I need to find the volume of ellipsoid: $$5x^2 + {y^2\over25} + {3z^2\over4} = 1$$
if the ellipsoid is bounded by $z={-1\over2}$ and $z=1$ planes.
I was able to find the total volume of the ellipsoid using the formula $V={4\pi\over3}*abc={40\pi\over3\sqrt15}$. But I don't think i can use this in any way to find the volume of the ellipsoid that is bounded by 2 planes.
To find the actual volume, I'm pretty sure I need to solve this: $V=\int_{-1\over2}^1S(x)dx$, where $S(x)$ is the area of the cross section of the ellipsoid, which is an ellipse. Now I think that the right move here would be to get the cross sections that are parallel to the $z$-axis. However my question is, how can I find these areas of the ellipses to plug into the above formula?
Any suggestions, would be greatly appreciated!
Best Answer
The map $$D:\quad(\xi,\eta,\zeta)\mapsto (x,y,z):=\left({1\over\sqrt {5}}\xi, \ 5\eta, \ {2\over\sqrt{3}}\zeta\right)$$ maps the spherical zone $$Z:=\left\{(\xi,\eta,\zeta)\biggm| \xi^2+\eta^2+\zeta^2\leq 1, \ -{\sqrt{3}\over4}\leq\zeta\leq{\sqrt{3}\over2}\right\}$$ onto the ellipsoid zone $E$ in question. For ${\rm vol}(Z)$ there are elementary formulas in the books. Now $${\rm vol}(E)={\rm det}(D)\>{\rm vol}(Z)=2\sqrt{{5\over3}}\>{\rm vol}(Z)\ .$$