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

$$

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

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

How about $y=x^4$?

Then,

$$

V = \int^1_0 \pi x^8 = \pi\left[\frac{x^9}{9}\right]^1_0 = \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^1_0 \pi x^{2a} = \pi\left[\frac{x^{2a+1}}{2a+1}\right]^1_0 = \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!

$$

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

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

How about $y=x^4$?

Then,

$$

V = \int^1_0 \pi x^8 = \pi\left[\frac{x^9}{9}\right]^1_0 = \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^1_0 \pi x^{2a} = \pi\left[\frac{x^{2a+1}}{2a+1}\right]^1_0 = \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!