Let $C$ be curve $(x-1)^2 + (y-2)^2 =4$ and $z=4$ orientated counterclockwise when viewed from high on the z-axis. Let $$\mathbf{F}(x,y,z)=(z^2 +y^2 +\sin x^2)\mathbf{i}+(2xy+xz)\mathbf{j}+(xz+2yz)\mathbf{k} $$ Evaluate $\oint_C \mathbf{F}\bullet d\mathbf{r}$
My work:
Normal $\mathbf{\hat{N}}$ is $\mathbf{k}$, $dS=dA$ and the projection to $xy$-plane is a circle with area of $4\pi$ and curl $\mathbf{F}=(2z-x)\mathbf{i}+z\mathbf{j}+\mathbf{k} $.
By Stokes theorem, $$\begin{align} \oint_C \mathbf{F}\bullet d\mathbf{r}&=\iint_S ((2z-x)\mathbf{i}+z\mathbf{j}+z\mathbf{k})\bullet\mathbf{k}\; dA \\ &=\int_0^{2\pi}\int_0^2 z r \;drd\phi,\; \text{$z=4$ on $S$ } \\ &=4\times4\pi=16\pi\end{align}$$ Does my work seem correct? Im not that sure if the fact that the circle is not origo centered changes something.
Your answer is correct, I believe, but you should be careful that you understand what happened when you evaluated the surface integral. Normally we can't directly evaluate surface integrals so we project them onto a plane (normally the plane $z=0$). The projected image would be centred on $(1,2)$, and so you wouldn't be able to use the simple polar transformation you used. Luckily, the curl dotted with the normal gives a function dependent only on $z$, a constant, so it doesn't matter where you evaluated this particular integral, only the area of the projection - which is in this case just the area of the circle.