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Flux integral over ellipsoidal surface

  1. Jan 19, 2016 #1
    1. The problem statement, all variables and given/known data

    I am having trouble with part iii) of the following problem:

    Verify the divergence theorem for the function ##\vec{u} = (xy,- y^2, + z)## and the surface enclosed by the three parts:
    (i) ##z = 0, x^2+y^2 < 1##,
    (ii) ##x^2+y^2 = 1, 0 \le z \le 1## and
    (iii) ##(a^2-1)x^2+(a^2-1)y^2+z^2 = a^2 , 1 \le z \le a, a > 1##.

    2. Relevant equations

    Divergence theorem:

    ##\int_V\vec{ \nabla}\cdot\vec{u}~dV=\oint_{\partial v}\vec{u}\cdot\vec{dS}##

    3. The attempt at a solution

    The boundaries in iii) describe part of an ellipsoid. Rewrite the equation as follows:
    ##\frac{x^2}{(\frac{a}{\sqrt{a^2-1}})^2}+\frac{y^2}{(\frac{a}{\sqrt{a^2-1}})^2}+\frac{z^2}{a^2}=1##

    LHS## = \int_0^{2\pi}\int_1^a\int_0^{\frac{z^2-a^2}{1-a^2}}\vec{ \nabla}\cdot\vec{u}~\rho~ d\rho ~dz ~d\theta##
    ##~~~~ = \int_0^{2\pi}\int_1^a\int_0^{\frac{z^2-a^2}{1-a^2}}\rho - \rho^2\sin{\theta}~ d\rho ~dz ~d\theta##
    ##~~~~=\frac{4\pi}{1-a^2}##

    I think that is correct... maybe someone can check my setup.

    I'm stuck with the RHS.. I tried to use cylindrical coordinates and it got ugly quickly, I tried to use generalised spherical coordinates but it got ugly equally as fast. I expect the process to be somewhat ugly, but mine is so much I feel for certain i'm doing something wrong. I'll give a brief overview of my strategy:

    For cylindrical coordinates I tried to parameterize it as follows: ##\vec{r}(\theta,z)=(\frac{z^2-a^2}{1-a^2}\cos{\theta},\frac{z^2-a^2}{1-a^2}\sin{\theta},z)##. For starters im not even sure if this is right. I then get the normal vector to this by computing: ##\vec{n}=\frac{\partial \vec{r}}{\partial \theta}\times\frac{\partial \vec{r}}{\partial z}##. Computing this and normalising i get: ##\hat{n}=\frac{(\cos{\theta},\sin{\theta},\frac{-2z}{1-a^2})}{\sqrt{1+\frac{4z^2}{(1-a^2)}}}##.Plugging all of this into the flux integral looks extremely messy and I don't think I will be able to do the integral. So I abort and resort to spherical coordinates...

    For spherical coordinates I tried to parameterize it as follows: ##\vec{r}(\theta,\phi)= (\frac{a}{\sqrt{a^2-1}}\sin{\theta}\cos{\phi},\frac{a}{\sqrt{a^2-1}}\sin{\theta}\sin{\phi},a\cos{\theta})## and the rest of the story goes like the cylindrical one..Any help would be greatly appreciated!
     
    Last edited: Jan 19, 2016
  2. jcsd
  3. Jan 19, 2016 #2

    LCKurtz

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    I don't understand your use of ##s##. What is part (ii) ##s = 1, 0 \le z \le 1## supposed to describe?
     
  4. Jan 19, 2016 #3
    I find the notation confusing too, this is my lecturers not mine. I assumed ##s^2=x^2+y^2## though out and hence ii) describes a cylinder of radius 1 and height 1.
     
  5. Jan 19, 2016 #4

    LCKurtz

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    If part (ii) is supposed to be ##x^2+y^2 = 1,~0\le z \le 1 ## then write it that way. Then while you are at it, write part (iii) out so it makes sense. Is part (i) supposed to be ##0\le x^2 + y^2 \le 1,~z = 0## ?
     
  6. Jan 19, 2016 #5
    Apologies, I thought I might have been interpreting it in the wrong way so I left it just in case. I have amended the original post. And yes i) is the unit disk in the xy plane.
     
  7. Jan 19, 2016 #6

    LCKurtz

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    OK, that makes more sense. So for part (iii) if you let ##b^2 = \frac{a^2}{a^2-1}## your equation is$$
    \frac{x^2}{b^2}+\frac{y^2}{b^2}+\frac{z^2}{a^2} = 1$$I haven't worked it out, but what I would try is$$
    x = b\sin\phi\cos\theta,~y = b\sin\phi\sin\theta,~z = a\cos\phi$$

    Edit, added: @pondzo: I realized after I posted this that it is what you tried for generalized spherical coordinates. For what it's worth, I calculated the integral with Maple and got$$
    \frac{2\pi}{3}\left(\frac{a^6\sqrt{2a^2-1}}{(a^2-1)(1-2a^2)^2} \right)$$for (iii).
     
    Last edited: Jan 19, 2016
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