# Multivariable Calculus surface integral over a square

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.

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 $$A=\frac{A_\text{sphere}-4A_c}{2}=2\pi(\sqrt{2}-1).$$

• I'm having a little trouble understanding the last step Apr 29 '18 at 11:43
• @chickenpie If you cut the sphere by four vertical planes along the square sides, you will get four identical spherical caps off. The remaining area consists of the northern part and the southern part over the square with equal areas. You need the northern part only. Thus if you know the cap area that you cut off, you can calculate the rest. Apr 29 '18 at 11:54

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:

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