📂Calculus

Proof of Green's Theorem

Theorem¹

Let the curve $\mathcal{C}$ be a simple, smooth, closed path in the plane $S = [a,b] \times [c,d]$, moving counterclockwise. If the function $P,Q : \mathbb{R}^2 \to \mathbb{R}$ is continuous on $\mathcal{C}$ and its derivative is also continuous,

$$\int_{\mathcal{C}} (Pdx + Qdy) = \iint_{S} (Q_{x} - P_{y}) dx dy$$

Explanation

This can be thought of as a theorem that converts line integrals into surface integrals. It’s widely known as a corollary derived from the Kelvin-Stokes theorem specifically limited to the plane. Despite there being more generalized theorems, it retains its significance in numerous fields due to its namesake.

Proof

Suppose $$I_{1} := \int_{\mathcal{C}} P dx \\ \displaystyle I_{2} := \int_{\mathcal{C}} Q dy$$ then $$\int_{\mathcal{C}} (Pdx + Qdy) = I_{1} + I_{2}$$ Let’s first find $I_{1}$.

The region for calculating $I_{1}$ is depicted above. The region encased by $\mathcal{C}$ is $$S = \left\{ (x,y) \in \mathbb{R} \ | \ a \le x \le b, y_{1}(x) \le y \le y_{2}(x) \right\}$$ thus, $$\begin{align*} I_{1} =& \int_{\mathcal{C}} Pdx \\ =& \int_{a}^{b} P(x,y_{1} (x))dx + \int_{b}^{a} P(x,y_{2} (x)) dx \\ =& - \int_{a}^{b} \left\{ P(x,y_{2} (x))-P(x,y_{1} (x)) \right\} dx \\ =& - \int_{a}^{b} \int_{y_{1}(x)}^{y_{2}(x)} {{\partial P(x,y)} \over {\partial y}} dy dx \\ =& - \iint_{S} P_{y} dy dx \end{align*}$$ Next, let’s calculate $I_{2}$. Usually, such proofs might conclude with ’the same method can be applied’, but Green’s theorem requires a direct computation. The approach is similar, but it’s important to verify since the resulting signs are in opposite directions.

The region for calculating $I_{2}$ is depicted above. The region encased by $\mathcal{C}$ is $$S = \left\{ (x,y) \in \mathbb{R}^{2} \ | \ c \le y \le d, x_{1}(y) \le x \le x_{2}(y) \right\}$$ thus, $$\begin{align*} I_{2} =& \int_{\mathcal{C}} Qdy \\ =& \int_{d}^{c} Q(x_{1}(y),y) dy + \int_{c}^{d} Q(x_{2}(y),y) dy \\ =& \int_{c}^{d} Q(x_{2}(y),y) dy - \int_{c}^{d} Q(x_{1}(y),y) dy \\ =& \int_{c}^{d} \left\{ Q(x_{2}(y),y) dy - Q(x_{1}(y),y) \right\} dy \\ =& \int_{c}^{d} \int_{x_{1}(x)}^{x_{2}(x)} {{\partial Q(x,y)} \over {\partial x}} dx dy \\ =& \iint_{S} Q_{x} dx dy \end{align*}$$ Adding the results of $I_{2}$ and $I_{1}$ gives $$\int_{\mathcal{C}} (Pdx + Qdy) = I_{2} + I_{1} = \iint_{S} Q_{x} dx dy - \iint_{S} P_{y} dy dx$$

Fubini’s Theorem: Suppose $R : [a,b] \times [c,d]$. If $f(x,\cdot)$ is integrable over $[c,d]$, $f(\cdot,y)$ is integrable over $[a,b]$, and $f$ is integrable over $R$, then $$\iint _{R} f dA = \int_{a}^{b} \int_{c}^{d} f(x,y) dy dx = \int_{c}^{d} \int_{a}^{b} f(x,y) dx dy$$

Given that the derivative $P$ of the premise is also continuous, thereby integrable, Fubini’s theorem can be applied. By changing the order of integration as follows, $$\iint_{S} P_{y} dy dx = \iint_{S} P_{y} dx dy$$ and standardizing the integration order to $dx dy$ results in $$\int_{\mathcal{C}} (Pdx + Qdy) = \iint_{S} ( Q_{x} - P_{y} ) dx dy$$

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While demonstrated with the rectangle $S$, this can be specialized for a small square $[\alpha, \alpha + \varepsilon] \times [\beta, \beta + \varepsilon]$ and further generalized by dividing a bounded region $\mathcal{R}$ into small squares with side length $\varepsilon$, then taking the limit as $\varepsilon \to 0$ to obtain the generalized theorem simply.

Although conditions and expressions may vary, the essence is largely the same. Rather than focusing on the generalization, it’s sufficient to acknowledge and move on as different textbooks might vary in detail.

Generalization ¹

If the two functions $P,Q$ defined in $\mathcal{R}$ are differentiable over $\mathcal{R}$, $$\int_{\mathcal{C}} (Pdx + Qdy) = \iint_{\mathcal{R}} (Q_{x} - P_{y}) dx dy$$

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• The curve $C^{2}$ is twice differentiable, and its derivatives are all differentiable.

See Also

• Generalization of Green’s theorem