### Weekly Challenge 43: A Close Match

Can you massage the parameters of these curves to make them match as closely as possible?

### Weekly Challenge 44: Prime Counter

A weekly challenge concerning prime numbers.

### Weekly Challenge 46: the Jabber-notty

Can you invert this confusing sentence from Lewis Carrol?

# Weekly Challenge 28: the Right Volume

##### Stage: 5 Short Challenge Level:
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!