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Green’s Theorem in the Plane · Tutorial 814 of 999

Deriving Boundary Formulas for Area

See how three choices of circulation field turn Green’s theorem into boundary formulas for area, and why the boundary must be positively oriented.

Advanced 9 min read

What You'll Learn

  • State the hypotheses needed to apply Green’s circulation theorem to a planar area.
  • Choose vector fields that produce the area integrand in Green’s theorem.
  • Derive each of the three boundary formulas for area.
  • Explain how reversing the boundary orientation changes the sign of each integral.
  • Compute and check the formulas on a rectangle, an ellipse, and a triangle.

From Circulation to Area

Green’s theorem relates an integral around a closed curve to an integral over the region it encloses. A useful consequence is that the area of a planar region can be calculated using only a line integral along its boundary. The result is not just one formula: different choices of vector field give three equivalent boundary expressions.

The direction in which the boundary is traversed matters. Green’s theorem uses the positive orientation: for a simple region, the boundary is traversed counterclockwise, leaving the interior on the left. With this orientation, the boundary integrals below equal the positive area. Reversing the direction changes the sign of each integral.

Setting: Let \(D\) be a bounded planar region with a simple, closed, piecewise-smooth boundary \(C\), oriented positively. Assume the functions in the circulation form of Green’s theorem are continuously differentiable on an open set containing \(D\). Write \(A(D)=\iint_D 1\,dA\) for the area of \(D\).

The Circulation Form of Green’s Theorem

For continuously differentiable functions \(P\) and \(Q\) defined on a neighborhood of \(D\), Green’s circulation theorem states that

$$ \oint_C P\,dx+Q\,dy = \iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA. $$

The expression \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\) is the scalar curl density for the planar vector field \(\mathbf{F}=(P,Q)\). To obtain an area formula, choose \(P\) and \(Q\) so that this density is the constant \(1\). Then the double integral on the right is \(\iint_D 1\,dA=A(D)\).

For a parametrized piece of the boundary, \(x=x(t)\), \(y=y(t)\), the line integral is evaluated by substituting \(dx=x'(t)\,dt\) and \(dy=y'(t)\,dt\). In particular, \(x\,dy\) contributes \(x(t)y'(t)\,dt\), and \(y\,dx\) contributes \(y(t)x'(t)\,dt\). A piecewise-smooth boundary is handled by summing these contributions over its pieces.

Deriving the Three Boundary Formulas

Each identity follows by a different choice of \(P\) and \(Q\). The choices are valid because the coordinate functions \(x\) and \(y\), and their constant multiples, are continuously differentiable everywhere in the plane.

1
To obtain \(\oint_C x\,dy\).
Choose \(P=0\) and \(Q=x\). Then \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1-0=1\), while \(P\,dx+Q\,dy=x\,dy\).
2
To obtain \(-\oint_C y\,dx\).
Choose \(P=-y\) and \(Q=0\). Then \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0-(-1)=1\), while \(P\,dx+Q\,dy=-y\,dx\).
3
To obtain the symmetric expression.
Choose \(P=-y/2\) and \(Q=x/2\). Then \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1/2-(-1/2)=1\), while \(P\,dx+Q\,dy=\frac12(x\,dy-y\,dx)\).

Applying Green’s theorem to each choice proves the identities:

$$ A(D) = \oint_C x\,dy = -\oint_C y\,dx = \frac12\oint_C(x\,dy-y\,dx). $$
Theorem: If \(D\) and \(C\) satisfy the stated hypotheses and \(C\) has positive orientation, each of the three boundary integrals above equals the area \(A(D)\). If the same simple boundary is traversed in the reverse direction, each integral equals \(-A(D)\), not \(A(D)\).

Why Positive Orientation Is Required

Green’s theorem is stated for the positively oriented boundary. Reversing a parametrization reverses the sign of a line integral: the same boundary pieces are traversed in the opposite direction, so their differential increments \(dx\) and \(dy\) change sign. Thus the circulation integral of each selected field changes sign. The area integral \(\iint_D 1\,dA\), however, remains positive. This is why the displayed identities, as formulas for positive area, require positive boundary orientation.

The orientation condition is not a minor convention that can be ignored after choosing a formula. It is part of the hypotheses that connect the boundary integral to \(A(D)\). A negative result for one of these integrals on a clockwise traversal records the reversed orientation; it does not mean the enclosed region has negative geometric area.

Worked Examples

Worked Example: A Positively Oriented Rectangle

Let \(D=[0,4]\times[0,3]\), whose area is \(4\cdot3=12\). Traverse its boundary counterclockwise through \((0,0)\), \((4,0)\), \((4,3)\), \((0,3)\), and back to \((0,0)\). We compute each boundary expression by adding its contributions from the four edges.

For \(\oint_C x\,dy\), the bottom and top edges have \(dy=0\). On the right edge, \(x=4\) and \(y\) increases from \(0\) to \(3\), giving \(\int_0^3 4\,dy=12\). On the left edge, \(x=0\), so its contribution is zero. Therefore \(\oint_C x\,dy=12\).

For \(-\oint_C y\,dx\), the vertical edges have \(dx=0\). On the top edge, \(y=3\) and \(x\) decreases from \(4\) to \(0\), so \(\int_4^0 -3\,dx=12\). On the bottom edge, \(y=0\), so the contribution is zero. Hence \(-\oint_C y\,dx=12\).

For the symmetric formula, the bottom and left edges contribute zero to \(x\,dy-y\,dx\). On the right edge, \(x=4\), \(dx=0\), and \(y\) increases from \(0\) to \(3\), so the contribution is \(\int_0^3 4\,dy=12\). On the top edge, \(y=3\), \(dy=0\), and \(x\) decreases from \(4\) to \(0\), so the contribution is \(\int_4^0 -3\,dx=12\). Thus \(\frac12\oint_C(x\,dy-y\,dx)=\frac12(24)=12\). All three values agree with the rectangle’s area.

As a check of the edge calculations, the three boundary expressions have the required values \(12\), \(12\), and \(12\); the symmetric expression uses a total integral of \(24\) before multiplication by \(1/2\).

Worked Example: A Positively Oriented Ellipse

Consider the ellipse parametrized by \(x=2\cos t\), \(y=3\sin t\), for \(0\le t\le2\pi\). It is traversed counterclockwise: at \(t=0\) it is at \((2,0)\), and as \(t\) increases it initially moves upward. Differentiating gives \(dx=-2\sin t\,dt\) and \(dy=3\cos t\,dt\).

For the first formula,

$$ \oint_C x\,dy = \int_0^{2\pi}(2\cos t)(3\cos t)\,dt = 6\int_0^{2\pi}\cos^2t\,dt = 6\pi. $$

Here we used \(\cos^2t=(1+\cos 2t)/2\), so its integral from \(0\) to \(2\pi\) is \(\pi\). For the second formula,

$$ -\oint_C y\,dx = -\int_0^{2\pi}(3\sin t)(-2\sin t)\,dt = 6\int_0^{2\pi}\sin^2t\,dt = 6\pi, $$

since \(\sin^2t=(1-\cos 2t)/2\) also integrates to \(\pi\) over this interval. Finally,

$$ \frac12\oint_C(x\,dy-y\,dx) = \frac12\int_0^{2\pi} \left(6\cos^2t+6\sin^2t\right)dt = \frac12\int_0^{2\pi}6\,dt = 6\pi. $$

Each integral gives \(6\pi\), the area enclosed by an ellipse with semiaxes \(2\) and \(3\). Substitution into the boundary expressions verifies their agreement.

Worked Example: A Triangle with Three Boundary Pieces

Take the triangle with vertices \((0,0)\), \((4,0)\), and \((1,3)\), traversed in that order and then back to \((0,0)\). This order is counterclockwise. Its area is \(\frac12(4)(3)=6\). We calculate the line integrals directly, using \(0\le t\le1\) on each nonhorizontal edge.

On the first edge, \(x=4t\), \(y=0\), from \((0,0)\) to \((4,0)\). Since \(dy=0\) and \(y=0\), both \(x\,dy\) and \(-y\,dx\) contribute zero.

On the second edge, \(x=4-3t\), \(y=3t\), from \((4,0)\) to \((1,3)\), so \(dx=-3\,dt\) and \(dy=3\,dt\). Its contributions to the first two integrals are

$$ \int_0^1 x\,dy = \int_0^1(4-3t)3\,dt = \int_0^1(12-9t)\,dt = 12-\frac92 = \frac{15}{2}, $$

and

$$ \int_0^1-y\,dx = \int_0^1-(3t)(-3)\,dt = \int_0^1 9t\,dt = \frac92. $$

On the final edge, \(x=1-t\), \(y=3-3t\), from \((1,3)\) to \((0,0)\), so \(dx=-dt\) and \(dy=-3\,dt\). Its contributions are

$$ \int_0^1x\,dy = \int_0^1-3(1-t)\,dt = -\frac32, \qquad \int_0^1-y\,dx = \int_0^1(3-3t)\,dt = \frac32. $$

Adding the edges gives \(\oint_C x\,dy=\frac{15}{2}-\frac32=6\) and \(-\oint_C y\,dx=\frac92+\frac32=6\). On the second edge, \(x\,dy-y\,dx=(12-9t)-(-9t)=12\), whose integral is \(12\). On the final edge, \(x\,dy-y\,dx=-3(1-t)+(3-3t)=0\); the first edge also contributes zero. Therefore \(\frac12\oint_C(x\,dy-y\,dx)=\frac12(12)=6\). Each result matches the triangle’s area.

A Common Orientation Mistake

A common error is to compute a boundary integral with a clockwise parametrization and then report its value as positive area. For the unit circle, take \(x=\cos t\), \(y=-\sin t\), \(0\le t\le2\pi\). This parametrization runs clockwise. Since \(dy=-\cos t\,dt\),

$$ \oint_C x\,dy = \int_0^{2\pi}(\cos t)(-\cos t)\,dt = -\int_0^{2\pi}\cos^2t\,dt = -\pi. $$

The incorrect conclusion would be that the disk has area \(-\pi\), or that the formula fails. The correction is to recognize the clockwise orientation: the integral equals the negative of the disk’s area. Traversing the circle counterclockwise instead gives \(\oint_C x\,dy=\pi\), the positive area.

Takeaway: The formulas use the signed direction of traversal. First check that the boundary is positively oriented; then apply the appropriate boundary integral to obtain positive area.

Check Your Understanding

Use the choices of \(P\) and \(Q\) and the orientation condition to answer these questions.

  1. For \(\oint_C x\,dy\), what are \(P\) and \(Q\), and what is the resulting curl density?
  2. For \(-\oint_C y\,dx\), why does the choice \(P=-y\), \(Q=0\) give curl density \(1\)?
  3. What choice of \(P\) and \(Q\) produces \(\frac12\oint_C(x\,dy-y\,dx)\)?
  4. What sign should the boundary formulas give if the same boundary is traversed clockwise?
  5. For a parametrized curve, how is \(x\,dy\) converted into an ordinary integral in the parameter?

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