# Thread: volume of oil tanker

1. ## volume of oil tanker

Here's a neat integration problem. Not particularly difficult, but cool. Give it a go if you have a mind to.

"The tank of an oil truck is 18 feet long and has elliptical cross-sections that are 6 feet wide and 4 feet high. Find the volume of the oil in the tank in terms of the depth h of the oil".

This one is a little different in that the cross sections are elliptic rather than circular. We have all seen these types of tankers on the road, it got me to thinking about such a problem.

2. I thought someone would bite on this problem. Here's how I done it. It appears to check. I just made up those dimensions from looking at one of those types of tankers on the road. we could've derived a general case instead.

Since the ellipse is 6 feet wide and 4 feet high, the ellipse has the equation

$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$

Solve for x in terms of y:

$x=\frac{3}{2}\sqrt{4-y^{2}}$

$V=\int_{-2}^{-2+h}(2)(3/2)\sqrt{4-y^{2}}(18)dy=54\int_{-2}^{-2+h}\sqrt{4-y^{2}}dy$

Now, integrate and we get:

$V=27\left[4sin^{-1}(\frac{h-2}{2})+(h-2)\sqrt{4h-h^{2}}+2\pi\right]$

For the general case, from $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

$V=\frac{ahL\sqrt{2bh-h^{2}}}{b}+\frac{abL\pi}{2}-absin^{-1}(\frac{b-h}{b})L-aL\sqrt{2bh-h^{2}}$

Kind of long connected and probably simplifies, but that's all I'm doing.

3. Originally Posted by galactus
Here's a neat integration problem. Not particularly difficult, but cool. Give it a go if you have a mind to.
Originally Posted by galactus
I thought someone would bite on this problem...
I didn't notice it before... Not that I would have been able to do it yet anyway.

Patience Galactus, soon I will be answering all these questions... (This term actually We'll be starting with this kind of work.)