# Volume of bounded region between sphere and a plane

A solid spherical cap of radius $R$ is the solid delimited by a sphere of radius $R$ and a plane that intersects the sphere.

The height $h$ of the solid spherical cap is the largest distance from a point of the solid to the plane intersecting the sphere.

Show that the volume of a solid spherical cap is given by:

$$V = \frac{\pi h^2(3R-h)}{3}$$

It is enough to consider the solid spherical cap delimited by the sphere $x^2 + y^2 + z^2 = R^2$ and the plane $z = R-h$

use triple integrals by spherical or cylindrical coordinates. Give me a few min to try it out.

You can use Cartesian coordinates and integrate disks. Let the plane be the $xy$ plane and the center at $(0,0,h-R)$ Then your volume is $x^2+y^2+(z+R-h)^2 \le R^2,\ z \ge 0$, $x$ runs over $\pm \frac {\sqrt{R^2-(R-h)^2}}R$ and $y$ does the same with an $x^2$ under the radical.
Or you can use cylindrical coordinates. Again, put the center on the $z$ axis at $h-R$. Now you have $r^2+(z+R-h)^2 \le R^2, z \ge 0$. The $\theta$ integral is easy due to symmetry and $r$ runs from $0$ to $\arcsin \frac {\sqrt{R^2-(R-h)^2}}R$