Integrating over plane-sphere intersection I have the following problem:
Let $\Sigma$ be the portion of the plane $x+y+z=m$ interior to the sphere $x^2+y^2+z^2=1$. Verify that when $|m|\leq\sqrt{3}$,$\int\limits_\Sigma(1-(x^2+y^2+z^2))dA=\frac{\pi}{18}(3-m^2)^2$
I tried using spherical coordinates but that did not help me. If we rotated the sphere we could use cylindrical coordinate but i don't see how to integrate this. 
Does anyone out there can help me resolve this issue?
Thanks,
$O^3$
 A: $\Sigma$ is a disc. The plane $P$ containing $\Sigma$ is defined by the normal vector $u=(1,1,1)$ and the number $m$. So $P$ is the plane normal to $u$ containing the point $M$ of coordinates $(m,0,0)$. The distance between the center $C$ of $\Sigma$ and $P$ is therefore given by the orthogonal projection formula $h=d(C,P)=\vec{OM}\cdot u/\|u\|=m/\sqrt3$. The radius of $\Sigma$ is $\sqrt{1-h^2}=\sqrt{1-m^2/3}$. 
By Pythagoras' theorem, a point of $\Sigma$ at a distance $r=\sqrt{x^2+y^2+z^2}$ from $C$ is at distance $\rho=\sqrt{r^2-h^2}$ from the center of $\Sigma$. As the derivative $$\frac{\mathrm d\rho}{\mathrm dr}=\frac r\rho$$ the integral on $\Sigma$ rewrites
$$2\pi\int_0^{\sqrt{1-m^2/3}} (1-r^2)\rho\mathrm d\rho=
2\pi\int_h^1(1-r^2)r\mathrm dr=2\pi\left(\frac{1-h^2}2-\frac{1-h^4}4\right).$$
This gives the required result.
A: You don't need the (3D) spherical coordinates. The intersection is just a circle. You need to integrate along a circle, which is normal to the direction (1,1,1). Because $x^2+y^2+z^2$ is invariant to rotation of the coordinate frame, you can just reorient the integration domain to put the circle normal to the $z$ axis! In that case, you have a plane
$$z=\frac{1}{\sqrt{3}}m$$
intersecting the unit sphere. Then you can just use the regular polar coordinates to evaluate the integral. At height $z$, the radius of the circle is $R=\sqrt{1-z^2}$. Now the integral becomes
$$2\pi\int_0^R (1-(z^2+r^2))r\,dr=2\pi ((1-z^2)R^2/2-R^4/4)$$
$$=\frac{\pi R^4}{2}=\frac{\pi}{2}\left(1-\frac{m^2}{3}\right)^2$$
