Volume of ellipsoid bounded by two planes.

Question

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!

Answer 1

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 $1$

Answer 2

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.