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# Volume integral over a region

Posted 8 years ago
 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!
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Posted 8 years ago
 Thank you guys! Never thought of such an amazing help! So my problem is solved ;)
Posted 8 years ago
 ... deleted!
Posted 8 years ago
 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 likeand 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 8 years ago
 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.