The Flux Form of Green’s Theorem

From Boundary Flow to an Area Integral

A vector field can carry fluid, heat, or another quantity across a curve. The outward flux measures the net amount crossing the boundary in the outward direction. A positive value means more flow leaves the region than enters it; a negative value means the net flow is inward.

For a complicated boundary, adding up the flow along every piece can be inconvenient. Green’s theorem gives another route: under its hypotheses, the total outward flux is the double integral over the enclosed region of a simple expression formed from the field’s partial derivatives.

Central idea: The outward-flux form of Green’s theorem converts a boundary integral into an area integral of the divergence $P_x+Q_y$.

Flux Along an Oriented Boundary

Let $\mathbf{F}(x,y)=(P(x,y),Q(x,y))$ be a planar vector field, and let $C$ be a smooth curve with a chosen direction. At a point on the curve, the component of $\mathbf{F}$ perpendicular to the curve determines how much of the field crosses it. The outward flux is the integral of that perpendicular component along the boundary.

Suppose a smooth part of $C$ is parametrized by $\mathbf{r}(t)=(x(t),y(t))$, for $a\leq t\leq b$. If $C$ is traversed counterclockwise around its interior, the interior lies to the left as the curve is followed. The outward unit normal therefore points to the right. The tangent vector is $(x'(t),y'(t))$, so a right-pointing normal with the same length is $(y'(t),-x'(t))$. Its length is the speed $\sqrt{x'(t)^2+y'(t)^2}$.

Dividing that normal by the speed gives the outward unit normal; multiplying the unit normal by the arc-length element $ds=\sqrt{x'(t)^2+y'(t)^2}\,dt$ cancels the speed. Thus the flux element has a particularly simple expression:

\mathbf{n}\,ds=(y'(t),-x'(t))\,dt, \qquad \mathbf{F}\cdot\mathbf{n}\,ds =P\,dy-Q\,dx.

This also gives a useful direct formula for calculating outward flux along a positively oriented boundary. The minus sign before $Q\,dx$ comes from the outward normal, not from an arbitrary convention.

Definition: For a positively oriented, piecewise smooth boundary $C$, the outward flux of $\mathbf{F}=(P,Q)$ across $C$ is $$ \oint_C \mathbf{F}\cdot\mathbf{n}\,ds =\oint_C P\,dy-Q\,dx, $$ where $\mathbf{n}$ is the outward unit normal. The line integral is added over all smooth pieces of $C$.

The Flux Form of Green’s Theorem

Let $D$ be a bounded plane region whose boundary $C$ is a positively oriented, simple closed, piecewise smooth curve. Suppose $P$ and $Q$ have continuous first partial derivatives on an open set containing $D$ and its boundary. The theorem relates the outward flux across $C$ to an integral over $D$.

Theorem (outward-flux form of Green’s theorem): Under the conditions above, $$ \oint_C P\,dy-Q\,dx = \iint_D \left(P_x+Q_y\right)\,dA. $$ Equivalently, the outward flux of $\mathbf{F}=(P,Q)$ across $C$ equals the double integral over $D$ of $P_x+Q_y$.

The expression $P_x+Q_y$ is the divergence of the planar field. It measures the field’s local tendency to flow outward rather than inward. Green’s theorem says that adding this quantity over the region gives the net outward flow through the whole boundary. Positive and negative contributions within the region both matter; the theorem gives their total, not necessarily the flux at each individual boundary point.

The orientation and the outward-normal convention must agree. Positive orientation means counterclockwise traversal around the interior. If the same boundary is traversed clockwise, the corresponding line integral changes sign. The area integral in the theorem still uses the positive area element $dA$, so the stated equality uses the counterclockwise orientation.

Worked Examples

Worked Example: Radial Field Across a Circle

Find the outward flux of $\mathbf{F}(x,y)=(x,y)$ across the counterclockwise circle $x^2+y^2=4$, using Green’s theorem. Then verify it directly.

Here $P=x$ and $Q=y$, so

P_x=1,\qquad Q_y=1,\qquad P_x+Q_y=2.

The region $D$ is the disk of radius $2$, whose area is $\pi(2)^2=4\pi$. Therefore, Green’s theorem gives

\iint_D 2\,dA =2\cdot 4\pi =8\pi.

To check, parametrize the boundary counterclockwise by $x=2\cos t$, $y=2\sin t$, for $0\leq t\leq2\pi$. Then $dx=-2\sin t\,dt$ and $dy=2\cos t\,dt$. The boundary flux is

\begin{aligned} \oint_C P\,dy-Q\,dx &=\int_0^{2\pi}\left((2\cos t)(2\cos t) -(2\sin t)(-2\sin t)\right)\,dt\\ &=\int_0^{2\pi}4(\cos^2t+\sin^2t)\,dt\\ &=\int_0^{2\pi}4\,dt =8\pi. \end{aligned}

The direct boundary calculation agrees with the area integral.

Worked Example: Flux Across a Rectangle

Find the outward flux of $\mathbf{F}(x,y)=(x^2,3y)$ across the counterclockwise boundary of $D=[0,2]\times[0,1]$.

In this case, $P=x^2$ and $Q=3y$, so

P_x=2x,\qquad Q_y=3,\qquad P_x+Q_y=2x+3.

Integrate over the rectangle, with $0\leq x\leq2$ and $0\leq y\leq1$:

\begin{aligned} \iint_D(2x+3)\,dA &=\int_0^2\int_0^1(2x+3)\,dy\,dx\\ &=\int_0^2\left[(2x+3)y\right]_0^1\,dx\\ &=\int_0^2(2x+3)\,dx\\ &=\left[x^2+3x\right]_0^2\\ &=(4+6)-0=10. \end{aligned}

For a boundary check, the bottom edge contributes $0$, because $y=0$, $dy=0$, and $Q=0$. On the right edge, $x=2$, $dx=0$, and $P=4$; as $y$ runs from $0$ to $1$, the contribution is $\int_0^1 4\,dy=4$. On the top edge, $y=1$, $dy=0$, and $Q=3$; the positive orientation traverses it from $x=2$ to $x=0$, giving $\int_2^0-3\,dx=6$. On the left edge, $x=0$, $dx=0$, and $P=0$, so the contribution is $0$. The total is $0+4+6+0=10$, as required.

Worked Example: Flux Across a Triangle

Find the outward flux of $\mathbf{F}(x,y)=(xy,x+y^2)$ across the counterclockwise triangle with vertices $(0,0)$, $(2,0)$, and $(0,1)$.

Here $P=xy$, $Q=x+y^2$, and

P_x=y,\qquad Q_y=2y,\qquad P_x+Q_y=3y.

The sloping edge from $(2,0)$ to $(0,1)$ has equation $y=1-\frac{x}{2}$. Thus $0\leq x\leq2$ and $0\leq y\leq1-\frac{x}{2}$ describe the triangle. The area integral is

\begin{aligned} \iint_D 3y\,dA &=\int_0^2\int_0^{1-x/2}3y\,dy\,dx\\ &=\int_0^2\left[\frac{3}{2}y^2\right]_0^{1-x/2}\,dx\\ &=\frac{3}{2}\int_0^2\left(1-\frac{x}{2}\right)^2\,dx\\ &=\frac{3}{2}\int_0^2\left(1-x+\frac{x^2}{4}\right)\,dx\\ &=\frac{3}{2}\left[x-\frac{x^2}{2}+\frac{x^3}{12}\right]_0^2\\ &=\frac{3}{2}\left(2-2+\frac{8}{12}\right)=1. \end{aligned}

To verify directly, use the flux integral $P\,dy-Q\,dx$ on each edge. On the bottom edge, $y=0$, $dy=0$, $Q=x$, and $x$ runs from $0$ to $2$, so the contribution is $\int_0^2 -x\,dx=-2$. On the sloping edge, parametrize by $x=2-2t,\ y=t$, for $0\leq t\leq1$; then $dx=-2\,dt$ and $dy=dt$. Substitution gives

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

so this edge contributes $\int_0^1(4-2t)\,dt=[4t-t^2]_0^1=3$. On the vertical edge $x=0$, $dx=0$ and $P=0$, so its contribution is $0$, regardless of its direction. The total boundary flux is $-2+3+0=1$, matching the double integral.

Why the Boundary Formula Has This Sign

The sign in $P\,dy-Q\,dx$ can be derived directly from the orientation. For a counterclockwise parametrized curve, the tangent is $(x',y')$. Turning the tangent clockwise by a right angle gives $(y',-x')$, which points outward because the interior is on the tangent’s left. The field dotted with this normal, including the arc-length factor, is therefore

(P,Q)\cdot(y',-x')\,dt =(Py'-Qx')\,dt =P\,dy-Q\,dx.

This derivation explains both terms and their signs. Reversing the direction of traversal changes $dx$ and $dy$ to their negatives, and so reverses the boundary integral. In contrast, the theorem as stated pairs the counterclockwise boundary integral with the ordinary positive area element.

A Common Sign Error

A frequent mistake is to use $P\,dx+Q\,dy$ for outward flux. That expression uses the field’s components along the tangent, rather than the outward normal. For example, on the circle in the first worked example, the incorrect expression would give

\begin{aligned} \oint_C P\,dx+Q\,dy &=\int_0^{2\pi} \left((2\cos t)(-2\sin t)+(2\sin t)(2\cos t)\right)\,dt\\ &=\int_0^{2\pi}0\,dt=0. \end{aligned}

But the outward flux is $8\pi$, not $0$. The correction is to use the outward normal formula $P\,dy-Q\,dx$. On the same circle that gives $\int_0^{2\pi}4\,dt=8\pi$. A useful check is to ask whether the chosen differential expression represents a normal component or a tangential component before integrating.

When the Area Integral Is Useful

Green’s theorem is especially useful when $P_x+Q_y$ is simple and the region has convenient double-integral bounds. It can replace several separate boundary integrals with one area integral. Conversely, if the field’s divergence is complicated but the boundary is easy to parametrize, the boundary formula may be preferable.

Whatever route is used, keep the roles distinct: $\mathbf{F}=(P,Q)$ is the field, $P_x+Q_y$ is the integrand over the interior, and $P\,dy-Q\,dx$ is the outward-flux integrand along a positively oriented boundary. Applying the theorem requires its stated regularity and boundary conditions, not merely a closed curve drawn around a region.

Check Your Understanding

Use the definitions and theorem to answer these questions. For calculation questions, show the derivative or boundary-substitution steps.

For $\mathbf{F}=(P,Q)$, what expression is integrated over the region to compute outward flux?
Why does a counterclockwise parametrization use the normal element $(dy,-dx)$?
Find the divergence of $\mathbf{F}=(x^2,y^2)$.
If a positively oriented boundary integral gives outward flux $5$, what does the same line integral give when the boundary is traversed clockwise?
For what purpose does the expression $P\,dx+Q\,dy$ differ from the outward-flux integral?

The Flux Form of Green’s Theorem

What You'll Learn

From Boundary Flow to an Area Integral

Flux Along an Oriented Boundary

The Flux Form of Green’s Theorem

Worked Examples

Worked Example: Radial Field Across a Circle

Worked Example: Flux Across a Rectangle

Worked Example: Flux Across a Triangle

Why the Boundary Formula Has This Sign

A Common Sign Error

When the Area Integral Is Useful

Check Your Understanding

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