We let $S$ be the surface of the paraboloid $2z = x^2+y^2$ bounded by $z=2$, and let $C$ be the boundary of $S$ at $z=2$. We want to use Stokes' Theorem to evaluate
$$\oint_C \vec{v} \cdot d\vec{r}$$
where $\vec{v} = 3y \hat{i} - xz \hat{j} + yz^2 \hat{k}$
Note this is not for an assignment, but for exam practice. This is my attempt:
We have
$$\nabla \times \vec{v} = (z^2+x)\hat{i} + 0\hat{j} + (-z-3)\hat{k}$$
To find $\hat{n}$, we first have $f(x,y,z) = x^2+y^2-2z = 0$, then
$$\hat{n} = \frac{\nabla f}{||\nabla f||} = \frac{2x\hat{i}+2y\hat{j}-2\hat{k}}{\sqrt{4x^2+4y^2+4}} = \frac{2x\hat{i}+2\hat{j}-2\hat{k}}{\sqrt{8z+4}} = \frac{x}{\sqrt{2z+1}}\hat{i} + \frac{y}{\sqrt{2z+1}}\hat{j} - \frac{1}{\sqrt{2z+1}}\hat{k}$$
So
$$\oint_C \vec{v} \cdot d\vec{r} = \iint (\nabla \times \vec{V})\cdot\hat{n} dS = \iint \frac{x^2+xz^2+z+3}{\sqrt{2z+1}} dS$$
I then parameterized by $x = r\text{cos}(\theta), y=r\text{sin}(\theta), z =r$, where $0 \leq \theta \leq 2\pi$ and $0 \leq r \leq 2$. I get that the 'distortion' is $\sqrt{2}r$. I then substituted all the above into the integral and got something odd like
$$\frac{2}{15} (55\sqrt{5} - 41)\pi$$ from WolframAlpha, although the answer should be something like $20\pi$. Could someone please point me in the right direction? Thanks!