Area of smaller part of surface

Question

The plane $y=2$ cuts the sphere $x^{2}+y^{2}+z^{2}=8$ into two parts.

To find the area of the smaller part I notice that the intersection of the plane and the sphere gives $x^{2}+z^{2}=4$ so the region of integration is $x^{2}+z^{2}\leq 4$ and $y=\sqrt{8-x^{2}-z^{2}}$ which leads to the integral

$$\int\int\sqrt{y_{x}^{2}+y_{z}^{2}+1}=\int_{0}^{2\pi}\int_{0}^{2}r\sqrt{\frac{3r^{2}+8}{8-r^{2}}}drd\theta$$

(here $y_{x}$ denotes the derivative of $y$ with respect to $x$ and similarly for $y_{z}$, and I used polar coordinates in the $x,z$ plane)

This integral seems too hard which makes me think it is either wrong or there is an easier way to go about it?

Answer 1

Something is your with the integrand. $$y_x=-\frac{x}{\sqrt{8-r^{2}}}, \quad y_z=-\frac{z}{\sqrt{8-r^{2}}}\Rightarrow y_{x}^{2}+y_{z}^{2}+1=\frac{r^2}{8-r^{2}}+1=\frac{8}{8-r^{2}}.$$ Hence $$S=\int\int\sqrt{y_{x}^{2}+y_{z}^{2}+1}=\int_{0}^{2\pi}\int_{0}^{2}r\sqrt{\frac{8}{8-r^{2}}} dr d\theta\\ =-\pi\sqrt{8}\int_{0}^{2}\frac{d(8-r^2)}{\sqrt{8-r^{2}}} =-\pi\sqrt{8}\left[2\sqrt{8-r^{2}}\right]_{0}^{2} =8\pi(2-\sqrt{2}).$$ This result is confirmed by the formula for the area of the spherical cap $S=2\pi rh$, where $r=2\sqrt{2}$ and $h=r-2$.

Answer 2

A hint: This area is equal to the area of a cylinder circumscribed to the sphere and having the same height as your spherical cap.

Answer 3

Hint:

a simpler way is to use the fact that the area of a spherical cap is: $$ A=2\pi rh $$

where $r$ is the radius of the sphere and $h$ is the height of the cap.