I'm trying to understand a derivation that I found here.
At some point, the author splits the integral:
$$ C_d \int_{\Omega}\cos\theta \,\delta\omega $$
Into the double integral:
$$ C_d \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi/2} \cos\theta \sin\theta \,\delta\theta\delta\phi $$
and explains why by saying:
The extra $\sin\theta$ may seem a little confusing at first, but it's necessary to take into account the smaller area towards the polar region.
I don't understand why that is. I would expect the following step to simply be this instead:
$$ C_d \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi/2} \cos\theta \,\delta\theta\delta\phi $$
This is the problem of writing the solid angle in terms of polar coordinates. The solid angle element is given by the surface element divided by the square of the distance from the center. Let's calculate this for a sphere of radius $R$:$$\delta \omega=\frac{\delta s}{R^2}$$ Now lets assume that you express this area in terms of polar coordinates. $\theta$ is the polar angle. Changing it is equivalent of moving along a meridian. $\phi$ is the azimuthal angle, that is controlling the motion along a parallel to the equator. Now let's choose a point on the sphere, and draw a tiny surface around that. We can choose the bounds to be along the meridians and along the parallels to the equator. The length along the meridian will be $R\delta\theta$. The length along the parallel is going to be $r\delta\phi$. Notice that the $r$ here is the radius of the circle parallel to the equatorial plane. It's easy to see that this radius varies with the polar angle as $r=R\sin\theta$. It's going to be $0$ at the pole, and $R$ at the equator. Now for this tiny area element, we can approximate that the sides form a rectangle, so $$\delta s=R\delta\theta r\delta\phi=R^2\sin\theta\delta\theta\delta\phi$$ Just use the first formula to find the solid angle in polar coordinates: $$\delta\omega=\sin\theta\delta\theta\delta\phi$$