Calculating Flux through Ellipsoid

In summary, the conversation discusses a vector field defined by F=(-y,x,0) on a part of the surface of an ellipsoid, given certain conditions for the size of the ellipsoid. The question asks to explain why the integral of the vector field over the surface is zero in one case and to find the integral in another case. The answer is obtained by setting up the correct integral and evaluating it, resulting in a potentially complicated answer.
  • #1
bananabandana
113
5

Homework Statement


Let ## E ## be the ellipsoid:
$$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+z^{2}=1 $$
Let ## S ## be the part of the surface of ## E ## defined by:
$$ 0 \leq x \leq 1, \ 0 \leq y \leq 1, \ z > 0 $$
Let F be the vector field defined by $$ F=(-y,x,0)$$

A) Explain why ## \int_{S} \vec{F} \cdot \vec{dS} =0 ## in the case ##a=b##

B) In the case ## a \neq b##, find ## \int_{S} \vec{F} \cdot \vec{dS} ##

Homework Equations



The Attempt at a Solution


Part A)
This really isn't obvious to me. Can we make some kind of symmetry argument? I can see that ## \nabla \cdot \vec{F} =0 ## but this doesn't help us as the surface ## S## is not closed.

Part B)
It looks like I can do this ,but the answer I get is so ugly that it makes me very suspicious.

Can easily show (by calculating ## \nabla E(x,y,z) ## that:

(1) $$ \vec{dS} = \begin{pmatrix} \frac{x}{a^{2}z} \\ \frac{y}{b^{2}z} \\ 1 \end{pmatrix} dxdy $$

So:

(2) $$ I=\int \vec{F} \cdot \vec{dS} =\frac{a^{2}-b^{2}}{a^{2}b^{2}} \int_{x=0}^{x=1}\int_{y=0}^{y=1} \frac{xy}{z} $$

(3) Substituting in for ##z##:
$$ I = \frac{a^{2}-b^{2}}{a^{2}b^{2}} \int_{x=0}^{x=1}\int_{y=0}^{y=1} \frac{xy}{\sqrt{1-\bigg(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \bigg)}} $$

If I evaluate the inner integral by making the subtitution:
(4) $$ u=\sqrt{1-\bigg(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \bigg)} $$
and then evaluate the outer one with a similar job, I get the result:

(5) $$ I = \frac{a^{2}-b^{2}}{3} \bigg[ \bigg(1-(\frac{1}{a^{2}}+\frac{1}{b^{2}})\bigg)^\frac{3}{2} - \bigg(1-\frac{1}{b^{2}}\bigg)^{\frac{3}{2}} - \bigg(1-\frac{1}{a^{2}}\bigg)^{\frac{3}{2}}-1 \bigg] $$

Is this ridiculous? If so where did I go wrong? :P
Realize it is quite a tedious question, but has been very frustrating. Help would be much appreciated!

Thank you :D
 
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  • #2
Notice that the size of the ellipse is all that changes as z goes from zero to one. So you can fix z for one slice at a time.
Your equation 2 should be enough to see why it is zero when a=b.
Fix your bounds on you integrals so z goes from 0 to 1 and bounds on x and y are functions.
 
  • #3
Isn't the region of integration a square? ( I thought the definition of the region as ## 0 \leq x \leq 1 ## and ## 0 \leq y \leq 1 ## implied this?). Hence, no need for writing ## y=y(x)## in the limits...
and yes, I know I can get it from (2), but the question was phrased in a way that it suggested you were meant to see this intuitively before getting to (2), and I didn't know how to do that.

Thanks!
 
  • #4
It can't be the square, unless a and b are large enough. Look at the case where a=b=1. (1,1) which would be in the square region is not in your region of interest.
You have a few cases. Look at the simple one I just mentioned, then to see how a and b affect the outcome examine when they are both less than 1 (in magnitude), both more than one and mixed.
 
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  • #5
Ah, sorry. I forgot to say that the question stated ## a > \sqrt{2} ## , ## b >\sqrt{2} ## and ## a \neq b ##. Might this change things? (Though I admit, I don't know how they got these conditions).
 
  • #6
Okay, then the whole area of the square is in the region.
On ds, what would be the harm in leaving z in the z column?
Then the dot product would eliminate it.
 
  • #7
bananabandana said:
Ah, sorry. I forgot to say that the question stated ## a > \sqrt{2} ## , ## b >\sqrt{2} ## and ## a \neq b ##. Might this change things? (Though I admit, I don't know how they got these conditions).

That just guarantees your square lies completely inside the ellipse on the xy plane.
 
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  • #8
Oh, of course, because the diagonal of the square is ## \sqrt{2} ##. I see! Thanks. I think I am leaving ## z ## in the ##z## column though, but the vector field ## \vec{F} ## is ##\vec{F}=(-y,x,0) ## - or did you mean something else?

But am I correct in thinking now that the answer I first got is most likely right? :)
 
  • #9
bananabandana said:
$$ I = \frac{a^{2}-b^{2}}{a^{2}b^{2}} \int_{x=0}^{x=1}\int_{y=0}^{y=1} \frac{xy}{\sqrt{1-\bigg(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \bigg)}} $$

bananabandana said:
But am I correct in thinking now that the answer I first got is most likely right? :)

I didn't check your work any farther than your integral above. It is set up correctly.
 
  • #10
So long as it's set up right, I'm happy. Thanks for the help! :)
 

What is an ellipsoid?

An ellipsoid is a three-dimensional geometric shape that resembles a stretched out sphere. It is defined by three semi-axes, which are the three lengths that define its size and shape. The most commonly known example of an ellipsoid is the Earth, which is referred to as a geoid.

What is flux?

Flux is a measure of the flow of a physical quantity through a given surface. In other words, it is the amount of a substance or energy that passes through a surface or boundary in a given amount of time. Flux can be calculated for many different physical quantities, such as electric or magnetic fields, heat, or fluid flow.

How do you calculate flux through an ellipsoid?

To calculate flux through an ellipsoid, you need to use the formula for flux integration, which takes into account the shape and orientation of the ellipsoid, as well as the physical quantity being measured. This formula involves integrating over the surface of the ellipsoid and can be quite complex, so it is often done using computer software or specialized mathematical methods.

What factors affect the flux through an ellipsoid?

The flux through an ellipsoid is affected by several factors, including the size and shape of the ellipsoid, the orientation of the surface with respect to the direction of the quantity being measured, and the strength or magnitude of the quantity being measured. Additionally, external factors such as the surrounding environment or other nearby objects can also affect the flux through an ellipsoid.

What are some practical applications of calculating flux through an ellipsoid?

Calculating flux through an ellipsoid has many practical applications in various fields of science and engineering. For example, it can be used to analyze the flow of fluids or gases through a curved surface, to calculate the electric or magnetic field in a specific region, or to study heat transfer in different materials. It can also be used in geophysical studies to analyze the Earth's magnetic or gravitational fields.

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