The right volume

If we suppose that the curve $y=f(x)$ is integrable then the volume created will be

$$V={\int }_0^1\pi y^{2}dx.$$

To get a feel for the sort of curve we might need, first consider the special case $y=x$, which clearly passes through the two points. Then,

$$V={\int }_0^1\pi x^2=\pi {\left[\frac{x^3}{3}\right]}_0^1=\frac{\pi }{3}.$$

This is slightly larger than $1$, so we could consider a family of curves which are beneath $y=x$.

We could look for a curve like the blue one in the diagram below:

Image

 

This looks like a section of a polynomial which has a root at $0$.

How about $y=x^4$?

Then,

$$V={\int }_0^1\pi x^8=\pi {\left[\frac{x^9}{9}\right]}_0^1=\frac{\pi }{9}.$$

This is still not right, but I think a polynomial could work.

Let's try $y=x^a$ where $a$ is a real number.

Then,

$$V={\int }_0^1\pi x^{2a}=\pi {\left[\frac{x^{2a+1}}{2a+1}\right]}_0^1=\frac{\pi }{2a+1}.$$

 

If we let $a=(\pi-1)/2$ then the volume is $1$.

Therefore a solution is 

$$y=x^{(\pi -1)/2}.$$

There are, of course, others!