- #1

1up20x6

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## Homework Statement

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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.