Hey there!
I know a lot of basics in mathematica, but with the following equations I came to my limits:
Is there any good way to compute this in mathematica?
Thank you in advance!
The volume of an ellipsoid is equal to (4/3)Pi a b c. Cutting the ellipsoid in half by taking z>=0, volume reduces to (2/3)Pi a b c.
Condition on volume then implies (2/3)Pi a b c <= 1, which depends on values of a, b, c.
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.
… deleted!
Thank you guys!
Never thought of such an amazing help!
So my problem is solved 