Question: $f(x,y)=\langle y-\cos y,x\sin y\rangle$

$C$ is the circle $(x-3)^2 + (y+4)^2 = 4$ orientated clockwise.

**Relevant theorems:**

Green's theorem (this is under the Green's theorem section of our book).

**What I Have Tried**

Using Green's theorem, I have come up with the double integral $\iint ( \sin y - 1) \,dA$. Beyond here, I can't figure out what to do for my limits of integration. I have come to the conclusion that polar might be the way to go, but am at a loss of how to proceed with that, since I have a circle whose center is not the origin.

If it matters, this is out of Stewart Calculus Edition 7. It is chapter 16.4 question 13. The answer is $4 \pi$.

Answer: 
I must say that your differentiation went wrong near the beginning of the problem, and thus this problem is a little over complicated from that. The y derivative of the first term in the vector field is $ 1 + \sin y$, and the x derivative of the second term is $ \sin y $ the difference of these two is not $\sin y + 1$.
You're doing a double integral over $C$, so one option for your limits of integration will be:

$$\int\_1^5 \int\_{-4-\sqrt{4-(x-3)^2}}^{-4+\sqrt{4-(x-3)^2}} \dots dy dx$$

Explanation: The boundaries of the circle you're integrating over stretch from $1$ to $5$, since it's a circle of radius $2$ centered at $\langle 3, -4 \rangle$.

The $y$-coordinates of the boundary will vary with $x$: we have arrived at the bounds above by simply solving for $y$ from the equation of the circle.