Green's Theorem

Recall that if $y = f(x)$ is a single variable function that is continuous on the interval $[a, b]$ and differentiable on $(a, b)$, then both parts of The Fundamental Theorem of Calculus Part 1 and The Fundamental Theorem of Calculus Part 2 has that:

(1)

\begin{align} \quad \int_a^b \frac{d}{dx} f(x) \: dx = f(b) - f(a) \end{align}

We also so that if $C$ is a smooth curve parameterized by $\vec{r}(t)$ for $a ≤ t ≤ b$ and if $f$ is a two or three variable differentiable function whose gradient $\nabla f$ is continuous on $C$ then The Fundamental Theorem for Line Integrals has that:

(2)

\begin{align} \quad \int_C \nabla f \cdot \: d \vec{r} = f(\vec{r}(b)) - f(\vec{r}(a)) \end{align}

We will now look at the two dimensional analogue of The Fundamental Theorem of Calculus. This theorem is known as Green's Theorem.

Theorem 1 (Green's Theorem): Let $D$ be a regular and closed region in the $xy$-plane whose boundary is positively oriented, piecewise smooth, simple, and closed curve $C$. Let $\mathbf{F} (x, y) = P(x, y) \vec{i} + Q(x,y) \vec{j}$ be a smooth vector field on $\mathbb{R}^2$. Then $\oint_C P(x, y) \: dx + Q(x, y) \: dy = \iint_D \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \: dA$.

• Proof: Let $D$ be a regular and closed region in the $xy$ plane that is bounded by $C$. Since $D$ is a regular region, then we can divide $D$ into subregions that do not overlap each other that are both $x$-simple and $y$-simple regions.

• Now if we have two subregions of $D$, say $D_1$ and $D_2$ as depicted below, then the orientations of their common boundary curve are opposite as depicted below.

• From The Fundamental Theorem for Line Integrals, the line integrals from the initial point of this common boundary curve to the terminal point of this common boundary curve summed twice will cancel out. Therefore the sum of the line integrals of the boundaries of the subregions is equal to the line integral of the boundary of the whole region $D$. Furthermore, since the subregions do not overlap, we have that the subregions do not share any interior points with each other and so the sum of the double integrals over these subregions is equal to the sum of double integral of the whole region $D$. All we need to show is that the formula above holds for any region $R$ that is both $x$-simple and $y$-simple.

• In proving Green's Theorem, we will show that $\int_C P(x, y) \: dx = - \iint_D \frac{\partial P}{\partial y} \: dx \: dy$ and $\int_C Q(x, y) \: dy = \iint_D \frac{\partial Q}{\partial x} \: dx \: dy$

• Let $D$ be a region that is both $x$-simple and $y$-simple.

• Since $D$ is $y$-simple then this region can be expressed as $D = \{ (x, y) \in \mathbb{R}^2 : a ≤ x ≤ b, f(x) ≤ y ≤ g(x) \}$.

• Note that since the boundary of $D$ (recall the boundary of $D$ is the curve $C$) is positively oriented, we have that the function $y = f(x)$ is oriented left to right, while the function $y = g(x)$ is oriented right to left. Therefore by the Fundamental Theorem of Calculus we have that:

(3)

\begin{align} \quad - \iint_D \frac{\partial P}{\partial y} \: dx \: dy = - \int_a^b \int_{f(x)}^{g(x)} \frac{\partial P}{\partial y} \: dy \: dx = -\int_a^b \left [ P(x, g(x)) - P(x, f(x)) \right ] \: dx = \int_a^b \left [ P(x, f(x)) - P(x, g(x)) \right ] \: dx \end{align}

• Now note that the change in $x$, $dx$ is equal to $0$ as we traverse the vertical portion of the boundary of $D$, so the line integrals of $P$ along the left and right boundaries of $D$ equal zero. So the line integral along the boundary of $D$ (the curve $C$) with respect to $x$ is equal to the line integrals of $P$ along the lower boundary of $D$ and the line integral of $P$ along the upper boundary of $D$.

• We note that the function $y = f(x)$ (which represents the lower boundary of $D$) can be trivially parameterized as $x = x$ and $y = f(x)$. Similarly, the curve $y = f(x)$ (which represents the upper boundary of $D$) can be trivially parameterized as $x = x$ and $y = g(x)$. Thus:

(4)

\begin{align} \quad \oint P(x, y) \: dx = \underbrace{\int_a^b P(x, f(x)) \: dx}_{\mathrm{line \: integral \: along \: y = f(x)}} - \underbrace{\int_a^b P(x, g(x)) \: dx}_{\mathrm{line \: integral \: along \: y = g(x)}} = \int_a^b \left [ P(x, f(x)) - P(x, g(x)) \right ] \: dx \end{align}

• Therefore we have that:

(5)

\begin{align} \oint P(x, y) \: dx = - \iint_D \frac{\partial P}{\partial y} \: dx \: dy \quad (*) \end{align}

• Now similarly, since $D$ is an $x$-simple region, we have that $D$ can be expressed as $D = \{ (x, y) \in \mathbb{R}^2 : f(y) ≤ x ≤ g(y), c ≤ y ≤ d \}$.

• Since $D$ is positively oriented, the left boundary $x = f(y)$ is oriented top to bottom, and the right boundary $x = g(y)$ is oriented bottom to top. Therefore by the Fundamental Theorem of Calculus we have that:

(6)

\begin{align} \quad \iint_D \frac{\partial Q}{\partial x} \: dx dy = \int_c^d \int_{f(y)}^{g(y)} \frac{\partial Q}{\partial x} \: dx \: dy = \int_c^d \left [ Q(x, g(y)) - Q(x, f(y)) \right ] \: dy \end{align}

• Now the change in $y$, $dy$ is equal to $0$ as we traverse the horizontal portion of the boundary of $D$, so the line integrals of $Q$ along the top and bottom boundaries of $D$ equal zero. So the line integral along the boundary of $D$ with respect to $y$ is equal to the line integrals of $Q$ along the left boundary of $D$ and the line integral of $Q$ along the right boundary of $D$.

(7)

\begin{align} \quad \oint_C Q(x, y) \: dy = \underbrace{- \int_c^d Q(x, f(y)) \: dy}_{\mathrm{line \: integral \: along \: x = f(y)}} + \underbrace{\int_c^d Q(x, g(y)) \: dy}_{\mathrm{line \: integral \: along \: x = g(y)}} = \int_c^d \left [ Q(x, g(y)) - Q(x, f(y)) \right ] \: dy \end{align}

• Therefore we have that:

(8)

\begin{align} \quad \oint_C Q(x, y) \: dy = \iint_D \frac{\partial Q}{\partial x} \: dx \: dy \quad (**) \end{align}

• Combining $(**)$ with $(*)$ and we have:

(9)

\begin{align} \: \oint_C P(x, y) \: dx + Q(x, y) \: dy = \iint_D \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \: dA \quad \blacksquare \end{align}

Other Ways to Write Green's Theorem

Recall from The Divergence and Curl of a Vector Field In Two Dimensions page that if $\mathbf{F} (x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a vector field on $\mathbb{R}^2$ then the curl of $\mathbb{F}$ is defined to be:

(10)

\begin{align} \quad \mathrm{curl} (\mathbf{F}) = \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \vec{k} \end{align}

Thus $\mathrm{curl} (\mathbf{F}) \cdot \vec{k} = \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right )$. We can therefore restate the formula in Green's theorem as:

(11)

\begin{align} \quad \oint_C P(x, y) \: dx + Q(x, y) \: dy = \iint_D \mathrm{curl} (\mathbf{F}) \cdot \vec{k} \: dA \end{align}

Or as:

(12)

\begin{align} \quad \oint_C \mathbf{F} \cdot d \vec{r} = \iint_D \mathrm{curl} (\mathbf{F}) \cdot \vec{k} \: dA \end{align}