# Partial Surface Area of Oblate Spheroid?

• hbchao
In summary, the conversation discusses calculating the surface area of a horizontal tank with elliptical heads on each end. The heads are represented by a general oblate spheroid and the conversation explores various ways to calculate the surface area using parametrization and integration. The final solution involves setting u=Sinh(v) and integrating the expression to find the surface area. The conversation also mentions determining the limits of integration and the final solution yields half the sphere's surface area when a=c.
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.

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:
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!

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?

hbchao said:
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:
$$A=\pi(ac\frac{Sinh^{-1}\epsilon}{\epsilon}+a^{2}), \epsilon=\sqrt{\frac{a^{2}-c^{2}}{c^{2}}},a\geq{c})$$
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:

## 1. What is a partial surface area of oblate spheroid?

A partial surface area of an oblate spheroid refers to the area of a curved surface on an oblate spheroid that is not the entire surface. It is a measure of the amount of the spheroid's surface that is included within a certain boundary.

## 2. How is the partial surface area of oblate spheroid calculated?

The partial surface area of an oblate spheroid can be calculated using the formula: A = 2πa²[1+ (1-e²)/e * atan(e)] where A is the partial surface area, a is the equatorial radius, and e is the eccentricity of the spheroid.

## 3. What are some real-world applications of calculating partial surface area of oblate spheroid?

The calculation of partial surface area of oblate spheroid is used in various fields such as astrophysics, geodesy, and cartography. It is used in determining the shape and size of planets, measuring the Earth's gravitational field, and making accurate maps of the Earth's surface.

## 4. How is the partial surface area of oblate spheroid different from total surface area?

The partial surface area of oblate spheroid only considers a portion of the spheroid's surface, while the total surface area includes the entire surface. The partial surface area is usually calculated for a specific region or boundary of the spheroid, while the total surface area is a measure of the entire surface of the spheroid.

## 5. How does the partial surface area of oblate spheroid relate to its volume?

The partial surface area of oblate spheroid is directly related to its volume. As the partial surface area increases, so does the volume. This is because a larger partial surface area means a larger portion of the spheroid's surface is included, resulting in a larger volume. The relationship between partial surface area and volume is also affected by the spheroid's eccentricity.

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