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Green’s Theorem in the Plane · Tutorial 813 of 999

Area Identities from Green’s Theorem

Learn how a constant-curl vector field turns Green’s theorem into an area formula, and how to evaluate and check that formula on disks and rectangles.

Advanced 10 min read

What You'll Learn

  • State the conditions under which Green’s theorem gives an area identity.
  • Choose vector fields whose scalar curl is one.
  • Relate the area integral to circulation along a positively oriented boundary.
  • Verify the area identities by parametrizing circles and traversing rectangle edges.
  • Recognize how reversing boundary orientation changes the sign of the integral.

From Curl to Area

A double integral of \(1\) over a planar region gives its area. Green’s theorem lets us express that same quantity as a line integral around the boundary, provided we choose the vector field appropriately. The key is to make its scalar curl constant: then the area integral in Green’s theorem becomes a constant multiple of the region’s area.

This is useful when the boundary is easier to describe or compute along than the interior is to integrate over. A circle, for example, has a simple parametrization, and a rectangle can be handled one edge at a time.

Central idea: For a positively oriented planar region, a vector field with scalar curl \(1\) has circulation around the boundary equal to the region’s area.

Conditions and Notation

Let \(D\) be a bounded planar region with a simple, piecewise-smooth boundary \(C\). Assume \(P\) and \(Q\) have continuous first partial derivatives on an open set containing \(D\). We orient \(C\) positively: as we traverse the boundary, the region remains on our left. For a simple region with no holes, this is the counterclockwise orientation.

Green’s theorem in circulation form states that

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

The quantity \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\) is the scalar curl of the planar vector field \(\mathbf{F}=(P,Q)\). The area of \(D\), denoted by \(A(D)\), is \(\iint_D 1\,dA\).

Definition: The scalar curl of \(\mathbf{F}=(P,Q)\) is \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\). A vector field has constant scalar curl \(c\) on \(D\) if this quantity equals the same constant \(c\) at every point of \(D\).

Choosing Fields with Curl One

We want the integrand in Green’s theorem to be \(1\), so the most direct choice is any field whose scalar curl is \(1\). Three convenient choices are

$$ \mathbf{F}_1(x,y)=\left(-\frac{y}{2},\frac{x}{2}\right), \qquad \mathbf{F}_2(x,y)=(0,x), \qquad \mathbf{F}_3(x,y)=(-y,0). $$

Check their curls directly. For \(\mathbf{F}_1\), \(P=-y/2\) and \(Q=x/2\), so \(Q_x=1/2\) and \(P_y=-1/2\). For \(\mathbf{F}_2\), \(P=0\) and \(Q=x\), so \(Q_x=1\) and \(P_y=0\). For \(\mathbf{F}_3\), \(P=-y\) and \(Q=0\), so \(Q_x=0\) and \(P_y=-1\). In each case \(Q_x-P_y=1\).

Green’s theorem therefore gives the following identities for any such field and any region satisfying the stated conditions:

$$ A(D) = \oint_C \left(-\frac{y}{2}\right)dx+\frac{x}{2}dy = \frac{1}{2}\oint_C(x\,dy-y\,dx), $$
$$ A(D)=\oint_C x\,dy, \qquad A(D)=\oint_C -y\,dx. $$

These are alternative ways to calculate the same area. They need not give equally convenient calculations for every boundary. The symmetric expression often treats the coordinates evenly, while either of the other expressions can be especially simple when some boundary segments have constant \(x\) or constant \(y\).

Why the Area Identities Hold

The reasoning is a direct application of Green’s theorem. For the symmetric field, its scalar curl is \(1\), so the theorem equates its boundary circulation to the integral of \(1\) over \(D\). That integral is the area. The other two fields have the same curl, so the same reasoning applies to them.

1
Choose a field.
Take \(P,Q\) with continuous first partial derivatives near \(D\) and \(Q_x-P_y=1\) throughout \(D\).
2
Apply Green’s theorem.
It gives \(\oint_C P\,dx+Q\,dy=\iint_D 1\,dA\).
3
Identify the area.
By the definition of area, \(\iint_D 1\,dA=A(D)\), so the boundary integral equals \(A(D)\).

More generally, if the scalar curl is a constant \(c\), then Green’s theorem gives

$$ \oint_C P\,dx+Q\,dy = \iint_D c\,dA = cA(D). $$

If \(c\ne 0\), this can be rearranged to \(A(D)=\frac{1}{c}\oint_C P\,dx+Q\,dy\). If \(c=0\), the equation only says that the circulation is zero; it does not determine the area. Choosing curl \(1\) avoids the extra division and gives the identities above immediately.

Worked Example: The Area of a Disk

Worked Example: The Area of a Disk

Let \(D\) be the disk of radius \(r>0\) centered at the origin. Its boundary is positively parametrized by \(x=r\cos t\), \(y=r\sin t\), with \(0\le t\le 2\pi\). Use the symmetric area identity.

Differentiate the coordinates with respect to \(t\):

$$ dx=-r\sin t\,dt, \qquad dy=r\cos t\,dt. $$

Substitute these expressions into the integrand:

$$ x\,dy-y\,dx =(r\cos t)(r\cos t\,dt)-(r\sin t)(-r\sin t\,dt) $$
$$ =r^2\cos^2t\,dt+r^2\sin^2t\,dt =r^2\,dt. $$

Therefore,

$$ A(D) =\frac{1}{2}\int_0^{2\pi}r^2\,dt =\frac{r^2}{2}\bigl(2\pi-0\bigr) =\pi r^2. $$

This checks against the standard area of a disk. The line-integral calculation also shows how the symmetric field works: along the circular boundary, the combination \(x\,dy-y\,dx\) simplifies to the constant \(r^2\,dt\).

Worked Example: A Rectangle Using \(x\,dy\)

Worked Example: A Rectangle Using \(x\,dy\)

Take the rectangle with vertices \((0,0)\), \((w,0)\), \((w,h)\), and \((0,h)\), in that counterclockwise order, where \(w,h>0\). Use \(A(D)=\oint_C x\,dy\). On the bottom and top edges, \(dy=0\), so those edges contribute zero. On the left edge, \(x=0\), so it also contributes zero.

Only the right edge contributes. There, \(x=w\), and \(y\) increases from \(0\) to \(h\). Thus

$$ \int_{\text{right edge}}x\,dy = \int_0^h w\,dy = w[h-0] = wh. $$

Adding the contributions from all four edges gives \(A(D)=0+wh+0+0=wh\). This agrees with the rectangle’s width-times-height area. As a check on the boundary calculation, \(x\,dy\) vanishes on each of the three edges where either \(dy=0\) or \(x=0\); the only remaining edge contributes exactly \(wh\).

Worked Example: A Shifted Rectangle Using \(-y\,dx\)

Worked Example: A Shifted Rectangle Using \(-y\,dx\)

Consider the rectangle \([a,b]\times[c,d]\), where \(a<b\) and \(c<d\), with vertices traversed counterclockwise starting at \((a,c)\). Use \(A(D)=\oint_C -y\,dx\). On the two vertical edges, \(dx=0\), so their contributions are zero.

On the bottom edge, \(y=c\) and \(x\) increases from \(a\) to \(b\). Its contribution is

$$ \int_a^b -c\,dx=-c(b-a). $$

On the top edge, positive orientation traverses from \((b,d)\) to \((a,d)\). Here \(y=d\), and \(x\) decreases from \(b\) to \(a\). Its contribution is

$$ \int_b^a -d\,dx =-d(a-b) =d(b-a). $$

The total boundary integral is therefore

$$ -c(b-a)+d(b-a) =(d-c)(b-a). $$

This is the rectangle’s area: its width is \(b-a\), and its height is \(d-c\). The result also shows that the position of the rectangle does not affect its area; the coordinate offsets \(a\) and \(c\) cancel from the calculation.

Orientation Matters

The identities above assume positive orientation. Reversing the direction of traversal reverses the sign of a line integral, while the geometric area remains positive. The disk provides a concrete check. With clockwise parametrization \(x=r\cos t\), \(y=-r\sin t\), \(0\le t\le 2\pi\), we have \(dx=-r\sin t\,dt\) and \(dy=-r\cos t\,dt\). Then

$$ x\,dy-y\,dx =-r^2\cos^2t\,dt-r^2\sin^2t\,dt =-r^2\,dt, $$

so the symmetric boundary integral is \(\frac12\int_0^{2\pi}-r^2\,dt=-\pi r^2\). The negative result is not a negative geometric area; it signals that the boundary was traversed in the wrong direction for the stated identity.

Common pitfall: Using the clockwise boundary integral as the area gives the wrong sign. For the clockwise disk calculation, the integral is \(-\pi r^2\), not \(\pi r^2\). Reverse the direction of traversal, or negate the integral, to recover the positive area.

When These Identities Are Useful

A boundary formula for area can be convenient when the boundary has a simple parametrization or consists of a small number of easily integrated segments. It can also provide a check on an area calculation carried out by a double integral. The choice among \(x\,dy\), \(-y\,dx\), and \(\frac12(x\,dy-y\,dx)\) is a choice of computational convenience: under the hypotheses above and with positive orientation, each gives the same area.

These identities are not a replacement for checking the theorem’s conditions. Green’s theorem requires suitable differentiability and a boundary with the stated regularity and orientation. The integral must also be taken around the whole boundary of the region under consideration. When the hypotheses hold, the method is short: choose a field with curl \(1\), then evaluate its circulation.

Check Your Understanding

Use the area identities and their hypotheses to answer these questions.

  1. What is the scalar curl of \(\mathbf{F}(x,y)=(-y,x)\), and what factor would you need to convert its circulation into area?
  2. For which edges of the rectangle \([0,w]\times[0,h]\) does the integral of \(x\,dy\) vanish, and why?
  3. Why does the formula \(A(D)=\frac{1}{c}\oint_C P\,dx+Q\,dy\) require \(c\ne 0\) when the field has constant scalar curl \(c\)?
  4. What sign does the symmetric area integral give for a clockwise circle, and what causes that sign?
  5. For the rectangle \([a,b]\times[c,d]\), how do the bottom and top contributions to \(\oint_C-y\,dx\) combine to give its area?

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