Given is $\displaystyle \vec{v}=x\vec{u}_1+y\vec{u}_2+z\vec{u}_3$ and they ask me to calculate the flux if

$\displaystyle \left \{ \begin{array} \ z=x^2+y^2 \\ \ x^2+y^2 \leq 1 \end{array} \right $

Where $\displaystyle \vec{n}$ is in the up direction.

First I need to make a function who discripe the surface so $\displaystyle \left\{ \begin{array} \ x=x \\ y=y \\ z=x^2+y^2 \right \end{array} $ thus $\displaystyle \vec{F}=x\vec{u}_1 + y\vec{u}_2 + (x^2+y^2)\vec{u}_3$

and $\displaystyle \vec{F}_x=\vec{u}_1 +2x\vec{u}_3 $ and $\displaystyle \vec{F}_y=\vec{u}_2 +2y\vec{u}_3 $

If I calculate the cros product of the two I get $\displaystyle \vec{e}_3-2x\vec{e}_1-2y\vec{e}_2$ then $\displaystyle \left ( \begin{array} \ -2x \\ -2y \\ 1 \\ \end{array} \right ) \left ( \begin{array} \ x \\ y \\ z \\ \end{array} \right ) \right \begin{array} \ =-2x^2-2y^2+z \end{array} \left $

Then the integral $\displaystyle \int^{1}_{0} \int^{\sqrt{1-x^2}}_0-2x^2-2y^2+x^2+y^2 =\int^{1}_{0} \int^{\sqrt{1-x^2}}_0-x^2-y^2$

Is this methode oké? if I calculate the integral I get $\displaystyle \frac{\pi}{4}$ and not $\displaystyle \frac{\pi}{2}$

Who can help? Greets.