Projected Area of Circle onto the Side of a Cylinder I have encountered a problem that requires me to find the projected area of a beam of light (circular cross-section, with radius $R_1$) vertically onto the side of a cylinder with $R_2$ as its cross-section. The setting is therefore similar to the problem of two-cylinder Steinmetz Solid. However, what if $R_1 \neq R_2$, especially $R_1 \gt R_2$?
Thanks.
 A: Let $R$ and $r$ be the radius of the beam and the cylinder respectively. Let $\mu = \frac{r}{R}$. 
Choose the coordinate axis such that the beam is illuminating the cylinder
from the $+ve$ $x$-axis.
In cylindrical polar coordinates $(r,\theta, z)$,
$$(r,\theta,z) \quad\iff\quad (x,y,z) = (r\cos\theta, r\sin\theta, z)$$
The illuminated area on the cylinder are those points satisfying:
$$ |\theta | \le \frac{\pi}{2}\quad\text{ and }\quad  r^2\sin^2\theta + z^2 \le R^2$$
Since the area element on the cylinder is $r dz d\theta$, the area we want is
$$\verb/Area/ = \int_{-\theta_0}^{\theta_0} \int_{-\sqrt{R^2 - r^2\sin^2\theta}}^{\sqrt{R^2 - r^2\sin^2\theta}} rdz d\theta
= 4rR \int_0^{\theta_0} \sqrt{1 - \mu^2 \sin^2\theta} d\theta$$
where
$$\theta_0 = \begin{cases}
\frac{\pi}{2}, & \mu \le 1,\\
\sin^{-1}(\mu^{-1}), & \mu > 1
\end{cases}$$
When $\mu \le 1$, the cylinder is smaller than the beam. It is easy to see
$$\verb/Area/_{\mu \le 1} = 4rR E\left(\mu\right)$$
where
$$E(k) = \int_0^{\pi/2} \sqrt{1 - k^2\sin^2\theta} d\theta
= \int_0^1 \sqrt{\frac{1-k^2t^2}{1-t^2}} dt
$$
is the complete elliptic integral of the second kind.
When $\mu > 1$, the cylinder is larger than the beam. Change variable to $t = \mu\sin\theta$, we have
$$\begin{align}
\verb/Area/_{\mu > 1}
&= 4rR\mu^{-1} \int_0^1 \sqrt{\frac{1-t^2}{1-\mu^{-2}t^2}} dt\\
&= 4rR\mu^{-1} \int_0^1 \frac{1 - t^2}{\sqrt{(1-t^2)(1-\mu^{-2}t^2)}}dt\\
&= 4rR\mu \int_0^1 \frac{(1 - \mu^{-2}t^2) - ( 1 - \mu^{-2})}{
\sqrt{(1-t^2)(1-\mu^{-2}t^2)}}\\
&= 4r^2 \left[E(\mu^{-1}) - (1-\mu^{-2})K(\mu^{-1})\right]
\end{align}
$$
where 
$$K(k) = \int_0^{\pi/2} \frac{1}{\sqrt{1 - k^2\sin^2\theta}} d\theta
= \int_0^1 \frac{1}{\sqrt{(1-k^2t^2)(1-t^2)}} dt
$$
is the complete elliptic integral of the first kind.
Summary:
$$
\bbox[8pt,border:1px solid blue]{
\verb/Area/ = \begin{cases}
4rR E\left(\frac{r}{R}\right), & r < R\\
4r^2 \left[E\left(\frac{R}{r}\right) - (1-\frac{R^2}{r^2})K\left(\frac{R}{r}\right)\right], & r > R
\end{cases}
}
$$
