volume of partially truncated cylinder Automotive machine work (a real life problem).  A rotary cutter plunges only partially into a plane (flat aluminum piston top) at a specific angle (12 degrees off of vertical).  This produces a partial ellipse-shaped cutout.  What is the volume of the cutout as a function of cutting depth, particularly if the cutting depth is minor so as to not reveal the entire ellipse shape?  In my specific case, the angle of approach and the cutter diameter is fixed.  My goal is to cut deep enough to achieve at given volume.  So, how deep to cut?
Show me some mercy, I tried figuring this out with the help of my college calculus book but it's dated 1964....
 A: The shape of the cutout (the material you are removing)
is a cylindrical wedge,
assuming the cutter has not cut deep enough to cut the full outline of an ellipse in the piston top,
and assuming the cutter would cut a cylindrical hole with a flat circular bottom if you had cut straight down instead of at an angle.
Using the notation given on 
the Wolfram Alpha page,
the radius of the cutter is $R$ (that is, $R$ is half the diameter), and
the distance the cutter moves into the piston top after it first touches the piston top is $h.$
At one end of the cutout, the edge of the cutout is a straight line segment $PQ$ that connects the two ends of the unfinished ellipse.
Inside the cutout is a flat surface in the shape of a circular segment
(that is, the surface is bounded by the segment $PQ$ and a circular arc).
We let $a$ be half the length of that line segment,
and $b$ be the distance along the flat surface inside the cutout
from the segment $PQ$ to the deepest point of the cutout.
We let $\phi$ be half the angle of the arc of the circle inside the cutout.
Since the cutter is at an angle $\alpha = 12^\circ$ from vertical,
by trigonometry $h = b \tan\alpha.$
Also by trigonometry, $\phi = \arccos\left(1 - \frac bR\right)$
and $a = R \sin\phi.$
Starting with a theoretical plane through the segment $PQ$ (the straight edge of the cutout) parallel to the axis of the cutter, we can imagine a series of parallel planes slicing through the cutout volume. 
If we let $x$ represent the distance of one of these planes from the axis of the cutter, using a negative distance if the plane is closer to the edge $PQ$
(in fact, for the plane through $PQ$ itself we say $x = -(b-R)$),
then the part of cutout that the plane at distance $x$ slices through
is a rectangle with height $(x + b - R)\tan\alpha$
and width $2\sqrt{R^2 - x^2}.$
The area of this rectangle is $(2\tan\alpha)(x + b - R)\sqrt{R^2 - x^2}.$
The approximate volume of the cutout is the sum of the volume of a set of
slices of the cutout between these parallel planes.
The exact volume of the cutout is given by the integral
$$
\int_{R-b}^R ((x+b-R)\tan\alpha)(2\sqrt{R^2 - x^2})\,dx
= \tfrac13\tan\alpha \left(a(3R^2 - a^2) + 3R^2(b-R)\phi\right).
$$
To find the volume for any particular cutting depth $h,$
you find $a,$ $b,$ and $\phi$ for that value of $h$
using the formulas $b = h\cot\alpha,$ $a = \sqrt{b(2R-b)},$
and $\phi = \arccos\left(1 - \frac bR\right),$ and plug them into
the right-hand side of the formula
$$
V = \tfrac13\tan\alpha \left(a(3R^2 - a^2) + 3R^2(b-R)\phi\right)
$$
in order to get the volume $V.$
To actually do all this substitution and plugging-in, math software
(or at least a spreadsheet program) is very helpful.
It seems very unlikely that there would be any useful formula that would
let you plug in $V$ and directly compute the value of $h$ that is required.
Instead, you can use numerical methods, which come down to some kind of
guess-and-check procedure. (The main distinction of good methods from 
bad ones is how well they do the guessing.)
Some math software will let you define an equation like the one above and will find $h$ for you.
But if you have to do the guessing yourself,
a reasonable method is to just try a value of $h$;
if the resulting volume is too large, try a smaller $h,$
but if the resulting volume is too small, try a larger $h.$
Once you find two values of $h$ such that one is too large and one is too small, you can start honing in on the desired value of $h$
by trying $h$ halfway between the closest too-large and too-small values,
repeatedly, until you narrow the possible values of $h$ down to
an interval that is smaller than the largest error you can tolerate.
