Flux of a vector field over an elliptical region

[peacey]

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 {x}^{2}+2\,{y}^{2}+{z}^{2}=1
(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:
{u}^{2}+2\,{v}^{2}+ \left( 1-u \right) ^{2}=1
{u}^{2}+{v}^{2}=u

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

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

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

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!

 

[peacey]

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!