Solid angle: integration

Question

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.

Answer 1

$d\Omega$ 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 $0\le\phi\le 2\pi$, $0\le \theta\le\pi/4$.

Answer 2

Solid angle, $\Omega$, is a two dimensional angle in 3D space & it is given by the surface (double) integral as follows:

$$\Omega=\text{(Area covered on a sphere with a radius $r$)}/{r^2}=$$ $$=\dfrac{\displaystyle \iint_S r^2\sin\theta \ d\theta \ d\phi}{r^2}=\iint_S \sin\theta \ d\theta \ d\phi.$$

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

Answer 3

It is perhaps better to to get used to integral curvature as a unit of solid angle in its own right by virtue of its inseparable definition / association with Gauss curvature.

If it were asked to explain

the equivalence of length between integrating over the arc of a unit circle ( radius in the plane $R=1$ ) and integrating over the subtended ( spanning plane) angle $\theta $,

an answer would be like:

$$ s_2-s_1= \int_{s1}^{s2} ds= \int_{\theta_1}^{\theta_2}R d \theta =\int_{\theta_1}^{\theta_2}\frac{ d \theta }{\kappa}= \frac{\theta_2-\theta_1}{\kappa}= (\theta_2-\theta_1) R $$ Angles can be summed or difference found from end of arc to end of arc.

Your question in other words is

to explain the equivalence of patch area between integrating over the surface of a unit sphere ( radius of spherical patch R =1 ) and integrating over the covering / surrounding (solid) angle.

Element of solid angle is $ K dA = d\Omega$ and lumped/integrated dimensionless solid angle ( K= Gaussian curvature) is $ \Omega= \int K \;dA$ termed Integral Curvature by Gauss and measured in units of steradians...

It is $\pi $ for an octant, $ 2 \pi$ for a hemisphere, $ 4 \pi \; $ for a full sphere, $- 4 \pi $ for a full pseudosphere and so on.

$$ A_2-A_1= \int_{A_2}^{A_1} \frac{K dA }{K}=\int_{\Omega_2}^{\Omega_1} \frac{d \Omega }{K}=\int_{\Omega_2}^{\Omega_1} R^2 {d \Omega } = (\Omega_2- \Omega_1) R^2$$

Solid angles can be summed or difference found when common triangular areas of common sphere center vertex are merged.

Whereas an angle is defined as spanned between 2 arms in Euclidean geometry of the plane, a solid angle is spanned among 3 planes with three dihedral angles pairwise between them.

In spherical trigonometry on unit sphere we consider a spherical triangular patch enclosed between intersected segments of a great circle and two longitudes making 3 dihedral angles $(A,B,C)$ has enclosed area

$$ \text{ Integral Curvature = Spherical Deficit } = (A+B+C-\pi) $$

derived from the Gauss-Bonnet theorem.