We have two responses for you

Darcy,

The simplest way to approach an ellipsoid is as a 'stretched sphere'.

You have a hemisphere (which you know the formula for) but with stretches.
For the hemisphere, there is a term $R^3.$
Now replace $R^3$ by $a \times b \times c$ where these are the lengths of the axes.
Each stretch (one R goes to a, one R goes to b, one R goes to c) is a stretch of the volume.

So your final volume for the half-ellipsoid is the formula for the hemisphere with $R^3$ replaced by $a \times b \times c$!

You can apply the same type of reasoning to ellipses in the plane, etc. for the area.
This reasoning does not apply to surface area of ellipsoids in 3-space, as not all parts of the surface 'stretch' by the same amount.

Walter Whiteley

Volume of an ellipsoid is$\large \frac43 \times \pi \times abc$ where a,b,c are the three radii. So a is half the length, etc.

So volume of a half-ellipsoid is

\[\frac12 \times \frac43 \times \pi \times \frac{A}{2} \frac{B}{2} c = \frac{\pi}{6} ABc \sim 0.5236 ABc\]

where A is length, B width, c height. Remember, A,B are measured right across the mound, c is from the flat surface.

If you're working with soil, you are probably not working to a very high accuracy, so

\[\frac{\mbox{ length } \times \mbox{ width } \times \mbox{ height }}{2}\]

would be a pretty good rule of thumb.

Good Hunting!
RD