Problem with a dipstick 
The problem:
Some houses are heated by burning oil. The oil is stored in a horizontal elliptical cylinder that is lying underground. To measure the residual volume, one has to use a dipstick that is brought down towards the bottom of the tank. The oil level $h$ is read by the stick to calculate the volume of the oil.
Questions:
1) A certain tank has a circular section with the radius of $0.85$ metres and the length of $5.2$ metres. Decide an expression for the volume of the oil $V (h)$ as a function of the height $h$.
2) Observe a tank with the length $l$ and the elliptical section where the ellipticals major axis has the length of $2a$ and its eccentricity is $e$. Decide a simplified formula for $V(h)$.

 A: Hint for the set-up: Since the tank is basically a prism with an elliptical cross section, all you need to look at is the area of the cross-section that the oil covers. Once this is found, multiply by the length of the tank to get the total volume of oil.
So, your goal is to model the cross-section. You may as well think of the center of the cross section as being on the origin in the $x-y$ plane. The trick I will suggest to you is to think of the negative x-axis as "down" so that the surface of the leveled oil is actually a horizontal line which is distance $h$ above the "bottom" of the tank (which is the left-most extremity of the cross-section.
By symmetry, you just need to find an expression for the area under the circle (and then ellipsis in the second part), above the $x$-axis between the bottom of the tank (say at $x=b$) and the top of the oil level $b+h$, and then double that (to pick up the mirror image area beneath it.

Another way to approach the problem is as a two-part function for $V(h)$. You could find a function for the bottom half of the tank, and then integrate to find area below $h$ and above the bottom half. The problem is that this breaks down when $h$ gets to be higher than half of the tank, because the tank begins to curve back in on itself, and your "bottom function" does not include that. So, you would then have another integral to tell you how much volume is between the top  half and the middle, and then just add the (constant) volume of the already filled bottom half.
