volume of spherical cap and spherical sector

Theorem 1.  The volume of a spherical cap is  $\pi h^2\left(r-\frac{h}{3}\right)$,  when $h$ is its height and $r$ is the radius of the sphere.

Proof.  The sphere may be formed by letting the circle  ${(x-r)}^2+y^2=r^2$,  i.e. $y=(\pm )\sqrt{rx-x^2}$,  rotate about the $x$-axis.  Let the spherical cap be the portion from the sphere on the left of the plane at  $x=h$  perpendicular to the $x$-axis.

Then the for the volume of solid of revolution yields the volume in question:

$$V=\pi {\int }_0^h{(\sqrt{rx-x^2})}^2𝑑x=\pi {\int }_0^h(2rx-x^2)𝑑x=\pi \underset{x=0}{\overset{h}{/}}\left(r{x}^2-\frac{x^3}{3}\right)=\pi h^2\left(r-\frac{h}{3}\right).$$

Theorem 2.  The volume of a spherical sector is  $\frac{2}{3}\pi r^{2}h$,  where $h$ is the height of the spherical cap of the spherical sector and $r$ is the radius of the sphere.

Proof.  The volume $V$ of the spherical sector equals to the sum or difference of the spherical cap and the circular cone depending on whether    or  $h>r$.  If the radius of the base circle of the cone is $\varrho$, then

But one can see that both expressions of $V$ are identical.  Moreover, if $c$ is the great circle of the sphere having as a diameter the line of the axis of the cone and if $P$ is the midpoint of the base of the cone, then in both cases, the power of the point $P$ with respect to the circle $c$ is

$${\varrho }^2=(2r-h)h.$$

Substituting this to the expression of $V$ and simplifying give  $V=\frac{2}{3}\pi r^{2}h$,  Q.E.D.