I've just learned Green's theorem and I need a little help in solving a problem!
I need to calculate $\oint_c \vec{F} d \vec{r} $, when the vector field $ \vec{F} =( \frac{y}{x^2+y^2}+ \frac{y}{(x-2)^2+y^2})\hat{i}-(\frac{x}{x^2+y^2}+ \frac{x-2}{(x-2)^2+y^2}) \hat{j}$ . Curve $C$ has origin at $(0,0)$, and has radius of 10, and circulates counterclockwise.
My professor taught how to solve this, but I didn't quite get it. She told us to use Green's theorem. However, the circle with radius 10 has two holes- at (0,0) and at (2,0). So I'm not sure how to solve this. She said to divide the circle into parts, with the holes circulating clockwise…. but how do I do this???And also that I should think of $\vec{F}=\vec{F}_1+\vec{F}_2$, so $\vec{F}_1=\frac{y}{x^2+y^2}\hat{i}-\frac{x}{x^2+y^2}\hat{j}$ and $\vec{F}_2=\frac{y}{(x-2)^2+y^2}\hat{i}-\frac{x-2}{(x-2)^2+y^2} \hat{j}$
I know that if I use Green's theorem, the answer would be $0$ anyways because $\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}= 0$ but I would like to know how to use Green's theorem when the region has holes…
Best Answer
Maybe a picture helps here:
Notice that union of the black, green, red, green, blue and green gives a simply connected region on which the curl vanishes hence the totality of an integral around that region is zero. The same goes for the union of the curves bounding the lower simply connected region. Furthermore, when we add these two integrals together the green and orange cross-cuts cancel just leaving the black CCW curve and the inner CW oriented integrals around the red and blue circles which contain the singular points $(0,0)$ and $(2,0)$ for your vector field. Notice, the integral around the red and blue circles are easy because only the singular term nontrivially contributes.
This is why your teacher wanted CW-oriented circles. I do think you can deform the circles to a point as Muphrid suggested, but, you can also follow my sketch.