Flux of a vector field over an elliptical region

In summary, the problem asks to find the flux of the vector field <-1, -1, -y> over the surface of the ellipsoid {x}^{2}+2\,{y}^{2}+{z}^{2}=1 where the surface is the part of the plane region z + x = 1 oriented in the positive z direction. The solution involves parametrizing the surface and finding the intersection of the plane and the ellipsoid, which is a circle of radius 1/2 centered at (1/2,0). Then, by converting to polar coordinates and finding the dot product of the vector field and the surface normal, the integral is solved and the flux is found to be -pi
  • #1
peacey
2
0

Homework Statement


Find the Flux of the Vector Field <-1, -1, -y> where the surface is the part of the plane region z + x = 1 that is on the ellipsoid [tex]{x}^{2}+2\,{y}^{2}+{z}^{2}=1[/tex]
(oriented in the +ve z direction)

Homework Equations


Surface Integral

The Attempt at a Solution


Parametrize the Surface:
<u, v, 1 - u>

The intersection of the plane and the ellipsoid is:
[tex]{u}^{2}+2\,{v}^{2}+ \left( 1-u \right) ^{2}=1[/tex]
[tex]{u}^{2}+{v}^{2}=u[/tex]

Which is a circle of radius 1/2 centered at (1/2,0)
Or the polar region [tex]0\leq r\leq \cos \left( \theta \right) [/tex] and [tex]0\leq \theta\leq 2\,\pi [/tex]

Then, ru x rv = <1, 0, 1>
Then dotting the vector field with the above vector = -1 - v

So the integral becomes:
[tex]\int \!\!\!\int \!-1-v{dv}\,{du}[/tex]
After converting to polar and limits for the circle:
[tex]\int _{0}^{2\,\pi }\!\int _{0}^{\cos \left( \theta \right) }\!-r-{r}^{
2}\sin \left( \theta \right) {dr}\,{d\theta}
[/tex]
Which gives me [tex]-1/2\,\pi [/tex]

But, when I try to find the flux with maple by using the Flux command, it gives me -pi/4

Am I doing it wrong? Could someone point out where I went wrong please?

Thank you!
 
Last edited:
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  • #2
Ooh nevermind, I found out why! Such a stupid mistake. I gave theta the range of 0 to 2pi, while the circle isn't defined after pi/2. I shouldve gone from -pi/2 to pi/2. Works then!
 

Related to Flux of a vector field over an elliptical region

What is the definition of flux of a vector field over an elliptical region?

The flux of a vector field over an elliptical region is the measure of the flow of the vector field through the boundary of the region. It represents the amount of "flow" or "flow rate" of the vector field through the region.

How is the flux of a vector field over an elliptical region calculated?

The flux of a vector field over an elliptical region can be calculated using the formula ∫∫R F · n dS, where F is the vector field, n is the unit normal vector to the boundary of the region, and dS represents the infinitesimal surface area element.

What is the significance of the direction of the flux of a vector field over an elliptical region?

The direction of the flux of a vector field over an elliptical region indicates the direction of the flow of the vector field through the boundary of the region. A positive flux indicates a flow out of the region, while a negative flux indicates a flow into the region.

Can the flux of a vector field over an elliptical region be negative?

Yes, the flux of a vector field over an elliptical region can be negative. This indicates that there is a net flow into the region, as opposed to a net flow out of the region for a positive flux.

How is the flux of a vector field over an elliptical region related to the divergence of the vector field?

The flux of a vector field over an elliptical region is mathematically related to the divergence of the vector field through the Divergence Theorem. This theorem states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of the vector field over the volume enclosed by the surface.

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