Multivariable Calculus surface integral over a square

Question

Thanks for any help in advance.

I'm currently working on a question which is as follows:

Find the area of the part of the sphere of radius a at the origin which is above the square in the (x,y) plane bounded by: $$ x = \frac{a}{\sqrt{2}} , x = -\frac{a}{\sqrt{2}} , y = \frac{a}{\sqrt{2}} , y = -\frac{a}{\sqrt{2}} $$ Hint for evaluating the integral: change to polar coordinates and evaluate the $r$ integral first.

I have found the surface element in terms of spherical polar coordinates, $a^2\sin\theta$, where $\theta$ is the angle from the $z$ axis, but i am having difficulty projecting it onto the square in the $(x,y)$ plane which does not agree with spherical coordinates.

Answer 1

I would calculate it like this (and let $a=1$, just multiply the result be $a^2$ at the end): denote by $A_c$ the surface of the spherical cap over $z\ge 1/\sqrt{2}$. Then it is easy to calculate in spherical coordinates ($0\le\theta\le\pi/4$, $0\le\phi\le 2\pi$) that $$ A_c=2\pi\int_0^{\pi/4}\sin\theta\,d\theta=2\pi\Big(1-\frac{1}{\sqrt{2}}\Big). $$ Now the seeking area is simply $$ A=\frac{A_\text{sphere}-4A_c}{2}=2\pi(\sqrt{2}-1). $$

Answer 2

Thanks to the symmetry of the figure , we can use as limits for the angle $\varphi$ (in the $xy$ plane) : $0\le\varphi\le\frac{\pi}{4}$.

If $P$ is the point of intersection of the sphere with the plane $x=\frac{a}{\sqrt{2}}$ at the angle $\varphi$, the distance from the origin of the projection of $P$ on the $xy$ plane ( the point $D$ in this figure) is: $r=\frac{a}{\sqrt{2}\cos \varphi}$.

(in the figure: $\overline{AC}=a$; $\overline{AD}=r$; $\angle{DAB}=\varphi$)

The angle $\theta$ from the $z$ axis and the radius passing through $P$ is now expressed as: $\theta=\frac{\pi}{2}-\arccos \left( \frac{r}{a}\right)=\arcsin\left(\frac{1}{\sqrt{2}\cos \varphi} \right)$, and the total area is:

$$ A = 8\int_0^{\frac{\pi}{4}}\int_0^{\arcsin\left(\frac{1}{\sqrt{2}\cos \varphi} \right)}a^2 \sin \theta d\theta d \varphi $$