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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.

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  • $\begingroup$ Can the beam be off-centre? The Steinmetz Solid implies that the two axes coincide, but it doesn't hurt to be explicit! $\endgroup$ – TonyK Mar 1 '16 at 21:19
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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:

$$ \bbox[8pt,border:1px solid blue]{ \verb/Area/ = \begin{cases} 4rR E\left(\frac{r}{R}\right), & r < R\\ 4r^2 \left[E\left(\frac{R}{r}\right) - (1-\frac{R^2}{r^2})K\left(\frac{R}{r}\right)\right], & r > R \end{cases} } $$

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