# Green's theorem over an annulus

I need help with this problem:

Verify Green's Theorem in the plane where $$S$$ is the annulus $$\{(x,y)\in\mathbb{R^2}|a^2\leq x^2+y^2\leq b^2\}$$ and

1. $$F(x,y)=\left(\frac{-y}{\sqrt{x^2+y^2}},\frac{x}{\sqrt{x^2+y^2}}\right)$$
2. $$F(x,y)=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$$
3. $$F(x,y)=\left(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right)$$

I was able to compute the line integral $$\int_{\partial S^+} F\cdot d\mathbf{r}$$, but I'm having problems with the double integral $$\iint_S\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA$$.

My problem is that I don't know which limist should I use. I know that I could use polar coordinates, but this problem is from a chapter before change of variables, so I think I'm not supposed to solve it like that.

I just need help with getting the limits of integration right, since I think it is easy to compute the double integral after that.

• Are you sure that you haven't learned polar coordinates (or spherical or cylindrical)? Most textbooks do those before discussing the general changes of variables in a multiple integral. – Ted Shifrin May 1 at 22:40
• I'm not a hundred percent sure, but there was a similar problem on the following chapter that explicitly said to use polar coordinates, so I think I'm supposed to do it without them on this one. – davidllerenav May 3 at 16:20
• I truly do not think your textbook expects you to integrate $\dfrac1{\sqrt{x^2+y^2}}$ over a disk or over an annulus using cartesian coordinates. – Ted Shifrin May 3 at 16:30

Split the integral in four parts:

$$\iint_S\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA=$$

$$=\int_{-b}^{-a}\int_{-\sqrt{b^2-x^2}}^{\sqrt{b^2-x^2}}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dydx+\int_{-a}^{a}\int_{\sqrt{a^2-x^2}}^{\sqrt{b^2-x^2}}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dydx+$$

$$+\int_{a}^{b}\int_{-\sqrt{b^2-x^2}}^{\sqrt{b^2-x^2}}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dydx+\int_{-a}^{a}\int_{-\sqrt{b^2-x^2}}^{-\sqrt{a^2-x^2}}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dydx$$

Maybe this can work too:

$$\int_{-b}^{b}\int_{-\sqrt{b^2-x^2}}^{\sqrt{b^2-x^2}}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dydx-\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dydx$$

• probably the best way to do it without polar coordinates. still feel bad for them.. – Saketh Malyala May 1 at 8:30
• It looks that it will be pretty messy to compute. But it shouldn't be too hard, right? – davidllerenav May 1 at 8:39
• I checked the first function and does not seem hard as the sum of squares simplifies a little. – Rafa Budría May 1 at 8:53