The question is:
What is the mean radius $\overline{r}$ from the midpoint of a cylinder of radius $a$ and height $h$ to its boundary surface? Evaluate $\overline{r}$ for $a = h/2 = 10~\mathrm{cm}$.
This sounds like homework, but it's an example problem a professor gave. I have to use this concept on a much harder problem, but some parts of the question stem from knowing the mean radius. Any help?
I was given:
$\frac{1}{4 \pi} \int_0^{2\pi}\int_0^{\pi}r \sin\theta d\theta d\beta$
but I can't figure what to do with that. It's been years since multivariate :(
$\endgroup$ 21 Answer
$\begingroup$OK, you don't know what mean radius is, and neither do I, so let's start with the formula. You're integrating over two angle variables, one between $0$ and $2\pi$ and the other between $0$ and $\pi$. This indicates a variant of a spherical coordinate system, but a slightly uncommon one. In normal spherical coordinate systems I'd have latitude range from $-\frac\pi2$ to $\frac\pi2$. So apparently this angle $\theta$ is measured not from the plane of the equator, but from the polar axis.
You divide the whole thing by $4\pi$, which is the surface area of the unit sphere, or perhaps you'd better think of this as the solid angle of any complete sphere. Likewise, this term $\sin\theta\,\mathrm d\theta\,\mathrm d\beta$ appears to be a solid angle element, i.e. the infinitesimal solid angle subtended by the infinitesimal changes in your two angle variables.
So to sum it up, the formula you are given takes the distance of each point from the center of the coordinate system, and averages these weighted by solid angle.
So what is the mean radius of your cylinder? I'll simply assume that the origin of the coordinate system is located in the center of that cylinder, and furthermore that the polar axis agrees with the axis of rotational symmetry of that cylinder. The cylinder consists of a side and two caps. So you could integrate over these independently. For that we first need to know the angle $\theta$ where they meet. Since $a=\frac h2$ that switchover angle is $\frac\pi4$.
\begin{align*} \bar r&=\frac1{4\pi}\left(2\int_0^{2\pi}\int_0^{\frac14\pi} \frac{h}{2\cos\theta}\,\sin\theta\,\mathrm d\theta\,\mathrm d\beta +\int_0^{2\pi}\int_{\frac14\pi}^{\frac34\pi} \frac{a}{\sin\theta}\,\sin\theta\,\mathrm d\theta\,\mathrm d\beta\right) \\&=\frac1{4\pi}\left(4\pi\int_0^{\frac14\pi} \frac{h}{2}\,\tan\theta\,\mathrm d\theta +2\pi\int_{\frac14\pi}^{\frac34\pi} a\,\mathrm d\theta\right) \\&=\frac{h}{2}\int_0^{\frac14\pi}\tan\theta\,\mathrm d\theta +\frac a2\int_{\frac14\pi}^{\frac34\pi}\mathrm d\theta \\&=\frac{\log2}{4}h+\frac{\pi}4a \\&=\frac{\log2}{4}\cdot20\text{cm}+\frac{\pi}4\cdot10\text{cm} \approx 11.32\text{cm} \end{align*}
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