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Green’s Theorem in the Plane · Tutorial 809 of 999

Computing Outward Flux from the Divergence

Use the divergence \(P_x+Q_y\) and Green’s theorem to find outward flux over several common planar regions.

Advanced 10 min read

What You'll Learn

  • State the flux form of Green’s theorem and its conditions.
  • Relate outward flux on a counterclockwise boundary to \(P\,dy-Q\,dx\).
  • Set up divergence integrals over rectangular and circular regions.
  • Calculate flux for a region bounded by several line segments.
  • Check an area-integral result using the boundary flux.
  • Identify how reversing boundary orientation changes the flux-form integral.

Outward Flux and Divergence

For a planar vector field, outward flux measures how much of the field crosses a closed boundary in the outward direction. Directly integrating along every piece of that boundary can require several parametrizations. The flux form of Green’s theorem provides another route: integrate the field’s divergence over the region inside the curve.

For \(\mathbf F=(P,Q)\), the quantity \(P_x+Q_y\) is called the divergence of the field. In this tutorial, it will be used as an integrand for calculating total outward flux. The boundary must have the positive orientation, meaning that the region stays to the left as the boundary is traversed; for a simple closed curve, this is counterclockwise.

The flux form of Green’s theorem: Suppose \(D\) is a bounded planar region with a positively oriented, piecewise-smooth simple boundary \(C\), and \(P\) and \(Q\) have continuous first partial derivatives on an open set containing \(D\). Then $$ \oint_C P\,dy-Q\,dx = \iint_D (P_x+Q_y)\,dA. $$ The boundary integral on the left is the outward flux of \(\mathbf F=(P,Q)\). Thus the double integral on the right gives the same outward flux.

The smoothness condition ensures that the derivatives in the integrand exist and are continuous throughout the region. The orientation condition is essential for the stated sign: reversing the direction of travel around \(C\) changes the sign of the boundary integral.

Why the Boundary Formula Measures Outward Flux

Let a short piece of a positively oriented boundary have displacement \((dx,dy)\). Its tangent direction is \((dx,dy)\), so an outward normal displacement with the same length is \((dy,-dx)\): it points to the right of the direction of travel. The field’s contribution to outward flux along this piece is therefore \(P\,dy-Q\,dx\). Summing these contributions gives the boundary integral.

The area-integral identity follows directly from the circulation form of Green’s theorem. That theorem says that for a positively oriented boundary, \(\oint_C M\,dx+N\,dy=\iint_D(N_x-M_y)\,dA\), under the corresponding smoothness conditions. Set \(M=-Q\) and \(N=P\). Then the circulation integrand on the boundary becomes \(-Q\,dx+P\,dy\), and the area integrand becomes \(P_x-(-Q)_y=P_x+Q_y\). This yields the flux form.

$$ \oint_C P\,dy-Q\,dx = \iint_D(P_x+Q_y)\,dA. $$

This derivation explains why the partial derivatives appear in this particular combination. It is not enough to integrate either \(P\) or \(Q\) over the region: both contributions to the field’s divergence are included.

Worked Example: Flux Through a Rectangle

Worked Example: Flux Through a Rectangle

Find the outward flux of \(\mathbf F=(x^2+y,\;x+y^2)\) across the boundary of the rectangle \(D=\{(x,y):0\leq x\leq2,\ 1\leq y\leq3\}\).

The field components are \(P=x^2+y\) and \(Q=x+y^2\). Their derivatives are \(P_x=2x\) and \(Q_y=2y\). The divergence is therefore \(2x+2y\). The rectangle’s boundary is taken counterclockwise, and the field is continuously differentiable everywhere, so the flux form applies.

Integrate over the rectangle:

$$ \begin{aligned} \iint_D(2x+2y)\,dA &=\int_0^2\int_1^3(2x+2y)\,dy\,dx\\ &=\int_0^2\left[2xy+y^2\right]_{1}^{3}\,dx\\ &=\int_0^2\big((6x+9)-(2x+1)\big)\,dx\\ &=\int_0^2(4x+8)\,dx\\ &=\left[2x^2+8x\right]_0^2\\ &=8+16=24. \end{aligned} $$

As a check, compute the boundary integral on its four counterclockwise edges. On the bottom edge, \(y=1\) and \(x\) runs from \(0\) to \(2\), so \(dy=0\) and \(Q=x+1\). Its contribution is \(-\int_0^2(x+1)\,dx=-4\). On the right edge, \(x=2\) and \(y\) runs from \(1\) to \(3\), so \(P=4+y\); the contribution is \(\int_1^3(4+y)\,dy=12\).

On the top edge, \(y=3\) and \(x\) runs from \(2\) to \(0\). Here \(Q=x+9\), so its contribution is \(-\int_2^0(x+9)\,dx=20\). On the left edge, \(x=0\) and \(y\) runs from \(3\) to \(1\); here \(P=y\), so the contribution is \(\int_3^1 y\,dy=-4\). The sum is \(-4+12+20-4=24\), agreeing with the area integral.

Using Symmetry on a Circular Region

For a disk, it is often useful to describe the region by its center and radius, then choose coordinates suited to that description. Symmetry can simplify the divergence integral: terms that are odd in a coordinate centered at the disk’s center integrate to zero.

Worked Example: Flux Through an Offset Disk

Find the outward flux of \(\mathbf F=(x^2,y^2)\) across the circle bounding the disk centered at \((1,0)\) with radius \(2\).

The divergence is \(P_x+Q_y=2x+2y\). Use \(u=x-1\) and \(v=y\), so the disk is \(u^2+v^2\leq4\), and \(x=1+u,\ y=v\). The divergence becomes \(2+2u+2v\). The integrals of \(u\) and \(v\) over this disk vanish by symmetry: reflecting the disk across either coordinate axis changes the sign of the corresponding coordinate but leaves the region unchanged. The disk’s area is \(\pi(2)^2=4\pi\). Consequently, the flux is

$$ \iint_D(2+2u+2v)\,dA = 2(4\pi)+2(0)+2(0) = 8\pi. $$

Check the answer directly on the circle using \(x=1+2\cos t,\ y=2\sin t\), for \(0\leq t\leq2\pi\). This parametrization travels counterclockwise. Since \(dx=-2\sin t\,dt\) and \(dy=2\cos t\,dt\),

$$ \begin{aligned} P\,dy-Q\,dx &=2(1+2\cos t)^2\cos t\,dt +8\sin^3t\,dt\\ &=(2\cos t+8\cos^2t+8\cos^3t+8\sin^3t)\,dt. \end{aligned} $$

Over one full turn, \(\int_0^{2\pi}\cos t\,dt=0\), \(\int_0^{2\pi}\cos^3t\,dt=0\), and \(\int_0^{2\pi}\sin^3t\,dt=0\). Also, \(\int_0^{2\pi}\cos^2t\,dt=\pi\). Thus the boundary flux is \(8\pi\), as obtained from the divergence integral.

Worked Example: A Piecewise-Bounded Triangle

Worked Example: Flux Through a Triangle

Find the outward flux of \(\mathbf F=(x^2,xy)\) across the triangular region with vertices \((0,0)\), \((2,0)\), and \((0,1)\).

Here \(P_x=2x\) and \(Q_y=x\), so the divergence is \(3x\). The triangle can be described by \(0\leq x\leq2\) and \(0\leq y\leq1-\frac{x}{2}\). The top bound comes from the line joining \((2,0)\) to \((0,1)\). The area integral is

$$ \begin{aligned} \iint_D3x\,dA &=\int_0^2\int_0^{1-x/2}3x\,dy\,dx\\ &=\int_0^2 3x\left(1-\frac{x}{2}\right)\,dx\\ &=\int_0^2\left(3x-\frac{3}{2}x^2\right)\,dx\\ &=\left[\frac{3}{2}x^2-\frac{1}{2}x^3\right]_0^2\\ &=6-4=2. \end{aligned} $$

Check by adding the contributions from the three boundary segments in counterclockwise order. The bottom segment goes from \((0,0)\) to \((2,0)\); there \(dy=0\) and \(Q=0\), so its contribution is \(0\). For the sloping segment from \((2,0)\) to \((0,1)\), set \(x=2-2t,\ y=t\), where \(0\leq t\leq1\). Then \(dx=-2\,dt,\ dy=dt\), and

$$ \begin{aligned} P\,dy-Q\,dx &=(2-2t)^2\,dt -((2-2t)t)(-2\,dt)\\ &=\big((4-8t+4t^2)+(4t-4t^2)\big)\,dt\\ &=(4-4t)\,dt. \end{aligned} $$

Its contribution is \(\int_0^1(4-4t)\,dt=[4t-2t^2]_0^1=2\). The final segment goes from \((0,1)\) to \((0,0)\), where \(x=0\), so both \(P\) and \(Q\) vanish and its contribution is \(0\). The total boundary flux is \(0+2+0=2\), which checks the area-integral result.

Orientation and a Common Sign Error

The flux formula \(\oint_C P\,dy-Q\,dx\) gives outward flux only when \(C\) is positively oriented. If a circle is instead traversed clockwise, this same integral has the opposite sign. For example, on the offset disk above, a clockwise parametrization would give \(-8\pi\) for \(\oint_C P\,dy-Q\,dx\), not the outward flux. The outward flux remains \(8\pi\); reverse the parametrization or negate the clockwise integral to obtain it.

Common mistake: Using a clockwise boundary in \(\oint_C P\,dy-Q\,dx\) and treating the result as outward flux. The flux-form boundary integral assumes counterclockwise orientation. Check the direction before interpreting its sign.

For the area-integral method, the most important practical steps are to differentiate the correct components, describe the region with valid bounds, and maintain the positive orientation when checking on the boundary. In particular, the divergence is \(P_x+Q_y\), not \(P_y+Q_x\).

Check Your Understanding

Use the flux form of Green’s theorem and pay attention to the direction of boundary travel.

  1. For a positively oriented boundary, what double integral gives the outward flux of \(\mathbf F=(P,Q)\)?
  2. What hypotheses on the field and boundary are needed for the flux form used here?
  3. Why can the terms \(u\) and \(v\) be omitted when integrating \(2+2u+2v\) over a disk centered at the origin in \(u,v\)-coordinates?
  4. What sign does \(\oint_C P\,dy-Q\,dx\) acquire if the direction of a positively oriented boundary is reversed?
  5. For \(\mathbf F=(x,y)\), calculate the outward flux across the circle \(x^2+y^2=1\) using the divergence integral.

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