Calculate the surface area of the part of the sphere
Calculate the surface area of the part of the sphere $x^2+y^2+z^2=a^2$ cut by the cylinder $x^2+y^2=b^2$, where $b<a$.
Explanation of Method
For the upper hemisphere $z=\sqrt{a^2-x^2-y^2}$, the classical identity $1+z_x^2+z_y^2=\dfrac{a^2}{a^2-x^2-y^2}$ makes the surface-area integrand radially symmetric and elementary in polar coordinates. Doubling for both hemispheres gives the total (this is the well-known "spherical zone" formula, depending only on the zone's height).
- Compute the surface-area integrand for the upper hemisphere.
- Convert to polar coordinates over the disk of radius $b$.
- Evaluate and double for both hemispheres.
Step 1 · Surface-area integrand
$z_x=\dfrac{-x}{\sqrt{a^2-x^2-y^2}},\,z_y=\dfrac{-y}{\sqrt{a^2-x^2-y^2}}$, so $1+z_x^2+z_y^2=\dfrac{a^2}{a^2-x^2-y^2}$, giving integrand $\dfrac a{\sqrt{a^2-x^2-y^2}}$.
Step 2 · Polar coordinates
$$ S_{\text{upper}} = \int_0^{2\pi}\int_0^b\frac a{\sqrt{a^2-r^2}}\,r\,dr\,d\theta = 2\pi a\int_0^b\frac r{\sqrt{a^2-r^2}}\,dr. $$
Step 3 · Evaluate
With $u=a^2-r^2$: $\displaystyle\int_0^b\frac r{\sqrt{a^2-r^2}}\,dr = a-\sqrt{a^2-b^2}$. So $S_{\text{upper}}=2\pi a\big(a-\sqrt{a^2-b^2}\big)$, and doubling for both hemispheres:
$$ S = 4\pi a\big(a-\sqrt{a^2-b^2}\big). $$
Additional Notes
- This derivation matches the source exactly, and is the classical formula for the surface area of a spherical zone: it depends only on the sphere's radius $a$ and the zone's height $h=2\big(a-\sqrt{a^2-b^2}\big)$, i.e. $S=2\pi ah$.
- Verified numerically: direct polar double integration (e.g. $a=2.0,\,b=1.2$) gives exactly $4\pi a(a-\sqrt{a^2-b^2})$.
Result
The surface area is: