Volume of ellipsoid bounded by two planes. 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!
 A: 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
$${\rm vol}(E)={\rm det}(D)\>{\rm vol}(Z)=2\sqrt{{5\over3}}\>{\rm vol}(Z)\ .$$
A: The axis of the ellipsoid in the $z$ direction is $1<2/\sqrt(3)$
so both limiting planes intersecate it.
The section with a plane at constant $z$ provides an ellipse , with equation
$$
{{x^{\,2} } \over {\left( {{{\sqrt 5 } \over 5}\sqrt {1 - {{3z^2 } \over 4}} } \right)^{\,2} }} + {{y^{\,2} } \over {\left( {5\sqrt {1 - {{3z^2 } \over 4}} } \right)^{\,2} }} = 1
$$
so with an area equal to
$$
A(z) = \pi ab = \pi \sqrt 5 \left( {1 - 3{{z^{\,2} } \over 4}} \right)
$$
Therefore the requested  volume is:
$$
\eqalign{
  & V = \int\limits_{ - 1/2}^1 {A(z)dz}  = \pi \sqrt 5 \int\limits_{ - 1/2}^1 {\left( {1 - 3{{z^{\,2} } \over 4}} \right)dz}  =   \cr 
  &  = \pi \sqrt 5 \left. {\left( {z - {{z^{\,3} } \over 4}} \right)} \right|_{z =  - 1/2}^{\;1}  = \pi \sqrt 5 {{39} \over {32}} \cr} 
$$
which confirms Ahmed's answer.
A: You can directly integrate to find the volume $V$ like in the following way . The only thing that changes is that now you will integrate from $z=-0.5$ to $z=1$
$$ V = \int\int\int (5x^2+\frac{y^2}{25}+\frac{3z^2}{4})dxdydz$$
limits of $x$ are $-\sqrt{1-\frac{y^2}{25}-\frac{3z^2}{4}}$ to $\sqrt{1-\frac{y^2}{25}-\frac{3z^2}{4}}$
limits of $y$ are $-\sqrt{1-\frac{3z^2}{4}}$ to $\sqrt{1-\frac{3z^2}{4}}$
limits of $z$ are $-0.5$ to $1$
A: We have,
$$(\frac{x}{\frac{1}{\sqrt{5}}})^2+(\frac{y}{5})^2+\frac{3}{4}z^2=1$$
Perform a "modified cylindrical coordinates" change of variables. Let $x=\frac{1}{\sqrt{5}} r \cos (\theta)$ and  $y= 5 r \sin (\theta)$. With $r \in [0,\sqrt{1-(3/4)z^2}]$ because we have $r^2+\frac{3}{4}z^2=1$. And let $\theta \in [0,2\pi]$ so that we close the ellipses. This gives a Jacobian of $\sqrt{5}r$. Then we have your volume is,
$$V=\int_{-0.5}^{1} \int_{0}^{2\pi} \int_{0}^{\sqrt{1-(3/4)z^2}} \sqrt{5}r dr d\theta dz$$
$$=\frac{39}{32}\pi \sqrt{5}$$
