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 so 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 at beneath $y=x$ with enough flexibility to all us to vary the final value of the volume. A simple choice is parabolas $y=Ax(x-1)$ for some multiplicative constant. These give
$$V = \int^1_0 \pi Ax^2(x-1)^2dx$$
Now that I see it, I'm not too keen on doing this integral, so I'm going to consider instead $y=A\sqrt{|x(x-1)|}$
This gives a volume
$$V = \int^1_0 \pi A x(1-x) dx = \pi A\left[\frac{x^2}{2}-\frac{x^3}{3}\right]^1_0 = \pi A \left[\frac{1}{2}-\frac{1}{3}\right] = \frac{\pi A}{6}$$
Thus, a curve which has the required properties is
$$y = \frac{6}{\pi}\sqrt{|x(x-1)|}$$
There are, of course, others!