Interval of stokes theorem

I have the vectorfield $F=(y e^x,x^2+e^x,z^2)$ and the curve $r(t)=(1+cos(t),1+sin(t),1-cos(t)-sin(t))$

I have to find the line integral of this using stokes theorem, however in this case I'm only interested in the interval. And I'm confused about how to define the $r$ region interval.

When the curve is projected on the x-y plane it creates a circle with center in (1,1) and r=1. Here $\theta$ goes from $0$ to $2\pi$. But how should I interpret $r$? Does the height of $y$ matter? because if it does not then the $r$ goes from 1 to 2 so that the interval becomes,

$\int_{0}^{2\pi} \int_{1}^{2} \,rdr d\theta$

But I'm confused if this is correct because I have only previously solved problems where the sphere had center in the origin. So do things change when the center moves away from the origin.

Solution 1:

I assume you are trying to evaluate line integral over closed curve with $0 \leq t \leq 2\pi$ and with positive orientation.

If you are applying Stokes' theorem to evaluate the line integral, you can consider any surface with the given curve as its boundary. The easiest in this cases will be

$\phi(r, t) = (1 + r \cos(t), 1 + r \sin(t), 1 - r \cos(t) - r \sin(t))$
$0 \leq r \leq 1, 0 \leq t \leq 2\pi$

$ \vec n = \phi_r \times \phi_t = (r - r \cos t, r - r \sin t, r)$

Also note, $\nabla \times \vec F = (0, 0, 2x) = (0, 0, 2 + 2 r \cos t)$

Can you take it from here?