+(y+2)^2+z^2=4\end{aligned}\right.$$

Using spherical coordinates seemed too hard for this question, so I tried to use polar coordinates around the $y$ axis:

$$\begin{cases} x=r\cos\theta\\ y=y\\ z=r\sin\theta \end{cases}$$

I got the following inequalities:

$$\left\{\begin{aligned}-\sqrt{4-r^2}\leq \ &y\leq\sqrt{4-r^2}\\ -2-\sqrt{4-r^2}\leq \ &y\leq-2+\sqrt{4-r^2}\end{aligned}\right.$$

But I don't know how to integrate them together. Maybe there's an easier way to do this? I thought that maybe because the volume is enclosed by two spheres, it is actually a two dimensional circle rotating about some axis. Problem is, I couldn't figure out which one; also I am not very experienced solids of revolution, and the real goal here is to practice on triple integrals.

Thanks!

Answer 1

EDIT

After having posted my solution I found that he problem is well known (at least since 1948, but most probably much longer), see e.g. http://mathworld.wolfram.com/Sphere-SphereIntersection.html

Original post

If your aim was to calculate the volume of the intersection of the two spheres (of radius $r=2$) you were very close to a solution as will be shown here.

Changing to cylindrical coordinates $x\to r \cos(\phi)$, $y\to y$, $z\to r \sin(\phi)$, $dx dy dz \to r dr dy d\phi $ the conditions transform to

$$4\geq x^2+y^2+z^2 = r^2+y^2 \\\to y^2 \leq 4-r^2 $$

giving

$$ -\sqrt{4-r^2}\leq y \leq \sqrt{4-r^2} \tag{1}$$

and

$$4\geq (x^2+(y+2)^2+z^2= r^2+(y+2)^2 \\\to (y+2)^2 \leq 4-r^2$$

giving

$$ -\sqrt{4-r^2}-2\leq y\leq \sqrt{4-r^2}-2\tag{2}$$

Combining carefully $(1)$ and $(2)$ taking each time the stronger inequality for $y$ we get

$$-\sqrt{4-r^2}\leq y\leq \sqrt{4-r^2}-2\tag{3}$$

From $(3)$ we deduce that the $y$-integral is just the length of the $y$-interval, i.e.

$$i_y = 2 \sqrt{4-r^2}-2$$

The range for $r$ follows from $(3)$ to be $0\leq r\leq \sqrt{3}$.

The $\phi$-integrals is just $2 \pi$, and we are left with the integral

$$v = 2 \pi \int_0^\sqrt{3} r i_y \, dr = 4 \pi \int_0^\sqrt{3} r ( \sqrt{4-r^2}-1) \, dr = \frac{10}{3}\pi \simeq 10.472$$

The volume of the union of the two spheres is then $2 \frac{4 \pi}{3} 2^3 - \frac{10}{3}\pi = 18 \pi$.

Answer 2

It's not clear whether you want the union or the intersection of the two spheres. At any rate: draw a figure! The two spheres intersect in a lens shaped object. One half of the lens is given by $$H=\bigl\{(x,y,z)\bigm| x^2+y^2+z^2\leq4, \ -2\leq y\leq-1\bigr\}\ .$$ Intersecting $H$ with a plane $y={\rm const.}$ gives a circular disc in the $(x,z)$-plane of radius $r(y)=\sqrt{4-y^2}$. Therefore we can write $${\rm vol}(H)=\pi \int_{-2}^{-1}r^2(y)\>dy=\ldots\quad.$$