Find the outward flux of a vector field across an ellipsoid

[1up20x6]

Homework Statement

[/B]
Find the outward flux of the vector field $\vec F = y^2e^{z^2+y^2} i + x^2 e^{z^2+x^2} j + z^2 e^{x^2+y^2} k$ across that part of the ellipsoid
$$ x^2 + y^2 + 4z^2 = 8$$ which lies in the region $0 ≤ z ≤ 1$
(Note: The two “horizontal discs” at the top and bottom are not a part of the ellipsoid.)
(Hint: Use the Divergence Theorem, but remember that it only applies to a closed surface, giving the total flux outwards across the whole closed surface)

Homework Equations

$\iint_S \vec F \cdot \vec n \, dS = \iiint_D \text{div}\ \vec F \, dV$

The Attempt at a Solution

I know that since the top and bottom discs aren't part of the ellipsoid, I'll have to subtract their fluxes from the final result. I also think I found the fluxes of those, at $z = 0$ the flux is 0 since $$ \iint_S \vec F \cdot \vec n \, dS = \iint_S <y^2e^{z^2+y^2}, x^2 e^{z^2+x^2}, z^2 e^{x^2+y^2}> \dot <0, 0, -1> \, dS = \iint_S -z^2 e^{x^2+y^2} dS$$ which evaluates to 0.
At $z = 1$, $$ \iint_S \vec F \cdot \vec n \, dS = \iint_S <y^2e^{z^2+y^2}, x^2 e^{z^2+x^2}, z^2 e^{x^2+y^2}> \dot <0, 0, 1> \, dS = \iint_S z^2 e^{x^2+y^2} dS$$ which evaluates to $$\iint_S e^{r^2} rdrd\theta = 18\pi e^4$$

When I try to find the divergence of the ellipsoid, I get
$$ \iint_S \vec F \cdot \vec n \, dS = \iiint_D \text{div}\ \vec F \, dV = \iiint_D 2ze^{x^2+y^2} \, dV$$

But at this point I'm lost. I thought I might have to convert to spherical coordinates, but that gets me a really ugly and long parameter where $$r_u \text{x}\ r_v = <-4\cos u \sin^2 v , -4\sin u \sin^2 v , -8\sin v \cos v>$$
and computing the dot product of this with $\vec F$ converted to spherical coordinates gives me
$$(-32 \cos u \sin u \sin^4 v)(\sin u e^{2\cos v + 8\sin^2 u \sin^2 v} + \cos u e^{2\cos v + 8\cos^2 u \sin^2 v}) - 16 \sin v \cos^2 v e^{8\cos^2 u \cos^2 v + \sin^2 u \sin^2 v}$$

So I don't think that's right.

 

[LCKurtz]

Homework Statement

[/B]
Find the outward flux of the vector field $\vec F = y^2e^{z^2+y^2} i + x^2 e^{z^2+x^2} j + z^2 e^{x^2+y^2} k$ across that part of the ellipsoid
$$ x^2 + y^2 + 4z^2 = 8$$ which lies in the region $0 ≤ z ≤ 1$
(Note: The two “horizontal discs” at the top and bottom are not a part of the ellipsoid.)
(Hint: Use the Divergence Theorem, but remember that it only applies to a closed surface, giving the total flux outwards across the whole closed surface)

Homework Equations

$\iint_S \vec F \cdot \vec n \, dS = \iiint_D \text{div}\ \vec F \, dV$

The Attempt at a Solution

I know that since the top and bottom discs aren't part of the ellipsoid, I'll have to subtract their fluxes from the final result. I also think I found the fluxes of those, at $z = 0$ the flux is 0 since $$ \iint_S \vec F \cdot \vec n \, dS = \iint_S <y^2e^{z^2+y^2}, x^2 e^{z^2+x^2}, z^2 e^{x^2+y^2}> \dot <0, 0, -1> \, dS = \iint_S -z^2 e^{x^2+y^2} dS$$ which evaluates to 0.
At $z = 1$, $$ \iint_S \vec F \cdot \vec n \, dS = \iint_S <y^2e^{z^2+y^2}, x^2 e^{z^2+x^2}, z^2 e^{x^2+y^2}> \dot <0, 0, 1> \, dS = \iint_S z^2 e^{x^2+y^2} dS$$ which evaluates to $$\iint_S e^{r^2} rdrd\theta = 18\pi e^4$$

When I try to find the divergence of the ellipsoid, I get
$$ \iint_S \vec F \cdot \vec n \, dS = \iiint_D \text{div}\ \vec F \, dV = \iiint_D 2ze^{x^2+y^2} \, dV$$

But at this point I'm lost. I thought I might have to convert to spherical coordinates, but that gets me a really ugly and long parameter where $$r_u \text{x}\ r_v = <-4\cos u \sin^2 v , -4\sin u \sin^2 v , -8\sin v \cos v>$$
and computing the dot product of this with $\vec F$ converted to spherical coordinates gives me
$$(-32 \cos u \sin u \sin^4 v)(\sin u e^{2\cos v + 8\sin^2 u \sin^2 v} + \cos u e^{2\cos v + 8\cos^2 u \sin^2 v}) - 16 \sin v \cos^2 v e^{8\cos^2 u \cos^2 v + \sin^2 u \sin^2 v}$$

So I don't think that's right.

You have (abusing your notation slightly) $$\iiint_V = \iint_S + \iint_{D1} + \iint_{D2}$$where $D_1$ and $D_2$ are the top and bottom discs you have calculated. You are trying to get the value of $\iint_S$. All you need to calculate to get that is $\iiint_V$ of the divergence. That is a volume integral, not a surface integral. It's true you are going to have to do that in two pieces in "spherical" coordinates. I would suggest the change of variables $$x = \sqrt 8\rho\sin\phi\cos\theta,~y =\sqrt 8\rho\sin\phi\sin\theta,~z=\sqrt 2\cos\phi$$The Jacobian turns out pretty simple. I haven't worked all the details but that should get you going.