Can somebody explain the equivalence between integrating over the surface of a unit sphere and integrating over solid angle? I have been trying to understand the following transformation using a Jacobian but have been unsuccessful:
$$\iiint dr\ d\theta\ d\phi\ r^2 \sin \theta\ f(r,\theta,\phi) = \iint dr\ d\Omega\ r^2 f(r,\Omega)$$
I believe my confusion is that the solid angle is a surface area in a certain sense, and so I am confused as to how integrating over these surface area values recovers integrating over the full surface area of the sphere.
I am also confused because one typically sees a Jacobian determinant employed for such transformations but determinants are defined only for square matrices and the number of variables in these two integrals are not the same.
Best Answer
$d\Omega$ is representing the surface area element on the unit sphere, so, formally, $d\Omega = \sin\theta\,d\theta\,d\phi$. The solid angle is just the area subtended by the region on the unit sphere from the origin. The integral $\displaystyle\int_S d\Omega$ represents a surface integral over the appropriate portion of the unit sphere. So you still are integrating over a $3$-dimensional region, in toto.
EXAMPLE: Suppose our $3$-dimensional region is the interior of the cone $2\ge z\ge\sqrt{x^2+y^2}$. In spherical coordinates, we get the integral $$\int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\sec\theta} f(r,\theta,\phi)dr\,\sin\theta\,d\theta\,d\phi.$$ So we can rewrite this as $$\int_S \left(\int_0^{2\sec\theta} f(r,\theta,\phi)dr\right)d\Omega\,,$$ and here $S$ is the portion of the sphere given by $0\le\phi\le 2\pi$, $0\le \theta\le\pi/4$.