# Partial Surface Area of Oblate Spheroid?

1. Feb 9, 2009

### hbchao

Basically I have a horizontal tank with elliptical heads on each end. Given any liquid height, I would like to calculate the surface area which is in contact with the liquid.

The heads can be represented by a general oblate spheroid described by equation (x2 + y2)/a2 + z2/c2 = 1.

I can compute the partial surface area for z=XXX to YYY by taking the surface of revolution about the z-axis (ie for a vertical tank).

However how can I calculate the partial surface area for x=XXX to YYY?

I am not looking for a closed form equation, just something I can numerically integrate.

2. Feb 9, 2009

### arildno

Well, you could parametrize the oblate spheroid as follows:
$$x=a\sin\phi\cos\theta,y=a\sin\phi\sin\theta, z=c\cos\phi, 0\leq\phi\leq\frac{\pi}{2},0\leq\theta\leq{2\pi}$$
Note that these are tweaked spherical coordinates!

Thus, the surface $$\vec{S}$$ can be represented as:
$$\vec{S}(\phi,\theta)=a\sin\phi\cos\theta\vec{i}+a\sin\phi\sin\theta\vec{j}+c\cos\phi\vec{k}$$

Local tangent vectors are then given by the partial derivatives of $$\vec{S}$$:
$$\vec{T}_{\phi}=\frac{\partial\vec{S}}{\partial\phi}$$
$$\vec{T}_{\theta}=\frac{\partial\vec{S}}{\partial\theta}$$

The local, infinitesemal area element is then the area of the parallellogram spanned by the infinitesemal tangent vectors $$\vec{T}_{\phi}d\phi,\vec{T}_{\theta}d\theta$$; and that equals $$||\vec{T}_{\phi}\times\vec{T}_{\theta}||d\phi{d}\theta$$

We therefore get that the area A of one end can be calculated as:
$$A=\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}||\vec{T}_{\phi}\times\vec{T}_{\theta}||d\phi{d}\theta$$

which is some sort of elliptical integral, I think.
EDIT:
At least, it should be something like:
$$A=2ac\pi\int_{0}^{\frac{\pi}{2}}\sin\phi\sqrt{1+\frac{a^{2}-c^{2}}{c^{2}}\cos^{2}\phi}{d}\phi$$

Last edited: Feb 9, 2009
3. Feb 9, 2009

### arildno

Hmm, on further thought, this should be exactly solvable!
Set $$u=\epsilon\cos\phi,\epsilon=\sqrt{\frac{a^{2}-c^{2}}{c^{2}}}$$
Thus, we get:
$$du=-\epsilon\sin\phi{d\phi}$$, or :
$$A=-\frac{ac\pi}{\epsilon}\int_{\epsilon}^{0}\sqrt{1+u^{2}}du=\frac{ac\pi}{\epsilon}\int^{\epsilon}_{0}\sqrt{1+u^{2}}du$$
Setting $$u=Sinh(v)$$, and we readily integrate the expression!

4. Feb 9, 2009

### hbchao

Thanks for the reply...however it has been many years since my high school/college days and I'm having a bit of a hard time following the transformation...

If the liquid level in the tank is X, how would I go about determining the limits of integration?

5. Feb 9, 2009

### arildno

Well, one half of the oblate spheroid should, if I did this correctly (I forgot a factor of 2 from my next to last post), be:
$$A=\pi(ac\frac{Sinh^{-1}\epsilon}{\epsilon}+a^{2}), \epsilon=\sqrt{\frac{a^{2}-c^{2}}{c^{2}}},a\geq{c})$$
Note that this yields half the sphere's surface area when a=c.

I'm sure you can find out the rest of your answer by yourself.

Last edited: Feb 10, 2009