Say I have a solid $\Omega\subset \mathbb{R}^n$ and I randomly rotate it. I want to show that the expected $(n-1)$ volume of the projection on a plane is $\frac{\Gamma(\frac{n}{2})}{2\pi^\frac{1}{2}\Gamma(\frac{n+1}{2})}$ of its surface area.
I started out by analyzing the case for a planar surface element with area $K$ with normal $\hat{n}$. The area of the projection on the $x_2\dots x_{n}$ plane is $|\hat{n}\cdot \hat{e}_1|K$. Expressing $\hat{n}$ using uniform random variables $\Theta_1,\dots,\Theta_{n-1}$, we get
$$\frac{\text{Avg}}{K}=\frac{1}{\text{vol}(S^{n-1})}\int_{S^{n-1}}|\hat{n}\cdot \hat{e}_1|d\sigma=$$$$\frac{1}{\text{vol}(S^{n-1})} \int_{[0,\pi]^{n-2}\times[0,2\pi]} |\cos\theta_1| \prod_{i=1}^{n-2} \sin ^{n-1-i}\theta_i d\theta_1\dots d\theta_{n-1}=$$ $$\cdots=\frac{\Gamma(\frac{n}{2})}{\pi^\frac{1}{2}\Gamma(\frac{n+1}{2})}$$
Why did I get twice the result? Is it because the surface has 2 sides and it actually has area $2K$?
How can I use this to get the general result? In this case, area elements can cast overlapping "shadows" so I can't just sum infinitesimal $K$s. It doesn't seem that simple to generalize.