Extending the Fundamental Theorem of Calculus

Recall that the Fundamental Theorem of Calculus says that

[latex]\large{\displaystyle\int_a^b{F}^\prime(x)dx=F(b)-F(a)}[/latex].

As a geometric statement, this equation says that the integral over the region below the graph of [latex]F^\prime(x)[/latex] and above the line segment [latex][a,b][/latex] depends only on the value of [latex]F[/latex] at the endpoints a and b of that segment. Since the numbers [latex]a[/latex] and [latex]b[/latex] are the boundary of the line segment [latex][a,b][/latex], the theorem says we can calculate integral [latex]\displaystyle\int_a^b{F}^\prime(x)dx[/latex] based on information about the boundary of line segment [latex][a,b][/latex] (Figure 1). The same idea is true of the Fundamental Theorem for Line Integrals:

[latex]\large{\displaystyle\int_C\nabla{f}\cdot{d}{\bf{r}}=f({\bf{r}}(b))-f({\bf{r}}(a))}[/latex].

When we have a potential function (an “antiderivative”), we can calculate the line integral based solely on information about the boundary of curve [latex]C[/latex].

Green’s theorem takes this idea and extends it to calculating double integrals. Green’s theorem says that we can calculate a double integral over region [latex]D[/latex] based solely on information about the boundary of [latex]D[/latex] Green’s theorem also says we can calculate a line integral over a simple closed curve [latex]C[/latex] based solely on information about the region that [latex]C[/latex] encloses. In particular, Green’s theorem connects a double integral over region [latex]D[/latex] to a line integral around the boundary of [latex]D[/latex].

Circulation Form of Green’s Theorem

The first form of Green’s theorem that we examine is the circulation form. This form of the theorem 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.

Notice that Green’s theorem can be used only for a two-dimensional vector field [latex]{\bf{F}}[/latex]. If [latex]{\bf{F}}[/latex] is a three-dimensional field, then Green’s theorem does not apply. Since

[latex]\large{\displaystyle\int_CPdx+Qdy=\displaystyle\int_C{\bf{F}}\cdot{\bf{T}}ds}[/latex],

this version of Green’s theorem is sometimes referred to as 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 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].

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]

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

You can view the transcript for “CP 6.34” here (opens in new window).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.

In Example “Applying Green’s Theorem over an Ellipse”, we used vector field [latex]{\bf{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]{\bf{F}}(x,y)=\langle{P},Q\rangle=\left\langle-\frac{y}2,\frac{x}2\right\rangle[/latex]. Therefore, by the same logic as in Example “Applying Green’s Theorem over an Ellipse”,

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

It’s worth noting that if [latex]{\bf{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.