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I am told that I have a cylindrically symmetric surface that is bounded between two circles $r=a$ at $z=\pm b$. I'm meant to use the Euler-Lagrange equation, so I'm trying to a functional for the area of a given surface. I tried to find the surface area element of the cylindrical surface by taking the cross product of $\frac{\partial\mathbf{r}}{\partial\theta}$ and $\frac{\partial \mathbf{r}}{\partial r}$ but this gives me $dS=r\sqrt{2(z')^2+1}drd\theta$ which I don't think is right and doesn't really help me.

Guest195
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    There must be another constraint, otherwise, $r=0$ would give a trivial minimum. Oh, wait, is this supposed to be $r=a$ at both $z=b$ and $z=-b$? That will give something more interesting. – robjohn Dec 12 '15 at 12:20
  • Yes, r=a at z=b and z=-b. – Guest195 Dec 12 '15 at 12:24

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Using Pappus's Theorem or the formula for the Area of a Surface of Revolution, we get $$ A=\int_{-b}^b2\pi r\sqrt{r'^2+1}\,\mathrm{d}z $$ and $$ \begin{align} \delta A &=\delta\int_{-b}^b2\pi r\sqrt{r'^2+1}\,\mathrm{d}z\\ &=2\pi\int_{-b}^b\sqrt{r'^2+1}\,\delta r\,\mathrm{d}z +2\pi\int_{-b}^br\frac{r'\delta r'}{\sqrt{r'^2+1}}\,\mathrm{d}z\\ &=2\pi\int_{-b}^b\sqrt{r'^2+1}\,\delta r\,\mathrm{d}z -2\pi\int_{-b}^b\left(\frac{rr''+r'^2}{\sqrt{r'^2+1}}-\frac{rr'r'r''}{\sqrt{r'^2+1}^3}\right)\delta r\,\mathrm{d}z\\ &=2\pi\int_{-b}^b\sqrt{r'^2+1}\,\delta r\,\mathrm{d}z -2\pi\int_{-b}^b\frac{rr''+r'^2+r'^4}{\sqrt{r'^2+1}^3}\,\delta r\,\mathrm{d}z\\ &=2\pi\int_{-b}^b\frac{r'^2-rr''+1}{\sqrt{r'^2+1}^3}\,\delta r\,\mathrm{d}z\\ \end{align} $$ Thus, to get $\delta A=0$ for any $\delta r$, we need $r'^2-rr''+1=0\iff\frac1{r^2}=\left(\frac{r'}r\right)'$.

Let $cr=\cosh(t)$, then $$ \begin{align} \frac1{r^2}&=\left(\frac{r'}r\right)'\\ \frac{r'}{r^3}&=\left(\frac{r'}r\right)'\frac{r'}r\\ c^2-\frac1{r^2}&=\left(\frac{r'}r\right)^2\\ \sqrt{c^2r^2-1}&=r'\\ cz&=\int\frac{c\,\mathrm{d}r}{\sqrt{c^2r^2-1}}\\ cz+d&=t \end{align} $$ Therefore, $r=\frac1c\cosh(cz+d)$. To be symmetric about $z=0$, we let $d=0$. Thus, to have $r(b)=a$ we need to solve $ac=\cosh(bc)$, then $$ r=\frac1c\cosh(cz) $$

robjohn
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  • Yeah; I am fine with how to solve the problem once I've found the surface area element- I should be able to arrive at the first part of your solution without using Pappus's Theorem. – Guest195 Dec 12 '15 at 15:48
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    The cylindrical surface area element can be taken to be $$r\sqrt{r'^2+1},\mathrm{d}z,\mathrm{d}\theta$$ where $r$ is a function of $z$. In this case, this leads directly to Pappus's Theorem. – robjohn Dec 12 '15 at 16:37
  • Yeah, but I don't understand how to reach that expression. – Guest195 Dec 13 '15 at 11:21
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    @Guest195: Computing the area of a surface of revolution is usually handled before Calculus of Variations. Take a look at the link. – robjohn Dec 13 '15 at 12:01