Use the same method as for volume, but define 2D region instead and integrate over this 2D region.

I.M.

Ben,

There are several ways to do it, here is an example:

In[1]:= (* with assignment *)
v = a b c h ;
a = 1 ; b = 2 ; c = 3 ; h = 4 ;
v 

Out[3]= 24

In[107]:= (* with rules *)
Clear[v, a, b, c, h] ;
v = a b c h ;
v /. {a -> 1, b -> 2, c -> 3, h -> 4}

Out[109]= 24

In[4]:= (* with function *)
Clear[v] ;
v[a_, b_, c_, h_] := a b c h ;
v[1, 2, 3, 4]

Out[6]= 24

In[7]:= (* with pure function *)
Clear[v] ;
v = #1 #2 #3 #4  &;
v[1, 2, 3, 4]

Out[9]= 24

Also see this, where useful basic concepts are explained. In general you can get all you answers from Mathematica documentation.

I.M.

Great solution, Adam,

I was not enough patient to get Integrate[] result. In fact if you add a condition that h plane must intersect the ellipsoid than computation is almost immediate.

Volume[reg, Assumptions -> a > 0 && b > 0 && c > 0 && h > 0 && h < c]

and Integrate gives the same result as it should:

Integrate[1, {x, y, z} \[Element] reg, Assumptions -> a > 0 && b > 0 && c > 0 && h > 0 && h < c]

I.M.

Hi,

Since version 10 it's possible to integrate over specified domain with both Integrate[] and NIntegrate[] functions. Consult documentation for details. Here is an example with NIntegrate[]:

R = ImplicitRegion[
    x^2/1 + y^2 + z^2 <= 1 && -0.5 <= z <= 0.5,
    {x, y, z}
   ] ;
RegionPlot3D[R]
NIntegrate[1, {x, y, z} \[Element] R]

I.M.