# Math2111: Chapter 3: Line integrals. Section 4: Green’s theorem

In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

We now establish another generalisation of the fundamental theorem of calculus which is known as Green’s theorem (named after George Green).

In Section 3 we established the Fundamental theorem of line integrals which gives a relationship between line integrals of conservative vector fields and the values of the potential function at the endpoints of the curve. Green’s theorem on the other hand gives a relationship between double integrals and line integrals in $\mathbb{R}^2.$

Green’s theorem in normal form

Consider a simple closed curve $\mathcal{C}$ in $\mathbb{R}^2$, that is we assume that the starting point coincides with the end point of the curve and that the curve does not intersect itself (see also the Jordan curve theorem). Let $\boldsymbol{F}:\mathbb{R}^2\to\mathbb{R}^2$ be continuously differentiable in the region ${}R$ whose boundary is $\mathcal{C}$. If we interpret $\boldsymbol{F}$ as the velocity field of a fluid, then from the interpretation of line integrals in normal form we saw that $\int_{\mathcal{C}} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{s}$ can be understood as measuring the amount of fluid leaving the region ${}R$. On the other hand, divergence can be interpreted as being the amount of fluid leaving a small square around a point per unit area. Hence the double integral over the region ${}R$ of the divergence is the amount of fluid leaving the area ${}R$. In this interpretation, Green’s theorem states that

$\displaystyle \begin{array}{c} \mbox{the amount of fluid crossing the boundary } \mathcal{C} \\ = \\ \mbox{the amound of fluid leaving the area } {}R.\end{array}$

In mathematical terms, this is stated in the following theorem.

Green’s theorem in normal form
Let $\mathcal{C}$ be a simple closed curve which can be parameterised by a piecewise continuously differentiable function and let ${}R$ be the region in $\mathbb{R}^2$ enclosed by $\mathcal{C}$. Let $\boldsymbol{F}:R\to\mathbb{R}^2$ be continuously differentiable. Then

$\displaystyle \int_{\mathcal{C}} \boldsymbol{F} \cdot \widehat{\boldsymbol{n}} \, \mathrm{d} s = \iint_{R} \mathrm{div} \,\boldsymbol{F} \,\mathrm{d} A,$

where $\widehat{\boldsymbol{n}}$ is the normal vector pointing outwards.

If $\boldsymbol{F}= F_1 \widehat{\boldsymbol{i}} + F_2 \widehat{\boldsymbol{j}}$. Assume that $\mathcal{C}$ is oriented counterclockwise (or oriented positively). Then the tangent vector to $\mathcal{C}$ is given by $\boldsymbol{T}= \mathrm{d} x \widehat{\boldsymbol{i}} + \,\mathrm{d} y \widehat{\boldsymbol{j}}$ and the normal vector pointing outwards is therefore $\boldsymbol{n}= \mathrm{d} y \widehat{\boldsymbol{i}} - \,\mathrm{d} x \widehat{\boldsymbol{j}}$, which is obtained by turning $\boldsymbol{T}$ by $90^\circ$ clockwise. Then

$\displaystyle \boldsymbol{F} \cdot \widehat{\boldsymbol{n}} \,\mathrm{d}s = F_1 \,\mathrm{d} y - F_2 \,\mathrm{d} x.$

Hence the above formula can also be written as

$\displaystyle \int_{\mathcal{C}} F_1 \,\mathrm{d} y - F_2 \, \mathrm{d}x = \iint_R \left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y}\right) \,\mathrm{d}x \,\mathrm{d}y. \hspace{2cm} (1)$

Green’s theorem in tangential form

In this version of Green’s theorem the line integral is over the tangential component of the vector field rather then the normal component. It can be obtained by a simple substitution. Let $\boldsymbol{G}=G_1 \widehat{\boldsymbol{i}} + G_2 \widehat{\boldsymbol{j}}$ be a continuously differentiable vector field and assume that the simple closed curve $\mathcal{C}$ is oriented counterclockwise. The aim is to find a relationship between the line integral

$\displaystyle \int_{\mathcal{C}} \boldsymbol{G}\cdot \mathrm{d} \boldsymbol{s} = \int_{\mathcal{C}} G_1 \,\mathrm{d} x + G_2 \,\mathrm{d} y$

and some double integral.

We can obtain such a formula from (1) by using the substitution $F_1 = G_2$ and $F_2 = -G_1$. That is, given the vector field $\boldsymbol{G}$, we define a vector field $\boldsymbol{F} = (G_2, - G_1)$ which is obtained by turning $\boldsymbol{G}$ by $90^\circ$ clockwise. Then the line integral in tangent form over $\boldsymbol{G}$ is the same as the line integral in normal form with $\boldsymbol{n}$ pointing outwards over $\boldsymbol{F}$ (since $\mathcal{C}$ is oriented counterclockwise), that is

$\displaystyle \begin{array}{rcl} \int_{\mathcal{C}} \boldsymbol{G}\cdot \mathrm{d} \boldsymbol{s} & = & \int_{\mathcal{C}} G_1 \,\mathrm{d} x + G_2 \,\mathrm{d} y \\ && \\ & = & \int_{\mathcal{C}} -F_2 \,\mathrm{d} x + F_1 \,\mathrm{d} x \\ && \\ & = & \iint_{R} \left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} \right) \,\mathrm{d} x \,\mathrm{d} y \\ && \\ & = & \iint_{R} \left(\frac{\partial G_2}{\partial x} - \frac{\partial G_1}{\partial y} \right) \,\mathrm{d} x\,\mathrm{d} y. \end{array}$

Green’s theorem in tangent form
Let $\mathcal{C}$ be a simple closed curve oriented positively which can be parameterised by a piecewise continuously differentiable function and let ${}R$ be the region in $\mathbb{R}^2$ enclosed by $\mathcal{C}$. Let $\boldsymbol{G}:R\to\mathbb{R}^2$ be continuously differentiable. Then

$\displaystyle \int_{\mathcal{C}} \boldsymbol{G} \cdot \mathrm{d} \boldsymbol{s} = \iint_{R} \mathrm{scalar} \,\mathrm{curl} \,\boldsymbol{G} \,\mathrm{d} A.$

Let $\boldsymbol{G} = (G_1,G_2)$. As we have seen before, we can write the above formula also as

$\displaystyle \int_{\mathcal{C}} G_1 \,\mathrm{d} x + G_2 \,\mathrm{d} y = \iint_{R} \left(\frac{\partial G_2}{\partial x} - \frac{\partial G_1}{\partial y} \right) \,\mathrm{d} x \,\mathrm{d} y.$

Examples

Example
Verify Green’s theorem in normal and tangent form for the vector field $\boldsymbol{F}= x\widehat{\boldsymbol{i}} + (y-x) \widehat{\boldsymbol{j}}$ and the curve $\boldsymbol{r}(t)= \cos t \widehat{\boldsymbol{i}} + \sin t \widehat{\boldsymbol{j}}$ for $0 \le t \textless 2\pi.$ $\Box$

Example
Let $\mathcal{C}$ be a simple closed curve and let ${}R$ denote the region enclosed by $\mathcal{C}$. Using Green’s theorem, find a line integral along $\mathcal{C}$ which yields the area of ${}R$.

Find the area of the region bounded by the curve $x^{2/3} + y^{2/3} = a^{2/3},$ where $a \textgreater 0.$ $\Box$

Example
Let $\mathcal{C}$ be a circle with radius ${}1$ centered at $(0,0)$ and oriented counterclockwise. Calculate $\int_{\mathcal{C}} \mathrm{e}^{x^2/2} \,\mathrm{d} x + (x + \sin \cos \mathrm{e}^y) \,\mathrm{d} y.$ $\Box$

Exercise
Let

$\displaystyle \boldsymbol{F}(x,y)=\left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right).$

Note that $\boldsymbol{F}=\nabla f$ where $f(x,y)=\arctan(y/x).$

Let $D=\{(x,y): -1\le x \le 1, 1\le y \le 2\}$ and let $\mathcal{C}$ be the boundary of ${}D$ which is oriented counterclockwise. Calculate $\int_{\mathcal{C}} \boldsymbol{F}\cdot \mathrm{d}\boldsymbol{s}.$ $\Box$

For an application of Green’s theorem see here.