Green’s Theorem: Learn It 2

Circulation Form of Green’s Theorem

The first form of Green’s theorem that we examine is the circulation form. This form relates the vector line integral over a simple, closed plane curve [latex]C[/latex] to a double integral over the region enclosed by [latex]C[/latex]. Therefore, the circulation of a vector field along a simple closed curve can be transformed into a double integral and vice versa.

Green’s theorem, circulation form

Let [latex]D[/latex] be an open, simply connected region with a boundary curve [latex]C[/latex] that is a piecewise smooth, simple closed curve oriented counterclockwise (Figure 2). Let [latex]{\bf{F}}=\langle{P},Q\rangle[/latex] be a vector field with component functions that have continuous partial derivatives on [latex]D[/latex]. Then,

[latex]\displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\oint_CPdx+Qdy=\displaystyle\iint_D(Q_x-P_y)dA[/latex].

A vector field in two dimensions with all of the arrows pointing up and to the right. A curve C oriented counterclockwise sections off a region D around the origin. It is a simple, closed region.
Figure 2. The circulation form of Green’s theorem relates a line integral over curve [latex]C[/latex] to a double integral over region [latex]D[/latex].
Green’s theorem can only be used for a two-dimensional vector field [latex]\mathbf{F}[/latex]. If [latex]\mathbf{F}[/latex] is a three-dimensional field, then Green’s theorem does not apply.[latex]\\[/latex]Since [latex]\displaystyle\int_CPdx+Qdy=\displaystyle\int_C\mathbf{F}\cdot\mathbf{T}ds[/latex], this version of Green’s theorem is sometimes called the tangential form of Green’s theorem.

The proof of Green’s theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that [latex]D[/latex] is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which [latex]{\bf{F}}\langle{P},Q\rangle[/latex] is conservative. In this case,

[latex]\large{\displaystyle\oint_CPdx+Qdy=0}[/latex]

because the circulation is zero in conservative vector fields. By the cross-partial property of conservative fields theorem, [latex]{\bf{F}}[/latex] satisfies the cross-partial condition, so [latex]P_y=Q_x[/latex]. Therefore,

[latex]\large{\displaystyle\iint_D(Q_x-P_y)dA=\displaystyle\iint_D0dA=0=\displaystyle\oint_CPdx+Qdy}[/latex],

which confirms Green’s theorem in the case of conservative vector fields.

Proof

Let’s now prove that the circulation form of Green’s theorem is true when the region [latex]D[/latex] is a rectangle. Let [latex]D[/latex] be the rectangle [latex][a,b]\times[c,d][/latex] oriented counterclockwise. Then, the boundary [latex]C[/latex] of [latex]D[/latex] consists of four piecewise smooth pieces [latex]C_1[/latex], [latex]C_2[/latex], [latex]C_3[/latex], and [latex]C_4[/latex] (Figure 3). We parameterize each side of [latex]D[/latex] as follows:

[latex]\begin{aligned} C_1:{\bf{r}}_1(t)&=\langle{t},c\rangle,a\leq{t}\leq{b} \\ C_2:{\bf{r}}_2(t)&=\langle{b},t\rangle,c\leq{t}\leq{d} \\ -C_3:{\bf{r}}_3(t)&=\langle{t},d\rangle,a\leq{t}\leq{b} \\ -C_4:{\bf{r}}_4(t)&=\langle{a},t\rangle,c\leq{t}\leq{d} \end{aligned}[/latex].

A diagram in quadrant 1. Rectangle D is oriented counterclockwise. Points a and b are on the x axis, and points c and d are on the y axis with b > a and d > c. The sides of the rectangle are side c1 with endpoints at (a,c) and (b,c), side c2 with endpoints at (b,c) and (b,d), side c3 with endpoints at (b,d) and (a,d), and side c4 with endpoints at (a,d) and (a,c).
Figure 3. Rectangle [latex]D[/latex] is oriented counterclockwise.

Then,

[latex]\begin{aligned} \displaystyle\int_C{\bf{F}}\bullet{d}{\bf{r}}&=\displaystyle\int_{C_1}{\bf{F}}\bullet{d}{\bf{r}}+\displaystyle\int_{C_2}{\bf{F}}\bullet{d}{\bf{r}}+\displaystyle\int_{C_3}{\bf{F}}\bullet{d}{\bf{r}}+\displaystyle\int_{C_4}{\bf{F}}\bullet{d}{\bf{r}} \\ &=\displaystyle\int_{C_1}{\bf{F}}\bullet{d}{\bf{r}}+\displaystyle\int_{C_2}{\bf{F}}\bullet{d}{\bf{r}}-\displaystyle\int_{-C_3}{\bf{F}}\bullet{d}{\bf{r}}-\displaystyle\int_{-C_4}{\bf{F}}\bullet{d}{\bf{r}} \\ &=\displaystyle\int_a^b{\bf{F}}({\bf{r}}_1(t))\bullet{\bf{r}}_1(t)dt+\displaystyle\int_c^d{\bf{F}}({\bf{r}}_2(t))\bullet{\bf{r}}_2(t)dt-\displaystyle\int_a^b{\bf{F}}({\bf{r}}_3(t))\bullet{\bf{r}}_3(t)dt-\displaystyle\int_c^d{\bf{F}}({\bf{r}}_4(t))\bullet{\bf{r}}_4(t)dt \\ &=\displaystyle\int_a^bP(t,c)dt+\displaystyle\int_c^dQ(b,t)dt-\displaystyle\int_a^bP(t,d)dt-\displaystyle\int_c^dQ(a,t)dt \\ &=\displaystyle\int_a^b(P(t,c)-P(t,d))dt+\displaystyle\int_c^d(Q(b,t)-Q(a,t))dt \\ &=-\displaystyle\int_a^b(P(t,d)-P(t,c))dt+\displaystyle\int_c^d(Q(b,t)-Q(a,t))dt \end{aligned}[/latex].

By the Fundamental Theorem of Calculus,

[latex]\large{P(t,d)-P(t,c)=\displaystyle\int_c^d\frac{\partial}{\partial{y}}P(t,y)dy\text{ and }Q(b,t)-Q(a,t)=\displaystyle\int_a^b\frac{\partial}{\partial{x}}Q(x,t)dx}[/latex].

Therefore,

[latex]\begin{aligned} &-\displaystyle\int_a^b(P(t,d)-P(t,c)dt+\displaystyle\int_c^d(Q(b,t)-Q(a,t))dt \\ &=-\displaystyle\int_a^b\displaystyle\int_c^d\frac{\partial}{\partial{y}}P(t,y)dydt+\displaystyle\int_c^d\displaystyle\int_a^b\frac{\partial}{\partial{x}}Q(x,t)dxdt \end{aligned}[/latex].

But,

[latex]\begin{aligned} -\displaystyle\int_a^b\displaystyle\int_c^d\frac{\partial}{\partial{y}}P(t,y)dydt+\displaystyle\int_c^d\displaystyle\int_a^b\frac{\partial}{\partial{x}}Q(x,t)dxdt&=-\displaystyle\int_a^b\displaystyle\int_c^d\frac{\partial}{\partial{y}}P(x,y)dydx+\displaystyle\int_c^d\displaystyle\int_a^b\frac{\partial}{\partial{x}}Q(x,t)dxdy \\ &=\displaystyle\int_a^b\displaystyle\int_c^d(Q_x-P_y)dydx \\ &=\displaystyle\int \ \displaystyle\int_D(Q_x-P_y)dA \end{aligned}[/latex].

Therefore, [latex]\displaystyle\int_C{\bf{F}}\bullet{d}{\bf{r}}=\int\int_D(Q_{x}-P_{y})dA[/latex] and we have proved Green’s theorem in the case of a rectangle.

To prove Green’s theorem over a general region [latex]D[/latex], we can decompose [latex]D[/latex] into many tiny rectangles and use the proof that the theorem works over rectangles. The details are technical, however, and beyond the scope of this text.

[latex]_\blacksquare[/latex]

Calculate the line integral

[latex]\displaystyle\oint_Cx^2ydx+(y-3)dy[/latex]

where [latex]C[/latex] is a rectangle with vertices [latex](1, 1)[/latex], [latex](4, 1)[/latex], [latex](4, 5)[/latex], and [latex](1, 5)[/latex] oriented counterclockwise.

Calculate the work done on a particle by force field

[latex]{\bf{F}}(x,y)=\langle{y}+\sin{x},e^y-x\rangle[/latex]

as the particle traverses circle [latex]x^2+y^2=4[/latex] exactly once in the counterclockwise direction, starting and ending at point [latex](2, 0)[/latex].

In the preceding two examples, the double integral in Green’s theorem was easier to calculate than the line integral, so we used the theorem to calculate the line integral. In the next example, the double integral is more difficult to calculate than the line integral, so we use Green’s theorem to translate a double integral into a line integral.

Calculate the area enclosed by ellipse [latex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/latex] (Figure 6).

A horizontal ellipse graphed in two dimensions. It has vertices at (-a, 0), (0, -b), (a, 0), and (0, b), where the absolute value of a is between 2.5 and 5 and the absolute value of b is between 0 and 2.5.
Figure 6. Ellipse [latex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/latex] is denoted by [latex]C[/latex].

In the above example, we used vector field [latex]\mathbf{F}(x,y)=\langle P,Q\rangle=\left\langle-\frac{y}{2},\frac{x}{2}\right\rangle[/latex] to find the area of any ellipse. The logic of the previous example can be extended to derive a formula for the area of any region [latex]D[/latex]. Let [latex]D[/latex] be any region with a boundary that is a simple closed curve [latex]C[/latex] oriented counterclockwise. If [latex]\mathbf{F}(x,y)=\langle P,Q\rangle=\left\langle-\frac{y}{2},\frac{x}{2}\right\rangle[/latex], then by the same logic as in the previous example,

[latex]\text{area of }D=\displaystyle\iint_DdA=\frac12\displaystyle\oint_C-ydx+xdy[/latex].

 If [latex]\mathbf{F}=\langle P,Q\rangle[/latex] is any vector field with [latex]Q_x-P_y=1[/latex], then the logic of the previous paragraph works. So, the equation above is not the only equation that uses a vector field’s mixed partials to get the area of a region.