Green's Theorem with a circle not centered at the origin.

[kana021693]

Problem: Evaluate Integral F dot dr, where C is the boundary of the region R and C is oriented so that the region is on the left when the boundary is traversed in the direction of its orientation.
F(x,y)=(e^(-x)+3y)i+(x)j
C is the boundary of the region R inside the circle x^2+y^2=16 and outside the circle x^2-2x+y^2=3

Homework Equations

Integral F dot dr=DOuble integral over the region R, (dg/dx-df/dy)dA

The Attempt at a Solution

I started by completing the square of that circle that is not centered at the origin, and got (x-1)^2+y^2=4. So now I know the inner region's boundary is a circle of radius 2 centered at (1,0).
Also, I got the double integral over x^2+y^2=16 , Double integral 0to 2pi, 0 to 4 (-2)r dr dtheta and got -32pi.
But I don't know what to do from here. The circle x^2-2x+y^2=3 is giving me a hard time. Can you tell me how to do the rest of this problem?

 

[kurt.physics]

The solution to this problem is as follows: First, complete the square on the second circle to get (x-1)^2+y^2=4. Then, set up the double integral over the region R such that the inner boundary is the circle of radius 2 centered at (1,0) and the outer boundary is the circle of radius 4 centered at the origin. The double integral over R is then given by: Double integral 0to 2pi, 0 to 2 (-2e^(-r cos(theta))r - 6r sin(theta)) r dr dthetaEvaluating this double integral yields a result of -128π.