Ellipse Segment Area of tank How do you arrive at the formula 
area = (AB/4)[arccos(1 - 2h/A) - (1 - 2h/A)sqrt(4h/A - 4h2/A2)]
from this website http://www.had2know.com/academics/ellipse-segment-tank-volume-calculator.html?
Any ideas will be most helpful thanks.
 A: Let $a$ and $b$ be the ellipse semiaxes, with axis $a$ cut by the segment base at $\bar x=a-h$ from the center. Then segment area is:
$$
area=2\int_{\bar x}^{a}y\,dx=2ab\int_0^{\bar t}\sin^2t\,dt
=ab\left(\bar t-\sin\bar t\cos\bar t\right),
$$
where I have used the parameterization $x=a\cos t$, $y=b\sin t$ and $\bar x=a\cos \bar t$. From the last equality it follows that $\cos\bar t=\bar x/a=1-h/a$: plug this (and the easily derived expressions for $\sin\bar t$ and $\bar t$) into the above formula and you are done.
EDIT.
With the data given in the comment below, we have:
$$
a=A/2,\quad b=B/2,
$$
and in addition, from $\cos\bar t=1-h/a=1-2h/A$ we get
$$
\bar t=\arccos(1-2h/A),\quad \sin\bar t=\sqrt{4h/A-4h^2/A^2}.
$$
Plugging that into my formula above we finally obtain
$$
area={AB\over4}\left[\arccos\left(1-{2h\over A}\right)-\left(1-{2h\over A}\right)\sqrt{{4h\over A}-{4h^2\over A^2}}\,\right],
$$
which is exactly the formula you quoted.
You might have got confused by my choice of $x$ as "vertical" axis, but of course that does not change the final result.

