Green's Theorem
Theorem: let D be a simply connected region in the \mathbb{R}^2 plane whose boundary, denoted C = \partial D, is a piecewise smooth, simple closed curve. The orientation of C is taken to be positive, meaning it is traversed counter-clockwise, such that the region D always lies to the left.
Consider a vector field \mathbf{F}: U \to \mathbb{R}^2 defined on an open set U containing D. Let this field be given by its components \mathbf{F}(x,y) = P(x,y)\mathbf{i} + Q(x,y)\mathbf{j}. If the component functions P and Q have continuous first-order partial derivatives on U, then Green’s theorem states that:
\oint_C (P\,\mathrm{d}x + Q\,\mathrm{d}y) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \mathrm{d}A
The term:
\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)
can be recognized as the scalar component of the curl of a three-dimensional vector field \mathbf{G} = P\mathbf{i} + Q\mathbf{j} + 0\mathbf{k} perpendicular to the plane.
Proof: the demonstration is constructed by proving the theorem for a specific class of simple regions and then generalizing the result. The strategy is to split the identity into two separate equalities and prove each one using the fundamental theorem of calculus.
We seek to prove:
\begin{aligned} & \oint_C P\,\mathrm{d}x = -\iint_D \frac{\partial P}{\partial y} \mathrm{d}A \\ & \oint_C Q\,\mathrm{d}y = \iint_D \frac{\partial Q}{\partial x} \mathrm{d}A \end{aligned}
Adding these two results yields the theorem.
Let us consider a region D that is of Type I, meaning it can be described by the inequalities a \le x \le b and g_1(x) \le y \le g_2(x), where g_1 and g_2 are continuous functions on [a,b].
We begin by evaluating the double integral using an iterated integral and the fundamental theorem of calculus.
\begin{aligned} \iint_D \frac{\partial P}{\partial y} \mathrm{d}A &= \int_a^b \left( \int_{g_1(x)}^{g_2(x)} \frac{\partial P}{\partial y} \mathrm{d}y \right) \mathrm{d}x \\ &= \int_a^b P(x,y) \bigg|_{y=g_1(x)}^{y=g_2(x)} \mathrm{d}x \\ &= \int_a^b \left[ P(x, g_2(x)) - P(x, g_1(x)) \right] \mathrm{d}x \end{aligned}
Next, we evaluate the line integral:
\oint_C P\,\mathrm{d}x
The boundary C is composed of four segments: the bottom curve C_1 defined by y=g_1(x) from x=a to x=b; the vertical line segment C_2 at x=b; the top curve C_3 defined by y=g_2(x) from x=b to x=a; and the vertical line segment C_4 at x=a.
The line integral is the sum of the integrals over these segments.
On C_1, we can parametrize the curve as \mathbf{r}(x) = (x, g_1(x)) for x \in [a,b], \mathrm{d}x is simply \mathrm{d}x:
\int_{C_1} P\,\mathrm{d}x = \int_a^b P(x, g_1(x)) \mathrm{d}x
On C_3, the curve is parametrized as \mathbf{r}(x) = (x, g_2(x)), but the positive orientation requires traversing from x=b to x=a:
\int_{C_3} P\,\mathrm{d}x = \int_b^a P(x, g_2(x)) \mathrm{d}x = - \int_a^b P(x, g_2(x)) \mathrm{d}x
On the vertical segments C_2 and C_4, the variable x is constant, which means \mathrm{d}x=0. Consequently, the integrals over these segments are zero.
\begin{aligned} & \int_{C_2} P\,\mathrm{d}x = 0 \\ & \int_{C_4} P\,\mathrm{d}x = 0 \end{aligned}
Summing these contributions gives the total line integral.
\oint_C P\,\mathrm{d}x = \int_a^b P(x, g_1(x)) \mathrm{d}x - \int_a^b P(x, g_2(x)) \mathrm{d}x = -\int_a^b \left[ P(x, g_2(x)) - P(x, g_1(x)) \right] \mathrm{d}x
Comparing our two final expressions, we establish the first equality.
\oint_C P\,\mathrm{d}x = -\iint_D \frac{\partial P}{\partial y} \mathrm{d}A
A symmetric argument holds for the second equality.
We now consider a region D that is of Type II, described by c \le y \le d and h_1(y) \le x \le h_2(y).
The double integral becomes:
\begin{aligned} \iint_D \frac{\partial Q}{\partial x} \mathrm{d}A &= \int_c^d \left( \int_{h_1(y)}^{h_2(y)} \frac{\partial Q}{\partial x} \mathrm{d}x \right) \mathrm{d}y \\ &= \int_c^d Q(x,y) \bigg|_{x=h_1(y)}^{x=h_2(y)} \mathrm{d}y \\ &= \int_c^d \left[ Q(h_2(y), y) - Q(h_1(y), y) \right] \mathrm{d}y \end{aligned}
The line integral:
\oint_C Q\,\mathrm{d}y
is computed over the boundary segments. The integrals over the horizontal top and bottom segments are zero because y is constant, making \mathrm{d}y=0. We are left with the left curve C_1^\prime (x=h_1(y)) and the right curve C_2^\prime (x=h_2(y)).
On the right curve C_2^\prime, orientation is from y=c to y=d:
\int_{C_2^\prime} Q\,\mathrm{d}y = \int_c^d Q(h_2(y), y) \mathrm{d}y
On the left curve C_1^\prime, orientation is from y=d to y=c:
\int_{C_1^\prime} Q\,\mathrm{d}y = \int_d^c Q(h_1(y), y) \mathrm{d}y = - \int_c^d Q(h_1(y), y) \mathrm{d}y
Summing these gives:
\oint_C Q\,\mathrm{d}y = \int_c^d Q(h_2(y), y) \mathrm{d}y - \int_c^d Q(h_1(y), y) \mathrm{d}y = \int_c^d \left[ Q(h_2(y), y) - Q(h_1(y), y) \right] \mathrm{d}y
This confirms the second equality:
\oint_C Q\,\mathrm{d}y = \iint_D \frac{\partial Q}{\partial x} \mathrm{d}A
For a region that is simultaneously Type I and Type II, adding the two proven equalities completes the proof of Green’s theorem.
For a more general region, it can be decomposed into a finite union of simple regions. The sum of the line integrals over the boundaries of these sub-regions equals the line integral over the original boundary, as the contributions from the interior cuts cancel in pairs due to opposite orientations. The sum of the double integrals is clearly the double integral over the entire region.