A segment of a spherical surface is defined by the intersection of a sphere, centered at the origin, and a pair of planes that are parallel to the xz plane. The surface does not include the bases of the segment.
My approach to calculating the average distance of the surface to the z axis is to set up a surface integral for the problem.
I know the radius of the sphere, $\rho$ and the radii of the spherical segment bases, $r_1$ and $r_2$.
The bounds for $\theta$ and $\phi$ will therefore be;
$\sec (\frac{r_1}{\rho})\leqslant\theta\leqslant \sec (\frac{-r_1}{\rho})$
$\sec (\frac{r_1}{\rho})\leqslant\phi\leqslant \sec (\frac{-r_1}{\rho})$
The distance, d parallel to the xy plane from any point on the surface to the z axis will be; $d^2=x^2+y^2$
I convert that to spherical coordinates and get $d=\sqrt{\rho^2\sin^2\phi}$
I think the surface integral will be;
Average distance = $\cfrac{1}{Area(S)} \iint_{S} \sqrt{\rho^2\sin^2\phi}\ dS$
The total surface area of the segment $A =2\pi\rho h$ and the height of the segment, h will be $h =\sqrt {\rho^2-r_2^2} - \sqrt {\rho^2-r_1^2}$
The area element for my integral will be $dS=\rho^2\sin\phi\ d\phi\ d\theta$
So the surface integral becomes;
$\cfrac{1}{2\pi\rho h} {\displaystyle\int_{\sec (\frac{r_1}{\rho})}^{\sec (\frac{-r_1}{\rho})}}{\displaystyle\int_{\sec (\frac{r_1}{\rho})}^{\sec (\frac{-r_1}{\rho})}} \sqrt{\rho^2\sin^2\phi}\ \rho^2\sin\phi\ d\phi\ d\theta$
Is this the right approach to give me the desired result, and have I set up the integral correctly? Thanks.