Green's theorem also works for regions with holes, such as a ring between two circles, as long as the whole boundary is used with the right orientation. This version explains how a field with zero curl can still circulate: all of its spin can be hidden inside a hole. You need Green's theorem for regions without holes.
Idea clave
If $D$ is bounded by an outer curve $C_1$ and inner curves $C_2, \ldots$, the positively oriented boundary keeps $D$ on the left: the outer curve counterclockwise and each inner curve clockwise. With that orientation,
$$\oint_{\partial D} P\,dx + Q\,dy = \iint_D (Q_x - P_y)\,dA.$$
Consequence: if $Q_x = P_y$ throughout a ring, the circulations around the outer and inner circles, both counterclockwise, are equal.
Por qué funciona
Cut the ring with two straight slits into two pieces without holes. Apply Green's theorem to each piece and add. Each slit is traced twice, once in each direction, so those integrals cancel. What remains is the outer circle counterclockwise and the inner circle clockwise, since walking around each piece with the piece on your left sends you clockwise along the inner arcs.
For the consequence: $\oint_{C_1, \text{ccw}} + \oint_{C_2, \text{cw}} = 0$, and $\oint_{C_2, \text{cw}} = -\oint_{C_2, \text{ccw}}$.
For the consequence: $\oint_{C_1, \text{ccw}} + \oint_{C_2, \text{cw}} = 0$, and $\oint_{C_2, \text{cw}} = -\oint_{C_2, \text{ccw}}$.
Worked examples
Ejemplo resuelto: A ring
Find $\oint_{\partial D} -y\,dx + x\,dy$, where $D$ is the ring ${}1 \le x^2 + y^2 \le 4$, positively oriented.
- $Q_x - P_y = 1 + 1 = 2$, and the ring has area ${}4\pi - \pi = 3\pi$, so the answer is ${}6\pi$.
- Check: the outer circle counterclockwise gives ${}2 \cdot 4\pi = 8\pi$ (twice its area); the inner circle clockwise gives $-2\pi$. Total ${}6\pi$.
Respuesta: ${}6\pi$
Ejemplo resuelto: The whirlpool field
Find $\oint_C \frac{-y\,dx + x\,dy}{x^2 + y^2}$ around the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$, counterclockwise.
- The integrand is not defined at the origin, which is inside the ellipse, so Green's theorem cannot be used on the whole inside.
- Use the ring between the ellipse and a small circle $x^2 + y^2 = \varepsilon^2$. There $Q_x - P_y = 0$, so the ellipse has the same counterclockwise circulation as the small circle.
- On the small circle, $\frac{-y\,dx + x\,dy}{x^2 + y^2} = \frac{\varepsilon^2\,dt}{\varepsilon^2} = dt$, giving ${}2\pi$.
Respuesta: ${}2\pi$
Ejemplo resuelto: From inner to outer
In the ring ${}1 \le x^2 + y^2 \le 9$, a field has $Q_x - P_y = 1$. Its counterclockwise circulation around the inner circle is ${}5$. Find its counterclockwise circulation around the outer circle.
- $$\oint_{\text{outer}} - \oint_{\text{inner}} = \iint_{\text{ring}} 1\,dA = 9\pi - \pi = 8\pi$$ (both circles counterclockwise).
- So $\oint_{\text{outer}} = 5 + 8\pi$.
Respuesta: ${}5 + 8\pi$
Error común
Orienting the inner curve like the outer one. With both circles counterclockwise, the theorem reads $\oint_{\text{outer}} - \oint_{\text{inner}} = \iint (Q_x - P_y)\,dA$, with a minus sign. In the ring example, adding ${}8\pi$ and ${}2\pi$ would give ${}10\pi$ instead of the correct ${}6\pi$. Walk each boundary curve with the region on your left.