Find the Smallest Volume of Function When Revolved Around x axis

Question

My class is having a competition to see who can find the function that has the smallest volume when revolved around the x - axis.

The function must meet the following conditions:

$f(-1) = 1$, $f(1) = 1$, and $\int\limits_{-1}^1 f(x) dx = 1$

I already know that $y =\frac{1}{4} + \frac{3 x^2}{4}$ satisfies these conditions and has a volume of $\frac{3\pi}{5}$ but am curious if anyone knows a method to find a function with a smaller volume.

Any guidance would be appreciated

Thanks

Answer 1

Pretty good, one cannot do much better.

By symmetry, we want a function $f(x)$ which is defined on $[0,1]$, with $f(1)=1$ and $\int_0^1 f(x)\,dx=\frac{1}{2}$, and such that $\int_0^1 (f(x))^2\,dx$ is as small as possible.

Since continuity was not asked for, let $f(x)=\frac{1}{2}$ if $0\le x\lt 1$, and let $f(1)=1$.

If we want continuity, approximate the above function by choosing a very small positive $\epsilon$, say $10^{-4}$, and let $f(x)$ be a suitably chosen number a bit below $\frac{1}{2}$ up to $x=1-\epsilon$, and on $[1-\epsilon,1]$ let $f(x)$ climb linearly to $1$.

By the Cauchy-Schwarz Inequality, one cannot do better than the first function we described.

Answer 2

Assuming you are searching among functions which are continuous, there is no minimiser. First, we have via the Cauchy–Schwarz inequality: $$ 1 = \int_{-1}^1 f(x) \, dx \leqslant \left( \int_{-1}^1 \, dx \right) \left( \int_{-1}^1 (f(x))^2 \, dx \right) = \frac{2}{\pi} V(f), $$ where $V(f)$ is the volume under the solid of revolution made from $f$, $\pi \int_{-1}^1 (f(x))^2 \, dx$. Therefore the volume is larger than $\pi/2$.

The conditions for equality in Cauchy–Schwarz imply that the minimising function should be constant, and it is easy to see that that would be $f(x) = 1/2$, but this is disallowed by the boundary conditions. Instead, take $$ f_{a}(x) = \begin{cases} \dfrac{a}{a+1} & \lvert x \rvert \leqslant a \\ \dfrac{a}{a+1} + \dfrac{\lvert x \rvert-a}{1-a^2} & \lvert x \rvert > a \end{cases}, $$ where $0<a<1$. Then $f_a$ is continuous, has integral $1$ and value $1$ at the endpoints. Its volume can be evaluated as $$ \pi \int_{-1}^1 f_a^2 = \frac{2\pi(2a+1)}{3(a+1)^2}, $$ which decreases to $\pi/2$ as $a \to 1$ (you can check the derivative is negative if you don't believe me). Thus you can find a continuous function for any volume bigger than $\pi/2$, but cannot equal it.