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Volume integral over a region

Posted 9 years ago

Hey there!

I know a lot of basics in mathematica, but with the following equations I came to my limits: enter image description here

Is there any good way to compute this in mathematica?

Thank you in advance!

POSTED BY: Marcel Schwegler
4 Replies

Thank you guys! Never thought of such an amazing help! So my problem is solved ;)

POSTED BY: Marcel Schwegler

... deleted!

POSTED BY: Henrik Schachner

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

ImplicitRegion

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.

POSTED BY: Udo Krause

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.

POSTED BY: S M Blinder
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