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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. Barrel

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?

N3buchadnezzar
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  • Put $a=b$, then your formula gives $\pi h a^2 \frac{13}{60}$... The integration seems to be off – TZakrevskiy Jul 16 '13 at 21:29
  • Yeah, I thought that $a$ and $b$ were the diameters for the barrel, and not the radii. Scaling them solved the problem. – N3buchadnezzar Jul 16 '13 at 21:35
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    The calculation was perhaps unnecessarily heroic. Let $k=h/2$. Put the origin in the middle, where it is meant to be. Our (upper) parabola is then $y=b-\frac{b-a}{k^2}x^2$. Square, integrate from $0$ to $k$. and double. – André Nicolas Jul 16 '13 at 21:49
  • Great! Has the question about the partially filled barrel been asked on the site before? Eg how to measure how much liquid is in sideways barrel. (Put a hole in the middle and use a stick). – N3buchadnezzar Jul 16 '13 at 21:55
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    @N3buchadnezzar Actually... did it solve the problem? Your integral was missing an extra $2$ in the last bracket (i.e. $(a^2+2b^2)$ instead of $(a^2+b^2)$); then it would have been right. However, it would not be equal to the formula from the external site (even after correction for the radii vs. diameters mistake)! The one you quoted is only an easier-to-remember-approximation. The actual formula for parabolic barrel can be seen here. – Peter Košinár Jul 16 '13 at 22:04

3 Answers3

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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 $2$.

André Nicolas
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Note that in their case $a$, $b$ are radiuses, not diameters. Another important thing to say is that the formula from that external site is not exact!

For simplicity, take $h=1$. Let's define our function on $[-1/2,1/2]$ as $f(x) = 4(a-b)x^2+b$. We write

$$\frac{V}{\pi}= 2\int _{0}^{1/2}(4(a-b)x^2+b)^2 \mathrm dx= 16 (a-b)^2 \cdot \frac{1}{5\cdot 16}+ 8 b(a-b)\cdot \frac{1}{3\cdot 4} + b^2$$ $$= (a-b)^2/5 + 2 b(a-b)/3 + b^2.$$

TZakrevskiy
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Volume of a barrel

V = 0.5 * πQuadrat * (π/18 + 1/6) * (r1Quadrat + r2Quadrat) * h parabolic bend
V = 0.5 * πQuadrat * (π/17 + 1/6) * (r1Quadrat + r2Quadrat) * h elliptic bend

Best regards Hans-Jürgen Gläsel

Hans-Jürgen Dr. Gläsel
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