Partial Surface Area of Oblate Spheroid?

[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.

 

[arildno]

Well, you could parametrize the oblate spheroid as follows:
[tex]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}[/tex]
Note that these are tweaked spherical coordinates!

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

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

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

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

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

 

[arildno]

Hmm, on further thought, this should be exactly solvable!
Set [tex]u=\epsilon\cos\phi,\epsilon=\sqrt{\frac{a^{2}-c^{2}}{c^{2}}}[/tex]
Thus, we get:
[tex]du=-\epsilon\sin\phi{d\phi}[/tex], or :
[tex]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[/tex]
Setting [tex]u=Sinh(v)[/tex], and we readily integrate the expression!

 

[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?

 

[arildno]

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?

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:
[tex]A=\pi(ac\frac{Sinh^{-1}\epsilon}{\epsilon}+a^{2}), \epsilon=\sqrt{\frac{a^{2}-c^{2}}{c^{2}}},a\geq{c})[/tex]
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.