This is a famous calculus problem and is stated like this
Given a barrel with height $h$, and a small radius of $a$ and
large radius of $b$. Calculate the volume of the barrel
given that the sides are parabolic.
Now I seem to have solved the problem incorrectly because here it seems 2
that the volume should be
$ \displaystyle \hspace{1cm}
V(a,b,h) = \frac{h\pi}{3}\left(2b^2 + a^2\right)\,.
$
Below is my attempt. As in the picture I view the barrel from the side, and try to find a formula for the parabola. So i solve
$ \displaystyle \hspace{1cm}
f(x) := A x^2 + B x + C
$
given $f(0) = f(h) = a/2$ and $f(h/2) = b/2$. This yields
$ \displaystyle \hspace{1cm}
f(x) = \frac{2(a-b)}{h^2} \cdot x^2 –
\frac{2(a-b)}{h} \cdot x +
\frac{a}{2}
$
Using the shell method integrating now gives the volume as
$ \displaystyle \hspace{1cm}
V(a,b,h) := \pi \int_0^h \bigl[f(x)\bigr]^2\,\mathrm{d}x
= \frac{\pi}{60} \cdot h (a+2b)^2 + \frac{\pi}{30} \cdot h(a^2+b^2)
$
Alas according to the formula above this seems incorrect! Where is my mistake?
Best Answer
Let $k=h/2$, and put the origin in the middle, where symmetry asks it to be.
Then the equation of the upper parabola is $$y=b-\frac{b-a}{k^2}x^2.$$ The integral of $\pi y^2\,dx$ from $0$ to $k$ is $$\pi k\left(b^2-\frac{2}{3}(b-a)b+\frac{1}{5}(b-a)^2\right).$$ This simplifies to $$\frac{\pi k}{15}(3a^2+4ab+8b^2)$$ Replace $k$ by $h/2$ and multiply by $2$.