Q:

Volume of a Sphere Segment Cut Out by Zigzag Motion of Radius Vector Through Sphere Center

A closed loop curve on a sphere radius $R$ enclosing curved area $A$ subtends a solid angle $ A/R^2$ steridians at center of the sphere. Volume $V$ is enclosed by concurrent straight generators between the center and loop edge.
Prove that such a solid has volume $ V = A\, R/3. $

A:

I use the following conventions for spherical coordinates 
$\qquad \qquad \qquad \qquad$
Suppose that the origin of the coordinate system is located at the center of the sphere. Hence, by constructing a triple integral for the volume in spherical coordinates we have 
$$\begin{align}
V&=\iiint dv \\
 &=\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)}\int_{r=0}^{R}r^2 \sin \theta dr d\theta d\phi \\
&=\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} \left. \frac{r^3}{3}\right|_{0}^{R} \sin \theta d\theta d\phi\\
&=\frac{R^3}{3} \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} \sin \theta  d\theta d\phi \\
&=\frac{R}{3} \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} R^2 \sin \theta  d\theta d\phi \\
&=\frac{R}{3}\iint_{A}da \\
&=A\frac{R}{3}
\end{align}$$
That's it! :)

