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!

• If you know the major and minor radii of an ellipse a and b, you can compute the area as PIab. Fixing a plane z=c you can compute the major and minor radii from the equation of the ellipsoid in terms of z. – rVitale Dec 7 '13 at 0:28
• Does this look right? $${x^2\over({1\over\sqrt5}\sqrt(1-{3z_0^2\over4}))^2} + {y^2\over(5\sqrt(1-{3z_0^2\over4}))^2}=1$$ From this $S(z_0)=\sqrt5\pi*(1-{3z_0^2\over4})$. Now for the volume $$V=\int_{-1\over2}^1 \sqrt5\pi*(1-{3z^2\over4}) dz={39\sqrt5\over32}\pi$$ I tried using the triple integral solution given below by Trafalgar Law and wrote it to Wolfram Mathemetica, but it didn't return a valid answer so I can't exactly check. – Visna Dec 7 '13 at 1:13
• This looks good, that's what I computed. I don't think that Trafalgar's is correct actually. I don't see how the surface function for the ellipse should be the integrand of a volume integral. – rVitale Dec 7 '13 at 1:17
• Thanks! Is it possible to accept a comment as an answer? Also from where can I mark your comment as a useful comment? – Visna Dec 7 '13 at 1:24
• if you place your cursor left of the comment you can upvote, but no you can't accept a comment as an answer. – rVitale Dec 7 '13 at 1:37

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 $${\rm vol}(E)={\rm det}(D)\>{\rm vol}(Z)=2\sqrt{{5\over3}}\>{\rm vol}(Z)\ .$$

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.

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 $1$

• The integrand should just be $1$. – alex.jordan Mar 12 '17 at 16:14

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,

$$V=\int_{-0.5}^{1} \int_{0}^{2\pi} \int_{0}^{\sqrt{1-(3/4)z^2}} \sqrt{5}r dr d\theta dz$$

$$=\frac{39}{32}\pi \sqrt{5}$$