Flux Form of Green’s Theorem

Learning Objectives

Apply the flux form of Green’s theorem.

The circulation form of Green’s theorem relates a double integral over region

D

$D$ to line integral

\oint_{C} F \cdot T d s

$\displaystyle\oint_C{\bf{F}}\cdot{\bf{T}}ds$ , where

C

$C$ is the boundary of

D

$D$ . The flux form of Green’s theorem relates a double integral over region

D

$D$ to the flux across boundary

C

$C$ . The flux of a fluid across a curve can be difficult to calculate using the flux line integral. This form of Green’s theorem allows us to translate a difficult flux integral into a double integral that is often easier to calculate.

theorem: Green’s theorem, flux form

Let $D$ be an open, simply connected region with a boundary curve $C$ that is a piecewise smooth, simple closed curve that is oriented counterclockwise (Figure 1). Let ${\bf{F}}=\langle{P},Q\rangle$ be a vector field with component functions that have continuous partial derivatives on an open region containing $D$ . Then,

$\displaystyle\oint{\bf{F}}\cdot{\bf{N}}ds=\displaystyle\iint_DP_x+Q_ydA$ .

<img src="/apps/archive/20220422.171947/resources/5b401794be0561b0b57ebb66b433fc0bfb8d584a" data-media-type="image/jpeg" alt="A vector field in two dimensions. A generic curve C encloses a simple region D around the origin oriented counterclockwise. Normal vectors N point out and away from the curve into quadrants 1, 3, and 4." id="17">

Figure 1. The flux form of Green’s theorem relates a double integral over region $D$ to the flux across curve $C$ .

Because this form of Green’s theorem contains unit normal vector $\bf{N}$ , it is sometimes referred to as the normal form of Green’s theorem.

Proof

Recall that $\displaystyle\oint{\bf{F}}\cdot{\bf{N}}ds=\displaystyle\oint_C-Qdx+Pdy$ . Let $M=-Q$ and $N=P$ . By the circulation form of Green’s theorem,

$\begin{aligned} \displaystyle\oint_C-Qdx+Pdy&=\displaystyle\oint_CMdx+nDy \\ &=\displaystyle\iint_DN_x-M_ydA \\ &=\displaystyle\iint_DP_x-(-Q)_ydA \\ &=\displaystyle\iint_DP_x+Q_ydA \end{aligned}$ .

$_\blacksquare$

Example: Applying Green’s theorem for flux across a circle

Let $C$ be a circle of radius $r$ centered at the origin (Figure 2) and let ${\bf{F}}(x,y)=\langle{x},y\rangle$ . Calculate the flux across $C$ .

<img src="/apps/archive/20220422.171947/resources/3dcb3fb449528300042551a059b72ac2dd14edb1" data-media-type="image/jpeg" alt="A vector field in two dimensions. The arrows point away from the origin in a radial pattern. They are shorter near the origin and much longer further away. A circle with radius 2 and center at the origin is drawn." id="19">

Figure 2. Curve $C$ is a circle of radius $r$ centered at the origin.

Show Solution

Let $D$ be the disk enclosed by $C$ . The flux across $C$ is $\displaystyle\oint_C{\bf{F}}\cdot{\bf{N}}ds$ . We could evaluate this integral using tools we have learned, but Green’s theorem makes the calculation much more simple. Let $P(x, y)=x$ and $Q(x, y)=y$ so that ${\bf{F}}=\langle{P},Q\rangle$ . Note that $P_x=1=Q_y$ , and therefore $P_x+Q_y=2$ . By Green’s theorem,

$\displaystyle\int_C{\bf{F}}\bullet{\bf{N}}=\displaystyle\int \ \displaystyle\int_D2dA=2\displaystyle\int \ \displaystyle\int_DdA$ .

Since $\displaystyle\int_DdA$ is the area of the circle, $\displaystyle\int_DdA=\pi{r}^2$ . Therefore, the flux across $C$ is $2\pi{r}$ .

Example: applying green’s theorem for flux across a triangle

Let $S$ be the triangle with vertices $(0, 0)$ , $(1, 0)$ , and $(0, 3)$ oriented clockwise (Figure 3). Calculate the flux of ${\bf{F}}(x,y)=\langle{P}(x,y),Q(x,y)\rangle=\langle{x}^2+e^y,x+y\rangle$ across $S$ .

<img src="/apps/archive/20220422.171947/resources/2a4456ea6490e1da9ebf0f3adaaac3b7295512fb" data-media-type="image/jpeg" alt="A vector field in two dimensions. A triangle is drawn oriented clockwise with vertices at (0,0), (1,0), and (0,3). The arrows in the field point to the right and up slightly. The angle is greater the closer they are to the axis." id="21">

Figure 3. Curve $S$ is a triangle with vertices $(0, 0)$ , $(1, 0)$ , and $(0, 3)$ oriented clockwise.

Show Solution

To calculate the flux without Green’s theorem, we would need to break the flux integral into three line integrals, one integral for each side of the triangle. Using Green’s theorem to translate the flux line integral into a single double integral is much more simple.

Let $D$ be the region enclosed by $S$ . Note that $P_x=2x$ and $Q_y=1$ therefore, $P_x+Q_y=2x+1$ . Green’s theorem applies only to simple closed curves oriented counterclockwise, but we can still apply the theorem because $\displaystyle\oint_C{\bf{F}}\cdot{\bf{N}}ds=-\displaystyle\oint_{-S}{\bf{F}}\cdot{\bf{N}}ds$ and $-S$ is oriented counterclockwise. By Green’s theorem, the flux is

$\begin{aligned} \displaystyle\oint_C{\bf{F}}\cdot{\bf{N}}ds&=\displaystyle\oint_{-S}{\bf{F}}\cdot{\bf{N}}ds \\ &=-\displaystyle\iint_D(P_x+Q_y)dA \\ &=-\displaystyle\iint_D(2x+1)dA \end{aligned}$

Notice that the top edge of the triangle is the line $y=-3x=3$ Therefore, in the iterated double integral, the y-values run from $y=0$ to $y=-3x+3$ , and we have

$\begin{aligned} -\displaystyle\iint_D(2x+1)dA&=-\displaystyle\int_0^1\displaystyle\int_0^{-3x+3}(2x+1)dydx \\ &=--\displaystyle\int_0^1(2x+1)(-3x+3)dx=-\displaystyle\int_0^1(-6x^2+3x+3)dx \\ &=-\left[-2x^3+\frac{3x^2}2+3x\right]_0^1=-\frac52 \end{aligned}$ .

try it

Calculate the flux of ${\bf{F}}(x,y)=\langle{x}^3,y^3\rangle$ across a unit circle oriented counterclockwise.

Show Solution

$\frac{3\pi}2$ .

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 6.36” here (opens in new window).

Example: applying green’s theorem for water flow across a rectangle

Water flows from a spring located at the origin. The velocity of the water is modeled by vector field ${\bf{v}}(x,y)=\langle5x+y,x+3y\rangle$ m/sec. Find the amount of water per second that flows across the rectangle with vertices $(-1, -2)$ , $(1, -2)$ , $(1, 3)$ , and $(-1, 3)$ , oriented counterclockwise (Figure 4).

<img src="/apps/archive/20220422.171947/resources/416bb9f3ebaa663f2b8f7fe903756dcf624c4709" data-media-type="image/jpeg" alt="A vector field in two dimensions. A rectangle is drawn oriented counterclockwise with vertices at (-1,3), (1,3), (-1,-2), and (1,-2). The arrows point out and away from the origin in a radial pattern. However, the arrows in quadrants 2 and 4 curve slightly towards the y axis instead of directly out. The arrows near the origin are short, and those further away from the origin are much longer." id="24">

Figure 4. Water flows across the rectangle with vertices $(-1, -2)$ , $(1, -2)$ , $(1, 3)$ , and $(-1, 3)$ , oriented counterclockwise.

Show Solution

Let $C$ epresent the given rectangle and let $D$ be the rectangular region enclosed by $C$ . To find the amount of water flowing across $C$ , we calculate flux $\displaystyle\int_C{\bf{v}}\cdot{\bf{N}}ds$ .

Let $P(x, y)=5x+y$ and $Q(x, y)=x+3y$ so that ${\bf{v}}=(P,Q)$ . Then, $P_x=5$ and $Q_y=3$ . By Green’s theorem,

$\begin{aligned} \displaystyle\int_C{\bf{v}}\cdot{\bf{N}}ds&=\displaystyle\iint_D(P_x+Q_y)dA \\ &=\displaystyle\iint_D8dA \\ &=8(\text{area of }D)=80 \end{aligned}$ .

Therefore, the water flux is $80$ m²/sec.

Recall that if vector field ${\bf{F}}$ is conservative, then ${\bf{F}}$ does no work around closed curves—that is, the circulation of ${\bf{F}}$ around a closed curve is zero. In fact, if the domain of ${\bf{F}}$ is simply connected, then ${\bf{F}}$ is conservative if and only if the circulation of ${\bf{F}}$ around any closed curve is zero. If we replace “circulation of ${\bf{F}}$ ” with “flux of ${\bf{F}}$ ,” then we get a definition of a source-free vector field. The following statements are all equivalent ways of defining a source-free field ${\bf{F}}=\langle{P},Q\rangle$ on a simply connected domain (note the similarities with properties of conservative vector fields):

The flux $\displaystyle\oint_C{\bf{F}}\cdot{\bf{N}}ds$ across any closed curve $C$ is zero.
If $C_1$ and $C_2$ are curves in the domain of ${\bf{F}}$ with the same starting points and endpoints, then $\displaystyle\int_{C_1}{\bf{F}}\cdot{\bf{N}}ds=\displaystyle\int_{C_2}{\bf{F}}\cdot{\bf{N}}ds$ . In other words, flux is independent of path.
There is a for ${\bf{F}}$ . A stream function for ${\bf{F}}=\langle{P},Q\rangle$ is a function $g$ such that $P=g_y$ and $Q=-g_x$ . Geometrically, ${\bf{F}}(a,b)$ is tangential to the level curve of $g$ at $(a, b)$ . Since the gradient of $g$ is perpendicular to the level curve of $g$ at $(a, b)$ , stream function $g$ has the property ${\bf{F}}(a,b)\bullet\nabla{g}(a,b)=0$ for any point $(a, b)$ in the domain of $g$ . (Stream functions play the same role for source-free fields that potential functions play for conservative fields.)
$P_x+Q_y=0$

Example: finding a stream function

Verify that rotation vector field ${\bf{F}}(x,y)=\langle{y},-x\rangle$ is source free, and find a stream function for ${\bf{F}}$ .

Show Solution

Note that the domain of ${\bf{F}}$ is all of $\mathbb{R}^2$ , which is simply connected. Therefore, to show that ${\bf{F}}$ is source free, we can show any of items 1 through 4 from the previous list to be true. In this example, we show that item 4 is true. Let $P(x, y)=y$ and $Q(x, y)=-x$ . Then $P_x+0=Q_y$ , and therefore $P_x+Q_y=0$ . Thus, ${\bf{F}}$ is source free.

To find a stream function for ${\bf{F}}$ , proceed in the same manner as finding a potential function for a conservative field. Let $g$ be a stream function for ${\bf{F}}$ . Then $g_y=y$ , which implies that

$\large{g(x,y)=\frac{y^2}2+h(x)}$ .

Since $-g_x=Q=-x$ we have $h'(x)=x$ . Therefore,

$\large{h(x)=\frac{x^2}2+C}$ .

Letting $C=0$ gives stream function

$\large{g(x,y)=\frac{x^2}2+\frac{y^2}2}$ .

To confirm that $g$ is a stream function for ${\bf{F}}$ , note that $g_y=y=P$ and $-g_x=-x=Q$ .

Notice that source-free rotation vector field ${\bf{F}}(x,y)=\langle{y},-x\rangle$ is perpendicular to conservative radial vector field $\nabla{g}=\langle{x},y\rangle$ (Figure 5).

<img src="/apps/archive/20220422.171947/resources/e340a7e412b44f17cd4f3c841f75736d64b46ff4" data-media-type="image/jpeg" alt="Two vector fields in two dimensions. The first has arrows surrounding the origin in a clockwise circular pattern. The second has arrows pointing out and away from the origin in a radial manner. Circles with radii 1.5, 2, and 2.5 and centers at the origin are drawn in both. The arrows near the origin are shorter than those much further away. The first is labeled F(x,y) = <y, -x> and the second is labeled for the gradient, delta g = <x, -y>." id="26">

Figure 5. (a) In this image, we see the three-level curves of $g$ and vector field ${\bf{F}}$ . Note that the ${\bf{F}}$ vectors on a given level curve are tangent to the level curve. (b) In this image, we see the three-level curves of $g$ and vector field $\nabla{g}$ . The gradient vectors are perpendicular to the corresponding level curve. Therefore, ${\bf{F}}(a,b)\bullet\nabla{g}(a,b)=0$ for any point in the domain of $g$ .

try it

Find a stream function for vector field ${\bf{F}}(x,y)=\langle{x}\sin{y},\cos{y}\rangle$ .

Show Solution

$g(x,y)=-x\cos{y}$ .

Vector fields that are both conservative and source free are important vector fields. One important feature of conservative and source-free vector fields on a simply connected domain is that any potential function $f$ of such a field satisfies Laplace’s equation $f_{xx}+f{yy}=0$ . Laplace’s equation is foundational in the field of partial differential equations because it models such phenomena as gravitational and magnetic potentials in space, and the velocity potential of an ideal fluid. A function that satisfies Laplace’s equation is called a harmonic function. Therefore any potential function of a conservative and source-free vector field is harmonic.

To see that any potential function of a conservative and source-free vector field on a simply connected domain is harmonic, let $f$ be such a potential function of vector field ${\bf{F}}=\langle{P},Q\rangle$ . Then, $f_x=P$ and $f_x=Q$ because $\nabla{f}={\bf{F}}$ . Therefore, $f_{xx}=P_x$ and $f_{yy}=$ . Since ${\bf{F}}$ is source free, $f_{xx}+f_{yy}=P_x+Q_y=0$ , and we have that $f$ is harmonic.

Example: satisfying laplace’s equation

For vector field ${\bf{F}}(x,y)=\langle{e}^x\sin{y},e^x\cos{y}\rangle$ , verify that the field is both conservative and source free, find a potential function for ${\bf{F}}$ , and verify that the potential function is harmonic.

Show Solution

Let $P(x,y)={e}^x\sin{y}$ and $Q(x,y)=e^x\cos{y}$ . Notice that the domain of ${\bf{F}}$ is all of two-space, which is simply connected. Therefore, we can check the cross-partials of ${\bf{F}}$ to determine whether ${\bf{F}}$ is conservative. Note that $P_y=e^x\cos{y}$ , so ${\bf{F}}$ is conservative. Since $P_x={e}^x\sin{y}$ and $Q_y={e}^x\sin{y}, P_x+Q_y=0$ and the field is source free.

To find a potential function for ${\bf{F}}$ , let $f$ be a potential function. Then, $\nabla{f}={\bf{F}}$ , so $f_x={e}^x\sin{y}$ . Integrating this equation with respect to $x$ gives $f(x,y)={e}^x\sin{y}+h(y)$ . Since $f_y=e^x\cos{y}$ , differentiating $f$ with respect to $y$ gives $e^x\cos{y}=e^x\cos{y}+h^\prime(y)$ . Therefore, we can take $h(y)=0$ , and $f(x,y)={e}^x\sin{y}$ is a potential function for $f$ .

To verify that $f$ is a harmonic function, note that $f_{xx}=\frac{\partial}{\partial{x}}({e}^x\sin{y})={e}^x\sin{y}$ and $f_{yy}=\frac{\partial}{\partial{x}}(e^x\cos{y})=-{e}^x\sin{y}$ . Therefore, $f_{xx}+f_{yy}=0$ , and $f$ satisfies Laplace’s equation.

try it

Is the function $f(x,y)=e^{x+5y}$ harmonic?

Show Solution

Module 6: Vector Calculus