Choosing a Field for Flux or Area
Green’s Theorem can express the outward flux across a planar boundary as a double integral over the enclosed region. In a workshop problem, the main decision is often how to make that relationship useful: use the given field if its divergence is manageable, or choose a field whose divergence directly produces the quantity being sought.
For example, the constant function \(1\) integrates over a region to give its area. A vector field with divergence \(1\) therefore converts area into outward flux. The same idea applies to a desired area integrand \(g(x,y)\): when a convenient differentiable field with divergence \(g\) is available, Green’s Theorem can replace the double integral with a boundary calculation.
The required theorem is the flux form of Green’s Theorem. It applies when the field is continuously differentiable on an open set containing the region, and the boundary is piecewise smooth. Corners are allowed. For a region with holes, every boundary component is included with the orientation just specified.
Why the Flux Formula Has This Form
Along a positively oriented boundary, write a small displacement as \((dx,dy)\). The outward normal displacement multiplied by the boundary element of length corresponds to the vector \((dy,-dx)\): it points to the right of the direction of travel. Dotting \((P,Q)\) with that vector gives \(P\,dy-Q\,dx\). Green’s Theorem then equates the resulting boundary flux to the accumulated divergence inside the region.
There is also a short derivation from the circulation form of Green’s Theorem. That form states that \(\oint_{\partial D}(A\,dx+B\,dy)=\iint_D(B_x-A_y)\,dA\). Set \(A=Q\) and \(B=-P\). The boundary integral becomes \(\oint_{\partial D}(Q\,dx-P\,dy)\), while the integrand becomes \(-P_x-Q_y\). Multiplying both sides by \(-1\) gives the flux formula above. The same orientation convention is essential in both forms.
Area as Outward Flux
To calculate area, choose \(\mathbf{F}(x,y)=(x/2,y/2)\). Its divergence is \(1/2+1/2=1\). Applying the theorem gives the boundary formula below. The formula works for regions with holes as long as every boundary component has positive orientation; the clockwise inner boundaries subtract the areas of the holes.
Other useful choices give equivalent area formulas. The field \((x,0)\) has divergence \(1\), as does \((0,y)\). Their flux integrals show that \(\oint_{\partial D}x\,dy\) and \(-\oint_{\partial D}y\,dx\) also equal the area, provided the entire positively oriented boundary is included. The symmetric formula is often convenient because it treats the two coordinates evenly.
Worked Example: Outward Flux Across a Disk
Worked Example: Outward Flux Across a Disk
Let \(D\) be the disk \(x^2+y^2\le 9\), and let \(\mathbf{F}(x,y)=(x/2,y/2)\). Find its outward flux, and check the result directly on the circular boundary.
First calculate the divergence: \(\displaystyle P_x=\frac12,\quad Q_y=\frac12,\quad P_x+Q_y=1.\) The field is continuously differentiable everywhere, so Green’s Theorem applies. The disk has area \(\pi(3)^2=9\pi\), giving \(\displaystyle \iint_D 1\,dA=9\pi.\)
For a boundary check, parametrize the counterclockwise circle by \(x=3\cos t,\ y=3\sin t\), for \(0\le t\le 2\pi\). Then \(dx=-3\sin t\,dt\) and \(dy=3\cos t\,dt\). On the boundary, \(P=3\cos t/2\) and \(Q=3\sin t/2\), so \(\displaystyle P\,dy-Q\,dx=\frac{9}{2}\cos^2t\,dt+\frac{9}{2}\sin^2t\,dt=\frac{9}{2}\,dt.\) Therefore the boundary flux is \(\displaystyle \int_0^{2\pi}\frac92\,dt=\frac92(2\pi)=9\pi.\) The boundary calculation agrees with the area integral.
Worked Example: Flux and Area of a Triangle
Worked Example: Flux and Area of a Triangle
Let \(D\) be the triangle with counterclockwise vertices \((0,0)\), \((4,0)\), and \((0,3)\). For \(\mathbf{F}(x,y)=(x^2,xy)\), compute the outward flux using both sides of Green’s Theorem. Then use a boundary area formula to find the enclosed area.
The divergence is \(\displaystyle P_x=2x,\quad Q_y=x,\quad P_x+Q_y=3x.\) The triangle is described by \(0\le x\le4\) and \(0\le y\le3(1-x/4)\). Thus \(\displaystyle \iint_D3x\,dA=\int_0^4\int_0^{3(1-x/4)}3x\,dy\,dx =\int_0^4 9x(1-x/4)\,dx.\) Evaluating step by step gives \(\displaystyle 9\left[\frac{x^2}{2}-\frac{x^3}{12}\right]_0^4 =9\left(8-\frac{16}{3}\right)=24.\)
Now check the boundary flux edge by edge. On the bottom edge, \(x=t,\ y=0\), for \(0\le t\le4\). Here \(dy=0\) and \(Q=xy=0\), so \(P\,dy-Q\,dx=0\). On the sloping edge, use \(x=4-4t,\ y=3t\), for \(0\le t\le1\). Then \(dx=-4\,dt,\ dy=3\,dt\), \(P=(4-4t)^2\), and \(Q=12t(1-t)\). The flux integrand is \(\displaystyle P\,dy-Q\,dx=3(4-4t)^2+4\bigl(12t(1-t)\bigr)=48-48t.\) Its integral is \(\displaystyle \int_0^1(48-48t)\,dt=48-24=24.\) On the vertical edge, \(x=0\), so \(P=Q=0\), and its contribution is zero. The total boundary flux is \(0+24+0=24\), as required.
For the area, evaluate \(\frac12\oint_{\partial D}(x\,dy-y\,dx)\). The bottom and vertical edges contribute zero. On the sloping edge, \(\displaystyle x\,dy-y\,dx=3(4-4t)-3t(-4)=12-12t+12t=12.\) Its integral from \(0\) to \(1\) is \(12\), so the area is \(12/2=6\). This also matches the triangle area \((4)(3)/2=6\).
Worked Example: An Annulus and Its Inner Boundary
Worked Example: An Annulus and Its Inner Boundary
Let \(D\) be the annulus \(1\le x^2+y^2\le9\). Find its area and the outward flux of \(\mathbf{F}(x,y)=(x/2,y/2)\) across its complete boundary.
This field has divergence \(1\) and is smooth on the entire plane, so the flux equals the annulus area. The outer circle has radius \(3\), and the hole has radius \(1\). Therefore \(\displaystyle \operatorname{Area}(D)=9\pi-\pi=8\pi,\) and Green’s Theorem gives total outward flux \(8\pi\).
Check the two boundary contributions directly. On a circle of radius \(r\), set \(x=r\cos t,\ y=r\sin t\). Then \(dx=-r\sin t\,dt,\ dy=r\cos t\,dt\), and \(\displaystyle P\,dy-Q\,dx=\frac{r^2}{2}\cos^2t\,dt+\frac{r^2}{2}\sin^2t\,dt=\frac{r^2}{2}\,dt.\) The outer boundary is counterclockwise, so its flux is \(\int_0^{2\pi}(9/2)\,dt=9\pi\). The inner boundary must be clockwise. Reversing its direction makes its contribution \(-\int_0^{2\pi}(1/2)\,dt=-\pi\). The total is \(9\pi-\pi=8\pi\), which verifies the theorem’s answer.
Worked Example: A Composite L-Shaped Region
Worked Example: A Composite L-Shaped Region
Let \(D\) be the L-shaped polygon with counterclockwise vertices \((0,0),(3,0),(3,1),(1,1),(1,3),(0,3)\). Find its area and the outward flux of \(\mathbf{F}(x,y)=(x/2,y/2)\). Check the flux by adding the contributions along the six boundary segments.
Since the divergence is \(1\), the flux equals the area. Divide the L-shape into the rectangle \(0\le x\le1,\ 0\le y\le3\), of area \(3\), and the rectangle \(1\le x\le3,\ 0\le y\le1\), of area \(2\). Thus the area and predicted outward flux are \(3+2=5\).
For a boundary check, use \(P=x/2\), \(Q=y/2\), and the flux element \(P\,dy-Q\,dx\). On the bottom edge, \(y=0\) and \(dy=0\), so the contribution is \(0\). On the right edge, \(x=3\) and \(y\) runs from \(0\) to \(1\), giving \(\int_0^1(3/2)\,dy=3/2\). On the inward horizontal edge, \(y=1\), \(x\) runs from \(3\) to \(1\), and the contribution is \(\displaystyle \int_3^1-\frac12\,dx=1.\) On the inward vertical edge, \(x=1\), \(y\) runs from \(1\) to \(3\), giving \(\int_1^3(1/2)\,dy=1\). On the top edge, \(y=3\) and \(x\) runs from \(1\) to \(0\), so the contribution is \(\int_1^0-(3/2)\,dx=3/2\). The final left edge has \(x=0\), so its contribution is \(0\). Adding all six values gives \(\displaystyle 0+\frac32+1+1+\frac32+0=5,\) as predicted.
If the region is split into the two rectangles, the shared edge appears twice, once with each orientation. The field has the same value on both sides, so those two flux contributions cancel. Only the exterior boundary remains. This cancellation is useful whenever a composite region is easier to analyze in pieces.
Common Orientation Mistake
A frequent error on a region with a hole is to traverse both the outer and inner boundaries counterclockwise. In the annulus example, that incorrect choice would give \(9\pi+\pi=10\pi\), rather than the correct \(9\pi-\pi=8\pi\). The inner circle is not oriented according to the outward direction from the hole’s disk: it is oriented so the annular region stays on the left as the boundary is traversed. Consequently, its direction is clockwise, and its flux contribution is negative for the radial field used above.
Putting the Method to Work
A reliable calculation has three checks. First, verify that the field is defined and continuously differentiable on a neighborhood of the region; a singularity inside or on the boundary prevents direct application of Green’s Theorem. Second, check that the boundary orientation matches the region, especially for holes and piecewise boundaries. Third, compare the boundary result with the area integral or a geometric area calculation when one is available.
For a flux problem, integrate the divergence over the region when it is simple, or evaluate \(P\,dy-Q\,dx\) along boundary pieces when those are simpler. For area, the field \((x/2,y/2)\) reliably turns the area into an outward-flux integral. The best route is the one that keeps the field regular, the orientation clear, and the arithmetic verifiable.
Check Your Understanding
For each question, identify the field, region, and orientation facts needed before calculating.
- Why does the field \((x/2,y/2)\) convert an outward-flux calculation into an area calculation?
- What orientation should be used on the inner boundary of an annulus, and why?
- For the triangle with vertices \((0,0),(2,0),(0,5)\), which boundary area formula would you use to verify its area?
- When a composite region is split into two pieces, what happens to the flux contributions along a shared edge?
- What regularity condition on the vector field is needed to apply Green’s Theorem directly?