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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:

$$\int \int \int dr\ d\theta\ d\phi\ r^2 \sin \theta\ f(r,\theta,\phi) = \int \int dr\ d\Omega\ r^2 f(r,\Omega)$$

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.

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$d\Omega$ is representing the surface area element on the unit sphere, so, formally, $d\Omega = \sin\theta\,d\theta\,d\phi$. Solid angle is just the area subtended by the region on the unit sphere. 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 $$\int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\sec\theta} f(r,\theta,\phi)dr\,\sin^2\theta\,d\theta\,d\phi.$$ 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 $0\le\phi\le 2\pi$, $0\le \theta\le\pi/4$.

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  • $\begingroup$ Is the upper bound of the solid angle integral 4 pi? I have never actually seen it written out anywhere explicitly. $\endgroup$ – Sam Manzer Aug 20 '14 at 23:01
  • $\begingroup$ No, this is a surface integral, not a single integral. I'll add an example to my answer. $\endgroup$ – Ted Shifrin Aug 20 '14 at 23:21
  • $\begingroup$ Ah ok, I was confused when trying to read about Lebedev grids. So a numerical quadrature for integration over a unit sphere would still consist of both theta and phi points, but these could be chosen keeping in mind that we are actually doing a surface integral and not two separate generic 1D integrals. $\endgroup$ – Sam Manzer Aug 21 '14 at 15:34
  • $\begingroup$ It's possible that you'd have a region on the sphere that is not of the form $\theta_0\le\theta\le\theta_1$, $\phi_0\le \phi\le\phi_1$, but probably in practice it'll be of that form frequently. $\endgroup$ – Ted Shifrin Aug 21 '14 at 16:32
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Solid angle $(\Omega)$ is a two dimensional angle in 3-D space & it is given by the surface (double) integral as follows:

$\Omega$ = (Area covered on a sphere with a radius r)/radius^2 =

$\dfrac{\iint_S r^2\sin\theta d\theta d\phi}{r^2}=\iint_S \sin\theta d\theta d\phi.$

Now, applying the limits, $\theta=$ angle of longitude & $\phi=$ angle of latitude & integrating over the entire surface of a sphere, we get $$\Omega=\int_0^{2\pi} d \phi\int_0^{\pi} \sin\theta d\theta$$ $$\Omega=\int_0^{2\pi} d\phi [-\cos\theta]_0^{\pi}=2\int_0^{2\pi} d\phi=2[\phi]_0^{2\pi}=2[2\pi]=4\pi $$

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