Volume of the intersection between two spheres as an integral How can I find the volume of the intersection between two spheres with radii $r_1$ & $r_2$ by integration, I found a formula online but I need it in integral i.e. I want to know the limits of the integral in $r, \theta$, and $\phi$.
 A: Thanks to the symmetry of the problem, there is no need to use triple integrals and complicated coordinate parametrizations; indeed the overlapping volume is a solid of revolution. Here is how to proceed in such a situation.
Let $S_1,S_2$ be two spheres of radii $R$ and $r$ respectively and let $c\ge0$ be the distance between their centers. Without loss of generality (see Nota bene below), let's assume that $R \ge r$, that the first sphere is centered at the origin and that the center of the second one lies along the $x$-axis $-$ its center is thus at $(c,0,0)$. We would like to determine their overlapping volume $V$; three cases are possible :
a) If $c \ge R+r$ : the spheres don't overlap, whence $V=0$.
b) If $c \le R-r$ : the whole small sphere lies inside the greater one, whence  $V = \frac{4}{3}\pi r^3$.
c) If $R-r < c < R+r$ : the two spheres intersect in a circle in the $yz$-plane; let's find its $x$-coordinate; an intersection point $P(x,y,z)$  belongs to both circles, so that
$$
\begin{cases}
   R^2 = x^2+y^2+z^2 \\
   r^2\; = (x-c)^2+y^2+z^2
\end{cases}
\verb+ +\Rightarrow\verb+ +
   R^2-r^2 = x^2-(x-c)^2 = 2cx-c^2
\verb+ +\Rightarrow\verb+ +
   x_* := \frac{R^2-r^2+c^2}{2c}
$$
The spherical "profiles" of the two balls in the $Oxz$-plane are delimited by the graph of the following functions :
$$
\begin{cases}
   z = f_1(x) = \sqrt{R^2-x^2} \\
   z = f_2(x) = \sqrt{r^2-(x-c)^2}
\end{cases},
\verb+ +\mathrm{because}\verb+ + y=0 \verb+ +\mathrm{on\,this\,plane}.
$$
The overlapping volume is thus made of two solids of revolution around the $x$-axis :
$$
\begin{cases}
   \displaystyle 
   V_1 = \pi\int_{c-r}^{x_*}f_1(x)^2\mathrm{d}x = \pi\int_{c-r}^{x_*}R^2-x^2\mathrm{d}x = \pi\left[R^2x-\frac{1}{3}x^3\right]_{c-r}^{x_*} = \ldots \\
   \displaystyle 
   V_2 = \pi\int_{c-r}^{x_*}f_2(x)^2\mathrm{d}x = \pi\int_{x_*}^Rr^2-(x-c)^2\mathrm{d}x = \pi\left[r^2x-\frac{1}{3}(x-c)^3\right]_{x_*}^R = \ldots 
\end{cases}
$$
hence finally $V = V_1+V_2$, which is a quite messy polynomial expression in $R$, $r$ and $c$ that your favorite computational software will simplify for you.

Nota bene : if the centers of your spheres are already fixed, it is always possible to do a change of basis thanks to a linear transformation which will send the $x$-axis on the line passing through their centers.
