# Find the outward flux of a vector field across an ellipsoid

• 1up20x6
In summary, the outward flux of a vector field is the measure of the flow of the vector field through a closed surface. A vector field is a mathematical function that assigns a vector to each point in space, often used to describe physical quantities. An ellipsoid is a three-dimensional shape similar to a sphere, but with different axis lengths. To find the outward flux of a vector field across an ellipsoid, Gauss's Law for Flux can be used. This involves calculating a surface integral using the dot product between the vector field and the surface normal vector. Finding the outward flux of a vector field across an ellipsoid has many applications in fields such as physics, engineering, and mathematics, including analyzing fluid flow, electric and magnetic fields, solving boundary
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.

1up20x6 said:

## 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.

## What is an outward flux?

The outward flux of a vector field is the measure of the flow of a vector field through a closed surface. It represents the net amount of the vector field that is passing out of the surface.

## What is a vector field?

A vector field is a mathematical function that assigns a vector to each point in space. It is often used to describe physical quantities such as velocity, force, or electric and magnetic fields.

## What is an ellipsoid?

An ellipsoid is a three-dimensional geometric shape that is similar in shape to a sphere, but its axes have different lengths. It can be described as a stretched or compressed sphere.

## How do you find the outward flux of a vector field across an ellipsoid?

To find the outward flux of a vector field across an ellipsoid, you can use the Gauss's Law for Flux. This involves calculating the surface integral of the dot product between the vector field and the surface normal vector over the surface of the ellipsoid.

## What are some applications of finding the outward flux of a vector field across an ellipsoid?

The calculation of outward flux of a vector field across an ellipsoid has applications in various fields such as physics, engineering, and mathematics. It can be used to analyze fluid flow, electric and magnetic fields, and to solve boundary value problems. It also has applications in image processing and computer graphics.

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