## Solid Angle Formed By a Surface

The solid angle of a solid centred at the origin is defined as the area of the intersection of the shape with the unit sphere
$x^2 = y^2 +z^2 = 1$

Example. Consider the solid
$x^2 +y^2 \leq 5z^2 , z >= 0$

The intersection of the solid with the sphere is the soltion to
$x^2 +y^2 +z^1 =1, x^2 +y^2 = 5z^2$
.
The solution is
$6z^2 =1 \rightarrow z = \frac{1}, x^2 +y^2 = \frac{5}{6}{\sqrt{6}}$

For a surface of form
$f(x,y,z)=0$
the surface area is given by
$A = \int_{xy} \frac{ \sqrt{(\frac{\partial f}{\partial x })^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2}}{\frac{\partial f}{\partial z}} dxdy$

With
$f(x,y,z)=x^2+y^2+z^2-1=0$

$\frac{\partial f}{\partial x}=2x, \frac{\partial f}{\partial y}=2y, \frac{\partial f}{\partial z}=2z$

$A= \int_{xy} \frac{\sqrt{4x^2+4y^2+4z^2}}{2z} dxdy = \int_{xy} \frac{1}{z} dxdy = \int_{xy} \frac{1}{\sqrt{1-x^2-y^2}} dxdy$

Transform to polar coordinates then
$A= \int^{\sqrt{5/6}}_0 \int^{2 \pi}_0 \frac{r}{\sqrt{1-r^2}} d \theta dr = -2 \pi [ \sqrt{-r^2}]^1_0 =2 \pi (1- \sqrt{1/6})$