Find volume of cask I was given the following question:
A wine cask has a radius at the top of $30 cm$ and a radius at the middle of $40 cm$.
The height of the cask is $1m$.
What is the volume of the cask in litres, assuming the shape of the side is parabolic?
I have to come to parabolic function of
$$y = \frac{-1}{250}(x-50)^2+40$$
The derivative of $y$ is:
$$\frac{dy}{dx} = \frac{2}{5} - \frac{x}{125}$$
Then I integrate and end up with an expression the length of $\pi$. Am I on the right track?
 A: We place the centre of the cask at the origin, and assume that the cask is symmetrical about its mid-plane. 
Consider the parabola $y=40-kx^2$. At $x=50$ we want $y=30$. So $k=\frac{10}{2500}=\frac{1}{250}$.  
We will rotate this parabola about the $x$-axis. The resulting solid is the cask.
It is useful to integrate from $x=0$ to $x=50$, and double. By the usual formula, the volume, in cubic cm, is
$$2\int_0^{50}\pi\left(40-\frac{x^2}{250}\right)^2\,dx.$$
Remark: Kepler once wrote a little book about how to compute the volumes of Austrian wine casks. The book did not sell well, and Kepler had to go back to casting horoscopes and theorizing about the motions of the planets. 
A: All lengths should be in the same units, say metres.
Let the radius $r$ of the cask, at height $h$ above the ground have equation $r=ah^2+bh+c$.
When $h=0$ we want $r=\frac{3}{10}$. When $h=\frac{1}{2}$ we want $r=\frac{4}{10}$. When $h=1$ we want $r=\frac{3}{10}$.
Putting these into $r=ah^2+bh+c$ and solving the three simultaneous equations in $a,b,c$ gives
$$r = \frac{3}{10}+\frac{2}{5}h-\frac{2}{5}h^2$$
Using the formula for the volume of revolution, we get
$$V = \pi \int_0^1 \left(\frac{3}{10}+\frac{2}{5}h-\frac{2}{5}h^2\right)^{\!2} \mathrm{d}h = ~???~\mathrm{m}^3$$
