Deriving Green’s Theorem on Rectangles

From Boundary Circulation to an Area Integral

A line integral around a closed curve records the accumulated contribution of a vector field along its boundary. Green’s Theorem relates that boundary quantity to a double integral over the enclosed region. On a rectangle, the relationship can be derived directly: opposite edges produce differences of function values, and the Fundamental Theorem of Calculus converts those differences into derivatives.

This derivation makes the signs in the circulation form explicit. It also shows why the two partial derivatives occur in the combination $Q_x-P_y$, rather than in some other order.

Rectangle, Field, and Orientation

Let $a<b$ and $c<d$, and let $R=[a,b]\times[c,d]$. Consider a vector field $\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle$, where $P$ and $Q$ have continuous first partial derivatives on an open set containing $R$. The open-set condition ensures that these functions and derivatives are defined around the entire rectangle, including its boundary.

We traverse the boundary $\partial R$ counterclockwise: first along the bottom from left to right, then up the right edge, then along the top from right to left, and finally down the left edge. This is the positive orientation. The circulation of $\mathbf{F}$ along a piecewise-smooth curve is the line integral $\int P\,dx+Q\,dy$ over that oriented curve.

Theorem: If $P$ and $Q$ have continuous first partial derivatives on an open set containing $R=[a,b]\times[c,d]$, then the counterclockwise boundary circulation satisfies $$ \oint_{\partial R} P\,dx+Q\,dy = \iint_R (Q_x-P_y)\,dA. $$

Here $Q_x$ denotes the partial derivative of $Q$ with respect to $x$, and $P_y$ denotes the partial derivative of $P$ with respect to $y$. The theorem concerns a rectangular region and its positively oriented boundary; the orientation is part of the statement.

Deriving the Identity from the Four Edges

Each edge integral can be written as an ordinary one-variable integral. On a horizontal edge, $dy=0$, so only $P\,dx$ contributes. On a vertical edge, $dx=0$, so only $Q\,dy$ contributes. The direction of travel determines the limits of integration.

Write the boundary integral edge by edge.
The bottom and top edges are horizontal, while the right and left edges are vertical. On the top and left, the coordinates decrease as the positively oriented boundary is traversed.

Pair the opposite vertical edges.
The right-edge contribution minus the left-edge contribution is the difference $Q(b,y)-Q(a,y)$. The Fundamental Theorem of Calculus expresses this as an integral of $Q_x$.

Pair the opposite horizontal edges.
The bottom-edge contribution minus the top-edge contribution is $P(x,c)-P(x,d)$. The Fundamental Theorem of Calculus expresses this as minus an integral of $P_y$.

Combine the iterated integrals.
Continuity of the partial derivatives permits the resulting integrals to be written as a double integral over the rectangle.

In detail, the four contributions, in boundary order, are:

\begin{aligned} \oint_{\partial R} P\,dx+Q\,dy &= \int_a^b P(x,c)\,dx +\int_c^d Q(b,y)\,dy\\ &\quad+\int_b^a P(x,d)\,dx +\int_d^c Q(a,y)\,dy. \end{aligned}

Reverse the limits on the top and left edge to express all four integrals using increasing limits. Then group the two vertical contributions together and the two horizontal contributions together:

\begin{aligned} \oint_{\partial R} P\,dx+Q\,dy &= \int_c^d \bigl(Q(b,y)-Q(a,y)\bigr)\,dy\\ &\quad+\int_a^b \bigl(P(x,c)-P(x,d)\bigr)\,dx. \end{aligned}

For each fixed $y$, apply the one-variable Fundamental Theorem of Calculus to $x\mapsto Q(x,y)$. For each fixed $x$, apply it to $y\mapsto P(x,y)$. The hypotheses hold because the relevant partial derivatives are continuous.

\begin{aligned} Q(b,y)-Q(a,y)&=\int_a^b Q_x(x,y)\,dx,\\ P(x,c)-P(x,d)&=-\int_c^d P_y(x,y)\,dy. \end{aligned}

Substituting these equalities into the grouped boundary integral gives

\begin{aligned} \oint_{\partial R} P\,dx+Q\,dy &=\int_c^d\int_a^b Q_x(x,y)\,dx\,dy -\int_a^b\int_c^d P_y(x,y)\,dy\,dx\\ &=\iint_R Q_x\,dA-\iint_R P_y\,dA\\ &=\iint_R (Q_x-P_y)\,dA. \end{aligned}

The change in the order of integration in the first term is valid because $Q_x$ is continuous on the rectangle; the corresponding iterated integrals agree. The other term already integrates over the same rectangle. Linearity of integration then combines the two integrands. This proves the stated identity for $R$.

Worked Example: A Rotational Field

Take $P(x,y)=-y$, $Q(x,y)=x$, and $R=[0,2]\times[1,3]$. First calculate the boundary circulation in the positive orientation.

On the bottom edge, $y=1$, $x$ runs from $0$ to $2$, and $dy=0$. Thus the contribution is $\int_0^2 -1\,dx=-2$.

On the right edge, $x=2$, $y$ runs from $1$ to $3$, and $dx=0$. The contribution is $\int_1^3 2\,dy=2(3-1)=4$.

On the top edge, $y=3$, $x$ runs from $2$ to $0$. Its contribution is $\int_2^0 -3\,dx=(-3)(0-2)=6$.

On the left edge, $x=0$, $Q=0$, so its contribution is $\int_3^1 0\,dy=0$. Therefore the total circulation is $-2+4+6+0=8$.

Now $Q_x=1$ and $P_y=-1$, so $Q_x-P_y=1-(-1)=2$. The rectangle has width $2$ and height $3-1=2$, hence the area integral is $2(2)(2)=8$. The two sides agree.

Worked Example: Nonconstant Partial Derivatives

Let $P(x,y)=x^2y$, $Q(x,y)=xy^2$, and $R=[0,1]\times[0,2]$. On the bottom edge $y=0$, $P=0$, and $dy=0$, so the contribution is $0$. On the right edge $x=1$, the contribution is $\int_0^2 y^2\,dy=[y^3/3]_0^2=8/3$.

On the top edge $y=2$, $P=2x^2$, and $x$ runs from $1$ to $0$. Its contribution is $\int_1^0 2x^2\,dx=[2x^3/3]_1^0=-2/3$. On the left edge $x=0$, $Q=0$, so the contribution is $0$. The boundary total is therefore $0+8/3-2/3+0=2$.

For the area integral, $Q_x=y^2$ and $P_y=x^2$. Thus

\begin{aligned} \iint_R(Q_x-P_y)\,dA &=\int_0^1\int_0^2(y^2-x^2)\,dy\,dx\\ &=\int_0^1\left(\frac{8}{3}-2x^2\right)\,dx\\ &=\frac{8}{3}-\frac{2}{3}=2. \end{aligned}

The area integral equals the boundary total, as required.

Worked Example: Contributions That Cancel

Consider $P(x,y)=xy$, $Q(x,y)=x^2+y$, and $R=[-1,1]\times[0,2]$. On the bottom edge $y=0$, $P=0$, so the contribution is $0$. On the right edge $x=1$, the contribution is

\int_0^2 (1+y)\,dy =\left[y+\frac{y^2}{2}\right]_0^2 =4.

On the top edge $y=2$, $P=2x$ and $x$ runs from $1$ to $-1$. Its contribution is $\int_1^{-1}2x\,dx=[x^2]_1^{-1}=1-1=0$. On the left edge $x=-1$, $Q=1+y$ and $y$ runs from $2$ to $0$. Its contribution is $\int_2^0(1+y)\,dy=0-4=-4$. The boundary total is $0+4+0-4=0$.

Here $Q_x=2x$ and $P_y=x$, so $Q_x-P_y=x$. The area integral is

\iint_R x\,dA =\int_0^2\int_{-1}^1x\,dx\,dy =\int_0^2\left[\frac{x^2}{2}\right]_{-1}^1\,dy =\int_0^2 0\,dy =0.

This verifies the identity even though the right and left edge contributions are individually nonzero.

A Common Sign Error

The top edge is traversed from right to left, not from left to right. Replacing its actual contribution with an integral whose limits increase, without also changing its sign, can reverse the result.

For a concrete example, set $P(x,y)=y$, $Q(x,y)=0$ on $R=[0,2]\times[0,1]$. The bottom contribution is $0$, the vertical contributions are $0$, and the top edge has $P=1$ while $x$ runs from $2$ to $0$. Its correct contribution is $\int_2^0 1\,dx=-2$. The area integrand is $Q_x-P_y=0-1=-1$, so the area integral is $(-1)(2)(1)=-2$.

A mistaken calculation might write the top contribution as $\int_0^2 1\,dx=2$, ignoring the direction of travel. That would give a boundary total of $2$, contradicting the area integral. The correction is either to keep the oriented limits, $\int_2^0 1\,dx=-2$, or to reverse the limits and insert a minus sign, $-\int_0^2 1\,dx=-2$.

Why the Rectangular Derivation Matters

The proof relies on a simple but powerful pairing. The right and left edges compare values of $Q$ at $x=b$ and $x=a$; the Fundamental Theorem turns that comparison into an integral of $Q_x$ across the rectangle. The bottom and top edges compare values of $P$ at $y=c$ and $y=d$; their orientation makes that difference produce $-P_y$.

Continuity of the first partial derivatives is not a decorative assumption. It ensures that the derivatives used in the iterated integrals are continuous on the rectangle, so the fundamental theorem and the change between orders of integration apply. The calculation also shows the role of positive orientation: reversing the boundary direction negates every edge integral, while the area integral in the theorem is unchanged. The stated identity therefore uses the counterclockwise orientation.

This rectangle proof gives a concrete source for the circulation integrand $Q_x-P_y$. It establishes the identity for rectangles under the stated hypotheses; it does not, by itself, establish the corresponding theorem for other shapes.

Check Your Understanding

Use the edge orientations and the proof above to answer these questions.

Which two edge contributions combine to produce the integral involving $Q_x$?
Why does the pair of horizontal-edge contributions produce $-P_y$, rather than $P_y$?
What regularity assumptions on $P$ and $Q$ are used in the derivation?
For $P(x,y)=0$ and $Q(x,y)=x$ on $[0,3]\times[0,2]$, what is $Q_x-P_y$, and what value does the theorem give for the boundary circulation?
If the same rectangular boundary is traversed clockwise, how does its circulation compare with the counterclockwise circulation?

Deriving Green’s Theorem on Rectangles

What You'll Learn

From Boundary Circulation to an Area Integral

Rectangle, Field, and Orientation

Deriving the Identity from the Four Edges

Worked Example: A Rotational Field

Worked Example: A Rotational Field

Worked Example: Nonconstant Partial Derivatives

Worked Example: Nonconstant Partial Derivatives

Worked Example: Contributions That Cancel

Worked Example: Contributions That Cancel

A Common Sign Error

Why the Rectangular Derivation Matters

Check Your Understanding

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