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In what follows you may assume that the following notation applies

$$y=y(x),\ y'=\ds\frac{dy}{dx}.$$

You may also assume that, unless otherwise stated, $y$ is a
sufficiently continuously differentiable function.

{\bf Question}

The smooth curve $y(x)$ is defined for $-\log 2 \leq x \leq \log
2$ and is such that $y(\pm \log 2)=\ds\frac{5}{4}$. If the curve
is rotated about the $x$-axis, show that the area of the surface
of revolution thus generated is given by

$$A=2\pi\ds\int_{-\log 2}^{\log 2} y\sqrt{1+{y'}^2} \,dx$$

Hence show that the extremal $y$ satisfies

$$\ds\frac{y}{\sqrt{1+{y'}^2}}=const.$$

and thus the area is stationary if $y=\cosh x$.

Now consider a surface of rotation in the shape of a cylindrical
spool formed from two parallel discs of radius $\ds\frac{5}{4}$
placed at $x=\pm \log 2$, joined along the $x$ axis by an
infinitely thin rod. By simple geometry, show that the surface
area of this shape if given by $\ds\frac{25\pi}{8}$ and is thus
less than the apparent minimum value obtained with $y=\cosh x$.
How do you explain this apparent contradiction? (Hint: remember
the assumptions on differentiability that have been made in the
first part of the question.)


\medskip

{\bf Answer}

PICTURE \vspace{2in}

\newpage
Standard calculus gives

Surface area $\undb{dA}=\undb{2\pi y} \undb{ds}$

elemental surface area\ \ \ \ \ circumference\ \ \ \ \ element
width

Thus

\begin{eqnarray*} A & = & \ds\int 2\pi y \,ds\\ & = & 2\pi \ds\int
y \sqrt{1+{y'}^2} \,dx \end{eqnarray*} standard results, see
lecture notes

\un{$A=2\pi \ds\int_{x=-\log 2}^{x=+\log 2} y\sqrt{1+{y'}^2}
\,dx$} as required

$F=F(y,y')$ only, so E-L equation has first integral
$y'\ds\frac{\pl F}{\pl y'}-F=const$

\begin{eqnarray*} & & \ds\frac{\pl F}{\pl y'}=\ds\frac{2\pi
y y'}{\sqrt{1+{y'}^2}}\\ & \Rightarrow & \ds\frac{2\pi y
{y'}^2}{\sqrt{1+{y'}^2}}-2\pi y\sqrt{1+{y'}^2}=const\\ &
\Rightarrow & \ds\frac{2\pi y}{\sqrt{1+{y'}^2}}=const\\ &
\Rightarrow & \un{\ds\frac{y}{\sqrt{1+{y'}^2}}=const=\alpha}
\rm{say}
\end{eqnarray*}

Thus we have ${y'}^2=\ds\frac{y^2}{\alpha^2}-1$ which is solved
via standard integrals to give

$$y=\alpha \cosh \left(\ds\frac{x}{\alpha}+c\right)\ \rm{for\
constant}\ c,\ alpha$$

>From symmetry of boundary conditions, we need $c=0$. The other
condition is satisfied with $\alpha=1$. Thus \un{$y=\cosh x$} is
the extremal solution.

Surface of rotation

PICTURE \vspace{2in}

Surface area (\un{inside} $-\log 2 < x < \log 2$) is $2 \times
\left[\pi\times
\left(\ds\frac{5}{4}\right)^2\right]=\ds\frac{25\pi}{8}=9.817...
(A)$

Now on extremal $y=\cosh x$ we have

\begin{eqnarray*} A & = & 2\pi \ds\int_{-\log 2}^{+\log 2} \cosh^2 x \,dx\\ &
= & 2\pi\left[\log 2+\ds\frac{1}{2}\sinh(2\log 2)\right]\\ & = &
10.25\ \ (B) \end{eqnarray*}

Clearly $(B)>(A)$, but we have assumed that $y=\cosh x$ is a
minimum. Assuming is still is (can be confirmed by considering
second variation) there is an apparent contradiciton. The
resolution is that the disc case is \un{not} formed by the
rotation of a \un{smooth} function as we have assumed in the case
of $(B)$. Thus $y=\cosh x$ \un{is} the minimal result if we
restrict the solution to smooth functions as in the question, the
disc result is not a valid solution.Hence no paradox!

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