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

[latex]\displaystyle\int_C{\bf{F}}\bullet{\bf{N}}=\displaystyle\int \ \displaystyle\int_D2dA=2\displaystyle\int \ \displaystyle\int_DdA[/latex].

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

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

[latex]\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}[/latex]

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

[latex]\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}[/latex].

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

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

[latex]\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}[/latex].

Therefore, the water flux is [latex]80[/latex] m²/sec.

Note that the domain of [latex]{\bf{F}}[/latex] is all of [latex]\mathbb{R}^2[/latex], which is simply connected. Therefore, to show that [latex]{\bf{F}}[/latex] 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 [latex]P(x, y)=y[/latex] and [latex]Q(x, y)=-x[/latex]. Then [latex]P_x+0=Q_y[/latex], and therefore [latex]P_x+Q_y=0[/latex]. Thus, [latex]{\bf{F}}[/latex] is source free.

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

[latex]\large{g(x,y)=\frac{y^2}2+h(x)}[/latex].

Since [latex]-g_x=Q=-x[/latex] we have [latex]h'(x)=x[/latex]. Therefore,

[latex]\large{h(x)=\frac{x^2}2+C}[/latex].

Letting [latex]C=0[/latex] gives stream function

[latex]\large{g(x,y)=\frac{x^2}2+\frac{y^2}2}[/latex].

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

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

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