# Regarding volume of an ellipsoid bounded by 2 planar cutting planes

1. Sep 10, 2014

### Corse

1. The problem statement, all variables and given/known data

Hi I require to compute the volume of a ellipsoid that is bounded by two planes. The first horizontal (xy) plane is cutting directly along the mid-section of the ellipsoid. The second horizontal plane is at a z = h below the first horizontal plane. The volume of the ellipsoid between these two planes is the interest.

2. Relevant equations

I understand that the volume of the ellipsoid is 4/3 ∏abc and the area of the cutting plane is ∏ab. However for this problem, I believe the use of integrals may be necessary.

3. The attempt at a solution

From the problem statement, I realised that this can be reduced to a half-ellipsoid and the single horizontal cutting plane at z = h below the first cutting plane. In doing so it will just be a subtraction between 2/3 ∏abc and the volume not bounded by the 2 planes.

If a is the length of the major axis of the 2nd cutting plane in the x direction,

a = A x √(1-(h/C)²);

where A is the major axis of ellipsoid in X direction
C = Major axis of ellipsoid in Z direction
h = vertical distance in the z direction of 2nd plane (below 1st plane)

I'm not sure how should I continue from here though.

Help!

Regards
Corse

2. Sep 10, 2014

### HallsofIvy

Staff Emeritus
So your ellipsoid is given by
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}+ \frac{z^2}{c^2}= 1$$
and it is cut by z= 0 and z= h< 0?

At z= 0, the plane cuts the ellipsoid in the ellipse
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$$
and at z= h, the plane cuts the ellipsoid in the ellipse
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1- \frac{h^2}{c^2}$$

I think I would be inclined to put this into "elliptical-cylindrical coordinates" taking
$x= ar cos(\theta)$, $y= br sin(\theta)$, and $z= cz';$

In those coordinates the equation of the ellipsoid becomes $r^2+ z'^2= 1$, the upper elliptical boundary becomes $r^2= 1$ (r= 1 since r cannot be negative), and the lower elliptical boundary becomes $r^2= 1- h^2$.

Of course, you will also have to change the differential using the Jacobian:
$$\left|\begin{array}{ccc}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z'} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z'} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z'}\end{array}\right|$$$$= \left|\begin{array}{ccc} a cos(\theta) & -ar sin(\theta) & 0 \\ b sin(\theta) & br cos(\theta) & 0 \\ 0 & 0 & c\end{array}\right|= abcr$$
so that $dxdydz= abc r drd\theta dz'$

Then the integral from r= 0 to r= 1- h will be from z'= h up z'= 0 while the integral from r= 1- h to 1 will be from $z'= -\sqrt{1- r^2}$ up to $z'= 0$.

3. Sep 11, 2014

### Corse

Hi HallsofIvy

Thank you for your quick response! That was helpful to get me going. I understand the jacobian transformation from the cartesian to the elliptical-cylindrical space.

I have a couple of questions:

1) Elliptical-cylindrical coordinates was used where x=arcos(θ), y=brsin(θ), and z=cz′;

May I know why were the coefficients of a,b and c in the elliptical coordinates?

2) You mentioned that the integral from r= 0 to r= 1- h will be from z'= h up z'= 0 .

So for my case, I would be interested in r=0 to r = 1-h; I'm assuming this fact would be crucial in evaluating the triple integral to obtain the volume. Which integration limits would this correspond to?

Could you advise regarding the limits of integration if they are correct?

$Volume = \int_{h/c}^0 \int_{0}^{2∏} \int abcr \,dr\,dΘ\,dz'$

Would it be necessary to obtain the traces of rΘ, rz' and Θz' to get the limits of integration?

4. Sep 11, 2014

### HallsofIvy

Staff Emeritus
$\frac{x}{a}= r cos(\theta)$, $\frac{y}{a}= r sin(\theta)$ and $\frac{z}{c}= z'$ so that
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}+ \frac{z^2}{c^2}= r^2 cos^2(\theta)+ r^2 sin^(\theta)+ z'^2= r^2+ z'^2= 1$$
The equation of a sphere in cylindrical coordinates.

I said that r goes from 0 to 1- h. So the integral will be
$$\int_{z'= h}^0\int_{r= 0}^{1- h}\int_{\theta= 0}^{2\pi} abcr d\theta dr dz'$$

5. Sep 11, 2014

### Corse

Hi HallsofIvy

I have a couple of questions:

1) May I know how did you get limits for r to be r = 0 to 1-h? Especially for the part r = 1-h

2) For z', should the bottom limits be -h instead of h?

3) Also, does converting the problem to cylindrical coordinates make it more convenient to solve? Could it be done in a similar fashion in cartesian coordinates?

Thank you for helping me out!

Regards
Corse

Last edited: Sep 11, 2014