Extending the Proof to Simple Planar Regions

From Rectangles to Curved Regions

On a rectangle, Green’s theorem follows by integrating partial derivatives and applying the fundamental theorem of calculus. The same idea works when one pair of opposite sides is replaced by curved graphs: slice the region, apply the one-variable theorem on each slice, and account for the moving endpoints.

The main complication is that a curved boundary contributes terms from both its position and its direction of travel. Those signs matter. Once the identity is established on regions described by vertical or horizontal slices, a finite subdivision lets us combine the results. Contributions along edges inside the full region appear twice, with opposite orientations, and therefore cancel.

Statement and Conditions

A simple planar region here means the bounded region enclosed by a simple, positively oriented, piecewise-smooth closed curve. Positive orientation means that the region stays to the left as the boundary is traversed; for a simple boundary, this is counterclockwise. Assume the boundary can be divided into finitely many smooth arcs and the region can be partitioned into finitely many Type I or Type II pieces with piecewise-smooth edges. This is the usual setting for the slicing argument.

Let \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\), where \(P\) and \(Q\) have continuous first partial derivatives on an open set containing the region and its boundary. Green’s theorem in circulation form is

The left side is the line integral along the positively oriented boundary. The right side is the double integral of the scalar function \(Q_x-P_y\) over the region \(D\). The derivation below establishes this identity on the slice-shaped pieces and then explains how to assemble them.

Derivation on a Type I Region

Let \(D\) be Type I, so each vertical slice runs from \(y=g_1(x)\) to \(y=g_2(x)\). We derive the contributions of \(P\) and \(Q\) separately. The vertical sides have \(dx=0\), but generally contribute to the \(Q\,dy\) part of the boundary integral.

For the \(P\) term, the fundamental theorem of calculus applied on each vertical slice gives

The lower graph is traversed from left to right in positive orientation, so its \(P\,dx\) contribution is \(\int_a^b P(x,g_1(x))\,dx\). The upper graph is traversed from right to left, so its contribution is \(-\int_a^b P(x,g_2(x))\,dx\). The two vertical sides have \(dx=0\). Thus the displayed expression is exactly the boundary integral of \(P\,dx\).

For the \(Q\) term, define the accumulated integral along a vertical slice by

By the Leibniz rule for differentiating an integral with variable endpoints,

Integrating this equation from \(a\) to \(b\) and rearranging yields

The vertical side at \(x=b\) is traversed upward and contributes \(F(b)\); the side at \(x=a\) is traversed downward and contributes \(-F(a)\). On the upper graph, the boundary travels from right to left, giving \(-\int_a^b Q(x,g_2(x))g_2'(x)\,dx\). On the lower graph, it travels from left to right, giving \(+\int_a^b Q(x,g_1(x))g_1'(x)\,dx\). These graph contributions are the negatives of the last two endpoint terms in \(F'(x)\). Consequently, the right side is exactly the boundary integral of \(Q\,dy\).

Adding the two identities proves Green’s theorem on this Type I region. The argument works with piecewise-smooth graph bounds by applying the calculation on each smooth interval and adding the resulting identities.

Derivation on a Type II Region

Now let \(D\) be Type II: \(c\le y\le d\), with \(h_1(y)\le x\le h_2(y)\). Horizontal slices are convenient. First, the fundamental theorem of calculus in the \(x\)-variable gives

The right-hand boundary curve is traversed with \(y\) increasing, so its \(Q\,dy\) contribution is \(\int_c^d Q(h_2(y),y)\,dy\). The left-hand curve is traversed with \(y\) decreasing, so its contribution is \(-\int_c^d Q(h_1(y),y)\,dy\). The horizontal edges have \(dy=0\). Thus the expression equals the boundary integral of \(Q\,dy\).

To account for \(-P_y\), define

Differentiating by the Leibniz rule gives

After integrating and rearranging,

The bottom horizontal edge is traversed from left to right and contributes \(G(c)\); the top edge is traversed from right to left and contributes \(-G(d)\). The right curved edge contributes \(+\int_c^d P(h_2(y),y)h_2'(y)\,dy\), and the left curved edge, traversed in the reverse direction, contributes \(-\int_c^d P(h_1(y),y)h_1'(y)\,dy\). The expression is therefore the boundary integral of \(P\,dx\). Adding the \(P\) and \(Q\) identities proves Green’s theorem on the Type II region as well.

Assembling the Pieces

Suppose the full region \(D\) is partitioned into finitely many Type I or Type II pieces \(D_1,\ldots,D_n\). Apply the identity just proved to each piece and add the results. The area integrals add because the pieces cover \(D\) and overlap only along their boundaries, which have area zero.

For the line integrals, each edge on the outside boundary occurs once, with the positive orientation of \(\partial D\). Each interior edge is shared by two pieces. The positive boundary direction for one piece traverses that edge one way; the positive direction for the adjacent piece traverses it the other way. Reversing a curve changes the sign of a line integral, so the two contributions cancel.

This proves the stated identity for the region under the subdivision assumptions. The proof does not require a single global description of the full region as Type I or Type II; different pieces may use different slicing directions.

Worked Example: Checking the Identity on a Type I Region

Worked Example: Checking the Identity on a Type II Region

Worked Example: Cancellation Along an Interior Edge

A Sign Error to Avoid

A frequent mistake is to use the same direction of travel for a graph boundary and for the corresponding endpoint term in the Leibniz rule. For example, take \(Q(x,y)=xy\) on the Type I triangle \(0\le x\le1\), \(0\le y\le x\). Here \(Q_x=y\), and

The upper graph is traversed from right to left, so its actual \(Q\,dy\) contribution is \(-\int_0^1x^2\,dx=-1/3\), not \(+1/3\). The right vertical side contributes \(\int_0^1 y\,dy=1/2\). The correct boundary total is therefore \(-1/3+1/2=1/6\), matching \(\int_0^1\int_0^x y\,dy\,dx=1/6\). Using a positive sign for the upper-graph term would instead give \(1/3+1/2=5/6\), which is incorrect. The graph contribution is the negative of the upper-endpoint term in \(F'(x)\).

Why the Extension Matters

The rectangle argument reveals the fundamental mechanism, but many useful planar regions have sloping or curved boundaries. Slicing allows the same one-variable calculus to handle those boundaries without requiring the entire region to have one simple rectangular description. Finite subdivision extends the argument further: it makes local slice descriptions compatible with a global boundary integral.

The key bookkeeping principle is to distinguish external edges from interior edges. External edges survive in the final line integral. Interior edges do not: each is counted once in each direction. This is also why correct orientation is essential at every stage, not just when stating the final theorem.