Flux of a vector field over an elliptical region

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SUMMARY

The discussion focuses on calculating the flux of the vector field <-1, -1, -y> over a surface defined by the plane z + x = 1, constrained by the ellipsoid x² + 2y² + z² = 1. The correct parametrization of the surface leads to the integral setup, which initially yielded an incorrect flux value of -π/4 using Maple. The error was identified as an incorrect range for θ, which should be from -π/2 to π/2, resulting in the correct flux of -1/2π.

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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}^{<br /> 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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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!
 

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