volume of a truncated ellipsoid

[Richard Johnson]

In cardiology it is important to estimate the volume of the left ventricle
of the heart. It happens that this shape is similar to about 2/3 of a
prolate ellipsoid. Is there a formula for the volume of a truncated
ellipsoid?

Richard Johnson

[Walter Whiteley]

I assume the truncation is perpendicular to one of the axes of the ellipsoid.

If so, you can think of this as a 'stretched sphere' - stretched by
A, B, C, along the three axes. If you think of the same truncation on the
unit sphere, there are formulae for what the volume of the truncates sphere
is.
The corresponding volume of the ellipsoid is just the spherical volume
multiplied
by AxBxC. (If two of the axes stretch by the same amount, the set B=C etc.)

Now for the volume of the truncated sphere:
There are standard formula, but here is a 'picture' (related to the old
Greek formula for the volume of the sphere).

Take a hemisphere surrounded by a cylinder of the same diameter.
Now slice by slice (parallel to the base) the area of the ring outside the
sphere and inside the cylinder IS the area inside an inverted cone
with 'base' the top of the cylinder and point at the center of the
hemisphere.

The volume of the truncated hemisphere IS the volume of the truncated
cylinder
minus the volume of the top of the cone cut off by the same truncation.
This reduced the problem to truncating cones and cylinders which is
straightforward.

Richard Johnson wrote:

--
Walter Whiteley
Mathematics and Statistics
York University
whit...@mathstat.yorku.ca
416-726-2100 ext 33971

[Ken Pledger]

In article <7o4j0o$ga1$1...@bgtnsc03.worldnet.att.net>, "Richard Johnson"
<ri...@datatool.com> wrote:

> In cardiology it is important to estimate the volume of the left ventricle
> of the heart. It happens that this shape is similar to about 2/3 of a
> prolate ellipsoid. Is there a formula for the volume of a truncated

> ellipsoid?...

Yes; but some of your data are not quite clear. Is your "about 2/3"
a length measurement along an axis? I'll assume it is. Also, does your
"prolate ellipsoid" mean a prolate spheroid, i.e. having two axes equal
and the other longer (as for a rugby ball)? I shan't assume that, but it
comes out easily as a special case.

Suppose the three axes of the ellipsoid have lengths 2a, 2b, 2c.
I assume you want to cut the ellipsoid by a plane normal to the axis of
length 2a, keeping a fraction q of that axis. This means that 2qa
is the length of the solid. (You may want q = 2/3.) Then the volume
of the solid is

(4/3)(pi)q^2(3 - 2q)abc.

For example, q = 0 gives zero volume, q = 1 gives the standard
formula (4/3)(pi)abc for the volume of an ellipsoid, and q = 1/2 gives
just half of that, as you'd expect. If I understand you aright, you want
the case q = 2/3, which gives volume (80/81)(pi)abc, very close to
(3.1)abc.

If you want a prolate spheroid having b = c, then of course the
volume is (4/3)(pi)q^2(3 - 2q)ab^2. But always remember that the
notations a, b, c mean _half_ the lengths of the axes of the uncut
ellipsoid.

All this is just the answer without all the integrals to calculate
it, but if you want those I may try to type them out (in ascii, messily).

Ken Pledger.