Imagine a torus, I cutted in half:
It is quite easy to calculate the red area:
$$A_1(\alpha=0) = \pi ( R+r)^2 - \pi R^2$$
as well as the green area:
$$A_1(\alpha=\frac{\pi}{2}) = 2\pi R \cdot r$$
But how could one calculate the area at certain angle $\alpha$ in the range $[0,\pi]$?
I came up with something like
$$ A(\alpha) = 2\pi R r + \pi r^2 \cdot \cos(\alpha)$$
which would be true for $\alpha = 0,~ \pi/2,~ \pi$ - but how can I be sure that it is correct for all the other $\alpha$?
This cross section is equivalent to the lateral surface area of a truncated cone with radii $r_1,r_2$ and slant height $s$:
$$A(r_1,r_2,s) = \pi(r_1+r_2)s.$$
Here, the slant height is $r$. The height of the truncated cone is $r \sin \alpha$, so if one radius is $R$, the other is $R - r \cos \alpha$.
This gives your cross-sectional area:
$$A(R,r,\alpha) = \pi(2R - r \cos \alpha)r.$$
This is valid for all $\alpha$ between $0$ and $\pi$.