I used Stokes' theorem to evaluate a closed line integral on a vector field, and the projection of the region bounded by that closed line integral was shifted one unit to the left from the origin. The 3D shape was cylindrical in nature so I used cylindrical coordinates to describe it.
I struggled to find the bounds for the circular projection, but some friends of mine said that you could essentially just evaluate the double integral as if the circle was centered at the origin.
Is this right? For vector fields can you just shift the circle to be centered on the origin? I am almost positive this is false for scalar fields, but I suppose for vector fields the flux remains the same for shadows of the same area?
Edit: The specific problem I was working on was as follows:
Let F be a vector field $\mathbf{F}=<y, -x, y>$. Evaluate $\oint_C \mathbf{F} \cdot d \mathbf{r}$, where $C$ is the intersection between the paraboloid $z=x^2+y^2$ and the plane $z-2x=3$, oriented counter-clockwise as viewed from the positive Z-axis.
I parametrized $C$ as $\mathbf{r}=<rcos\theta,\ rsin\theta,\ 3+2rcos\theta>$
$\mathbf{r}_r \times \mathbf{r}_\theta = <-2r, 0, r>$ (this also indicates the integral is positive)
$curl(\mathbf{F}) = <1, 0, -2>$
Using Stokes'
$$\oint_C \mathbf{F} \cdot d \mathbf{r}=\iint_D curl(\mathbf{F}) \cdot \mathbf{r}_r \times \mathbf{r}_\theta\ drd\theta$$
Now at this point, I used bounds for D of $\theta$ goes $0$ to $2\pi$ and $r$ goes $0$ to $2$. I know that this circle has the same area as the shadow of the surface enclosed by C, but will it give the same answer?