Letting $S \in \mathbb{R}^3$ be the paraboloid which is the graph of $1-x^2-y^2$ over the unit disk $D$ centred at the origin in the $x,y$ plane. $S$ is parameterised by $$\textbf r(x,y)=(x,y,1-x^2-y^2), x^2+y^2\leq 1$$ Let $n$ be the unit normal to $S$ and $D$ which points out of $\Omega$ (the solid region enclosed by $S$ and $D$, and let $v$ be the vector field defined by $$v(x,y,z)=(x,y,1)$$ I want to evaluate $\int \int_{S}v\cdot n dA$ and $\int \int_{D}v\cdot ndA$
I started out by calculating the normal for the surface $S$ which got me $(2x,2y,1)$ and then turning this into the unit normal left me with $n=\frac{(2x,2y,1)}{\sqrt{4x^2+4y^2+1}}$, but then when I substitute in to calculate the flux I am left with the integral $$\int \int_{S} \frac{2x^2+2y^2+1}{\sqrt{4x^2+4y^2+1}}dxdy$$ which I tried to use polar co-ordinates to calculated but was unsuccessful, any help where I'm going wrong? Thanks.
First, you should write $\iint_D ...\,dx\,dy$ once you've written things in terms of the parametrization. Next, the serious error is that you've forgotten that $dS = \sqrt{4(x^2+y^2)+1}\,dx\,dy$, so the length of the normal vector cancels out and you should have just $$\iint_D (1+2(x^2+y^2))\,dx\,dy.$$
COMMENT: If you've had Gauss's Theorem (the Divergence Theorem), you can double-check your work because the flux upwards across the paraboloid minus the flux upwards across the disk should be equal to the triple integral of the divergence over the volume (which is $\pi$).