Calculating fluxintegral out of the surface $1-x^2-y^2$ I am trying to calculate the flux integral of the vector field
$$
\vec{F} = (x,y,1+z)
$$
Out of the surface $z=1-x^2-y^2$, $z\geq 0$
Answer : $\frac{5\pi}{2}$
I begin by defining a vector that traces out the surface and calculate the cross product of its derivate to get normal vector.
$$
\vec{r} = (x,y,1-x^2-y^2)\\
\vec{r}_x\times \vec{r}_y = (2x,2y,1)
$$
Next, I calculate the corresponding double integral using polar coordinates:
$$
\iint \vec{F} \cdot \hat{n}dS = \iint_D\vec{F}\cdot\vec{n}dxdy = \iint_D2+x^2+y^2dxdy\\
=\int_0^{2\pi}\int_0^{1}(2+r^2)*rdrd\theta \neq \frac{5\pi}{2}
$$
Where am I going wrong?
 A: Isn't $\vec F \cdot (\vec r_x\times \vec r_y) = 2x^2+2y^2+1+1-x^2-y^2 = 2+x^2+y^2$?
A: You should consider that $ds$=$\frac{dxdy}{\sigma}$ and thus projecting the whole surface on the XOY plane where $\sigma$=$|\vec n . \vec k|$ and $\vec n$ is the unit vector that is in this case $\vec n$ = $\frac{\vec grad(S)}{||\vec grad(S)||}$ = $\frac{2x\vec i + 2y\vec j + 2z\vec k}{\sqrt{4x^2 +4y^2 +4z^2}}$=$\frac{x\vec i + y\vec j + z\vec k}{\sqrt{x^2 +y^2 +z^2}}$=$\frac{x\vec i + y\vec j + z\vec k}{1}$ where $x^2 +y^2 +z^2=1$ and so you can evaluate your integral.
$$
\iint \vec{F} \cdot \hat{n}dS = \iint_D\vec{F}\cdot\vec{n}\frac{dxdy}{\sigma} = \iint_Dx^2+y^2+z(1-z)\frac{dxdy}{z}=\iint_D \frac{x^2+y^2-(1-x^2-y^2)+\sqrt{1-x^2-y^2}dxdy}{\sqrt{1-x^2-y^2}}=2 \iint_D \frac{x^2+y^2}{\sqrt{1-x^2-y^2}}dxdy -\iint_D \frac{1}{\sqrt{1-x^2-y^2}}dxdy + \iint_D dxdy$$. With D= {${{ (x,y) \in R^2 \ ,  x^2+y^2=1}}$} Now considering $x=rcos(\theta)$, $y=rsin(\theta)$ with the Jaacobian $|J|$ = $r$ you can solve the integral that will surely lead to $\frac{5\pi}{2}$
