Extending the Fundamental Theorem of Calculus

Recall that the Fundamental Theorem of Calculus says that

$\large{\displaystyle\int_a^b{F}^\prime(x)dx=F(b)-F(a)}$.

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

$\large{\displaystyle\int_C\nabla{f}\cdot{d}{\bf{r}}=f({\bf{r}}(b))-f({\bf{r}}(a))}$.

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

Figure 1. The Fundamental Theorem of Calculus says that the integral over line segment $[a,b]$ depends only on the values of the antiderivative at the endpoints of $[a,b]$.

Figure 2. The circulation form of Green’s theorem relates a line integral over curve $C$ to a double integral over region $D$.

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

$\large{\displaystyle\int_CPdx+Qdy=\displaystyle\int_C{\bf{F}}\cdot{\bf{T}}ds}$,

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 $D$ is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which ${\bf{F}}\langle{P},Q\rangle$ is conservative. In this case,

$\large{\displaystyle\oint_CPdx+Qdy=0}$

because the circulation is zero in conservative vector fields. By Cross-Partial Property of Conservative Fields Theorem, ${\bf{F}}$ satisfies the cross-partial condition, so $P_y=Q_x$. Therefore,

$\large{\displaystyle\iint_D(Q_x-P_y)dA=\displaystyle\iint_D0dA=0=\displaystyle\oint_CPdx+Qdy}$,

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

Then,

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

By the Fundamental Theorem of Calculus,

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

Therefore,

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

But,

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

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

To prove Green’s theorem over a general region $D$, we can decompose $D$ 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.

$_\blacksquare$

Example: applying Green’s Theorem over a rectangle

Calculate the line integral

$\displaystyle\oint_Cx^2ydx+(y-3)dy$

where $C$ is a rectangle with vertices $(1, 1)$, $(4, 1)$, $(4, 5)$, and $(1, 5)$ oriented counterclockwise.

Show Solution

Solution

Let ${\bf{F}}(x,y)=\langle{P}(x,y),Q(x,y)\rangle=\langle{x}^2y,y-3\rangle$. Then, $Q_x=0$ and $P_y=x^{2}$. Therefore, $W_x-P_y=-x^{2}$.

Let $D$ be the rectangular region enclosed by $C$ (Figure 4). By Green’s theorem,

$\begin{aligned} \displaystyle\oint_Cx^2ydx+(y-3)dy&=\displaystyle\iint_D(Q_x-P_y)dA \\ &=\displaystyle\int \ \displaystyle\int_D-x^2dA=\displaystyle\int_1^5 \ \displaystyle\int_1^4-x^2dxdy \\ &=\displaystyle\int_1^5-21dy = -84 \end{aligned}$.

Figure 4. The line integral over the boundary of the rectangle can be transformed into a double integral over the rectangle.

Analysis

If we were to evaluate this line integral without using Green’s theorem, we would need to parameterize each side of the rectangle, break the line integral into four separate line integrals, and use the methods from Line Integrals to evaluate each integral. Furthermore, since the vector field here is not conservative, we cannot apply the Fundamental Theorem for Line Integrals. Green’s theorem makes the calculation much simpler.

 

Example: Applying Green’s Theorem to Calculate work

Calculate the work done on a particle by force field

${\bf{F}}(x,y)=\langle{y}+\sin{x},e^y-x\rangle$

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

Show Solution

Let $C$ denote the circle and let $D$ be the disk enclosed by $C$. The work done on the particle is

$W=\displaystyle\oint_C({y}+\sin{x})dx+(e^y-x)dy$.

As with Example “Applying Green’s Theorem over a Rectangle”, this integral can be calculated using tools we have learned, but it is easier to use the double integral given by Green’s theorem (Figure 5).

Let ${\bf{F}}(x,y)=\langle{P}(x,y),Q(x,y)\rangle=\langle{y}+\sin{x},e^y-x\rangle$. Then, $Q_x=-1\text{ and }P_y=1$. Therefore, $Q_x-P_y=-2$.

By Green’s theorem,

$\begin{aligned} W&=\displaystyle\oint_C({y}+\sin{x})dx+(e^y-x)dy \\ &=\displaystyle\iint_D(Q_x-P_y)dA=\displaystyle\iint_D-2dA \\ &=-2(\text{area}(D))=-2\pi(2^2)=-8\pi \end{aligned}$.

Figure 5. The line integral over the boundary circle can be transformed into a double integral over the disk enclosed by the circle.

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

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.

Example: applying green’s theorem over an ellipse

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

Figure 6. Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is denoted by $C$.

Show Solution

Let $C$ denote the ellipse and let $D$ be the region enclosed by $C$. Recall that ellipse $C$ can be parameterized by

$x=a\cos{t},y=b\sin{t},0\leq{t}\leq2\pi$.

Calculating the area of $D$ is equivalent to computing double integral $\displaystyle\iint_DdA$. To calculate this integral without Green’s theorem, we would need to divide $D$ into two regions: the region above the x-axis and the region below. The area of the ellipse is

$\displaystyle\int_{-a}^a\displaystyle\int_0^{\sqrt{b^2-(bx/a)^2}}dydx+\displaystyle\int_{-a}^a\displaystyle\int_{\sqrt{b^2-(bx/a)^2}}^0dydx$.

These two integrals are not straightforward to calculate (although when we know the value of the first integral, we know the value of the second by symmetry). Instead of trying to calculate them, we use Green’s theorem to transform $\displaystyle\iint_DdA$ into a line integral around the boundary $C$.

Consider vector field

${\bf{F}}(x,y)=\langle{P},Q\rangle=\left\langle-\frac{y}2,\frac{x}2\right\rangle$.

Then, $Q_x=\frac12$ and $P_y=-\frac12$, and therefore $Q_x-P_y=1$. Notice that ${\bf{F}}$ was chosen to have the property that $Q_x-P_y=1$. Since this is the case, Green’s theorem transforms the line integral of ${\bf{F}}$ over $C$ into the double integral of 1 over $D$.

By Green’s theorem,

$\begin{aligned} \displaystyle\iint_DdA&=\displaystyle\iint_D(Q_x-P_y)dA \\ &=\displaystyle\int_C{\bf{F}}\bullet{d}{\bf{r}}=\frac12\displaystyle\int_C-ydx+xdy \\ &=\frac12\displaystyle\int_0^{2\pi}-b\sin{t}(-a\sin{t})+a(\cos{t})b\cos{t}dt \\ &=\frac12\displaystyle\int_0^{2\pi}ab\cos^2t+ab\sin^2tdt=\frac12\displaystyle\int_0^{2\pi}abdt=\pi{a}b \end{aligned}$.

Therefore, the area of the ellipse is $\pi{a}b$.

In Example “Applying Green’s Theorem over an Ellipse”, we used vector field ${\bf{F}}(x,y)=\langle{P},Q\rangle=\left\langle-\frac{y}2,\frac{x}2\right\rangle$ 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 $D$. Let $D$ be any region with a boundary that is a simple closed curve $C$ oriented counterclockwise. If ${\bf{F}}(x,y)=\langle{P},Q\rangle=\left\langle-\frac{y}2,\frac{x}2\right\rangle$. Therefore, by the same logic as in Example “Applying Green’s Theorem over an Ellipse”,

area of $D=\displaystyle\iint_DdA=\frac12\displaystyle\oint_C-ydx+xdy$.

It’s worth noting that if ${\bf{F}}=\langle{P},Q\rangle$ is any vector field with $Q_x-P_y=1$, 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.