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Calculating Flux through Ellipsoid

  1. Mar 28, 2015 #1
    1. The problem statement, all variables and given/known data
    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} ##
    2. Relevant equations

    3. 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
     
  2. jcsd
  3. Mar 28, 2015 #2

    RUber

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    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.
     
  4. Apr 4, 2015 #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!
     
  5. Apr 4, 2015 #4

    RUber

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    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.
     
  6. Apr 4, 2015 #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).
     
  7. Apr 4, 2015 #6

    RUber

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    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.
     
  8. Apr 4, 2015 #7

    LCKurtz

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    That just guarantees your square lies completely inside the ellipse on the xy plane.
     
  9. Apr 4, 2015 #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? :)
     
  10. Apr 4, 2015 #9

    LCKurtz

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    I didn't check your work any farther than your integral above. It is set up correctly.
     
  11. Apr 6, 2015 #10
    So long as it's set up right, I'm happy. Thanks for the help! :)
     
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