Solid angle: integration

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

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 $$\Omega=\int_0^{2\pi} d \phi\int_0^{\pi} \sin\theta d\theta$$ $$\Omega=\int_0^{2\pi} d\phi [-\cos\theta]_0^{\pi}=2\int_0^{2\pi} d\phi=2[\phi]_0^{2\pi}=2[2\pi]=4\pi $$