The integral separates
$\int_{M} z dx dy dz = \int_{0}^{c}z\int_{A(z)} dA dz$
with
$A(z)=\pi a b \left(1-\frac{z^2}{c^2}\right)$ as the area of the ellipse
$\frac{x^2}{a^2 \left(1-\frac{z^2}{c^2}\right)} + \frac{y^2}{b^2 \left(1-\frac{z^2}{c^2}\right)}=1$ at height
$z$. So
In[1]:= Integrate[\[Pi] a b z (1 - z^2/c^2), {z, 0, c}]
Out[1]= 1/4 a b c^2 \[Pi]
Is there any good way to compute this in mathematica?
A good way is always to strip off obvious parts and let do
$Mathematica$ the hard part of a problem - which does not exist here, anyway. Nevertheless you can do
In[11]:= Integrate[z, {x, y, z} \[Element] Ellipsoid[{0, 0, 0}, {a, b, c}]]
Out[11]= ConditionalExpression[0, a > 0 && b > 0 && c > 0]
primarily to check syntax because the result is obvious. And now
In[15]:= Integrate[
z, {x, y, z} \[Element]
ImplicitRegion[ x^2/a^2 + y^2/b^2 + z^2/c^2 <= 1 && 0 <= z <= c && a > 0 && b > 0 && c > 0, {x, y, z}]][[1]]
Out[15]= {{1/4 a b c^2 \[Pi], a > 0 && b > 0 && c > 0}}
the first element was only used to get the result into one line for the post. Without that it looks like
and agrees with the hand-made separation.
Is the usage of ImplicitRegion
a good way? If you say yes, then the answer to your question is yes.