Chapter 17 - Line Integrals and Green's Theorem

Vector fields

Introduction

A real-valued function $f(x,y)$ may be called a scalar field. The word ‘field’ indicates a mathematical structure that assigns a concrete object of a certain fixed type to every point in space. A function $f$ may be interpreted as assigning a scalar value, the output $f(x,y)$, to each point in space.

A vector field $\mathbf{F}$ assigns a vector value to each point of space. The data of a vector field can be conveyed as a collection of coordinate functions, where each coordinate function takes a point $p$ in space as input:

$$\mathbf{F}(x,y)=(F_1(x,y),F_2(x,y)).$$

(In 3D, the vector field has 3 coordinate functions.) We can also represent the same data as:

$$\mathbf{F}(p)=F_1(p){\mathbf{e}}_{\mathbf{i}}+F_2(p){\mathbf{e}}_{\mathbf{j}}.$$

Vector fields in 3D are harder to visualize:

Vector fields are often used to represent a fluid flow (the vector gives flow velocity) or a force field (vector gives force). Both types of data require the assignment of a vector to each point in space.

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Gradient

Gradient

Given a function $f(x,y)$, a natural vector field may be formed using its partial derivative functions. This is called the gradient of $f$, and $f$ is called the potential function of this vector field:

$$\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)=(f_x,f_y).$$

In 3D, the gradient of $f(x,y,z)$ has components $\nabla f=(f_x,f_y,f_z)$.

Recall that the partial $f_x$ is large when $f$ increases steeply in the $+x$ direction, and it is large and negative when $f_x$ increases steeply in the $-x$ direction. These facts generalize with the combined components: $\nabla f$ points in the direction of steepest increase of $f$. This can be understood by recalculating the formula for $D_{\mathbf{u}}(f)$:

Gradient and directional derivative

$$D_{\mathbf{u}}(f)=f_x\cdot h+f_y\cdot k=\nabla f\cdot \mathbf{u}$$

If we consider unit vectors $\mathbf{u}$ pointing in various directions, we see that the greatest value of $D_{\mathbf{u}}(f)$ occurs when $\mathbf{u}$ aligns with $\nabla f$. This means that $\nabla f$ points in the direction in which $D_{\mathbf{u}}(f)$ is largest, i.e. the direction of steepest incline.

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Exercise

01

Compute gradient

For each of the following functions, compute the gradient:

• (a) $f(x,y)=x^2-3xy$
• (b) $f(x,y)=\cos (y-x)$
• (c) $g(x,y,z,w)=xz{e}^{yw}$.

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Exercise

02

Derivative using gradient

Let $g(x,y,z)=z^2-x{y}^2$. Find $D_{\mathbf{v}}(f)$ at the point $(2,1,3)$, where $\mathbf{v}=(-1,2,2)$, by first computing the gradient and evaluating a dot product.

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Gradient (part 2)

The level curves of a function $f(x,y)$ are the sets of points satisfying $f(x,y)=c$ for various $c\in \mathbb{R}$. The gradient vector is perpendicular to the level curves:

Proposition: Gradient $\perp$ level curves

The level curves to $f$ are perpendicular to the gradient vector.

Proof: Let $\mathbf{r}(t)$ parametrize the level curve $f(x,y)=c$, so $f(\mathbf{r}(t))=c$ for all $t$. Then:

$$0=\frac{d}{dt}c=\frac{d}{dt}f(\mathbf{r}(t))=\nabla f\cdot {\mathbf{r}}^{\prime }(t),$$

so $\nabla f\perp {\mathbf{r}}^{\prime }(t)$.

This result can be generalized to surfaces. The level surfaces of a function $f$ are given by $f(x,y,z)=k$. Any curve $\mathbf{r}(t)$ in such a level surface satisfies $f(\mathbf{r}(t))=k$, and therefore $0=\frac{d}{dt}f(\mathbf{r}(t))=\nabla f\cdot {\mathbf{r}}^{\prime }(t)$. This means the tangent plane to a level surface $f(x,y,z)=k$ has normal vector given by $\mathbf{n}=\nabla f$ at any given point.

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Example

Tangent plane to a hyperboloid

Tangent plane to a hyperboloid

Problem: Find the tangent plane at $(2,1,3)$ to the one-sheeted hyperboloid given by $4x^2+9y^2-z^2=16$.

Solution: Consider the function $f(x,y,z)=4x^2+9y^2-z^2$, so the hyperboloid is the level curve $f(x,y,z)=16$. The normal vector is then given by the gradient $\nabla f=(8x,18y,-2z)$. At $(2,1,3)$ this takes the value $(16,18,-6)$. The tangent plane is perpendicular to this vector and passes through $(2,1,3)$:

$$\nabla f\cdot ((x,y,z)-(2,1,3))=0,16(x-2)+18(y-1)-6(z-3)=0.$$

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Exercise

03

Direction with given slope

You are hiking on a mountain with altitude in meters given by the function:

$$f(x,y)=2500+100(x+y^2)e^{-0.3y^2}.$$

You are at the point $(-1,-1)$. Which two directions could you head (as an angle $\theta$ from the $+x$-axis) in order to ascend at a 20% grade?

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Gradient (part 3)

If you follow the gradient vectors, you will go up or down a path of steepest ascent. This is also the path that crosses levels curves most quickly.

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Line integrals II: vector fields, aligned

Introduction

The aligned vector line integral is the integral of a vector field over a curve according to its alignment with the curve. It measures how much the vector field flows along the curve. It does not depend on the chosen parametrization (i.e. on the speed), but its sign depends on the direction traveled along the curve. It is given for a parametric curve $\mathbf{r}(t)$ that traverses its image $\mathcal{C}$ one time, together with a vector field $\mathbf{F}$, by the formula:

$${\int }_{\mathcal{C}}\mathbf{F}\cdot d\mathbf{s}={\int }_{\mathcal{C}}\mathbf{F}(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)dt.$$

Here we use the infinitesimal vector $d\mathbf{s}={\mathbf{r}}^{\prime }(t)dt$. This is the tangent vector scaled by $dt$; alternatively it is the vector of infinitesimal travel of the particle during $dt$.

The same line integral is sometimes written with a different notation:

$${\int }_{\mathcal{C}}\mathbf{F}\cdot d\mathbf{s}={\int }_{\mathcal{C}}{F}_{1}dx+F_{2}dy+F_{3}dz.$$

This notation is produced by writing $d\mathbf{s}=(dx,dy,dz)$ and expanding the dot product.

To evaluate a line integral in this notation, first write a parametrization $\mathbf{r}(t)=(x(t),y(t),z(t))$ that traverses $\mathcal{C}$ exactly once, then substitute $x^{\prime }(t)dt$ for $dx$, and $y^{\prime }(t)dt$ for $dy$, and $z^{\prime }(t)dt$ for $dz$, then factor out the $dt$ and evaluate the integral in $t$.

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Exercise

01

Vector line integral over a curve

Compute the line integral ${\int }_{\mathcal{C}}\mathbf{F}\cdot d\mathbf{s}$, where $\mathbf{F}=(x+y^2,xz,y+z)$ and $\mathcal{C}$ is traversed by the parametric curve $\mathbf{r}(t)=(t^2,t^3,-2t)$, $t\in [0,2]$.

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Example

Vector line integral

Vector line integral

Problem: Compute ${\int }_{\mathcal{C}}2ydx-3dy$, where $\mathcal{C}$ is an ellipse parametrized by $\mathbf{r}(\theta )=(5+4\cos \theta ,3+2\sin \theta )$.

Solution: First calculate that $x^{\prime }(t)=-4\sin \theta$ and $y^{\prime }(t)=2\cos \theta$. Plug everything into the integrand to obtain:

$$(5+4\cos \theta ,3+2\cos \theta )=(2(3+2\sin \theta )(-4\sin \theta )-3(2\cos \theta ))d\theta .$$

Simplify this and evaluate:

$${\int }_0^{2\pi }(2(3+2\sin \theta )(-4\sin \theta )-3(2\cos \theta ))d\theta =-{\int }_0^{2\pi }24\sin \theta +16{\sin }^2\theta +6\cos \theta =-16\pi .$$

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Exercise

02

Vector line integral over a segment

Compute the line integral ${\int }_{\mathcal{C}}\frac{-ydx+xdy}{x^2+y^2}$, where $\mathcal{C}$ is the line segment from $(1,0)$ to $(0,1)$.

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Curl

Introduction

In what follows we study a 2D version of the Fundamental Theorem of Calculus called Green’s Theorem. The 1D integral ${\int }_a^b{F}^{\prime }(x)dx$ over the $x$-axis is upgraded to a double integral ${\iint }_{\mathcal{D}}\nabla {\times }_z\mathbf{F}\cdot d\mathbf{s}$ over a 2D region $\mathcal{D}$. The derivative $F^{\prime }$ is replaced by a new kind of derivative $\nabla {\times }_z\mathbf{F}$ called ‘curl’. The difference $F(b)-F(a)$ taken on the boundary of the 1D interval $[a,b]$ is upgraded to the line integral ${\oint }_{\mathcal{C}}\mathbf{F}\cdot d\mathbf{s}$ taken around the boundary $\mathcal{C}$ of the 2D region $\mathcal{D}$.

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Curl as derivative operator

Curl as derivative operator

Start with a vector field $\mathbf{F}$ with vector components $\mathbf{F}=(F_1,F_2)$, where $F_1(x,y)$ and $F_2(x,y)$ are functions of points in space.

From this $\mathbf{F}$ we create a scalar field called the scalar curl of $\mathbf{F}$, written $\nabla {\times }_z\mathbf{F}$. This field is calculated with the formula:

$$\nabla {\times }_z\mathbf{F}=\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}.$$

One way to remember this formula is to create a vector-derivative-operator with components $\nabla =(\frac{\partial }{\partial x},\frac{\partial }{\partial y})$, and then take the cross product “$\nabla \times \mathbf{F}$”, applying $\frac{\partial }{\partial x}$ and $\frac{\partial }{\partial y}$ to the function components of $\mathbf{F}$. (Interpret $\nabla$ and $\mathbf{F}$ as 3D vectors with zero $z$-component.) The result is a vector with only a $z$-component. The $z$ subscript indicates that we are just taking the scalar value of that $z$-component.

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Example

Computing scalar curl

Computing scalar curl

Suppose $\mathbf{F}=(xy,e^x)$. Then $\nabla {\times }_z\mathbf{F}=e^x-x$, considered as a function of $(x,y,z)$.

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Exercise

01

Computing scalar curl

Compute $\nabla {\times }_z\mathbf{F}$, where $\mathbf{F}=(e^{x+y},e^{x-y})$.

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Exercise

02

Computing scalar curl

Compute $\nabla {\times }_z\mathbf{F}$, where $\mathbf{F}=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$.

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Exercise

03

Understanding curls

• (a) Draw the vector field $\mathbf{F}=(-y,x)$.
• (b) Compute the curl $\nabla {\times }_z\mathbf{F}$, where $\mathbf{F}=(-y,x)$.
• (b) Draw the vector field from 12A-02.

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Exercise

04

Line integral and vector field

What is the vector field being integrated in this line integral?

$$\oint x^{3}dy-e^{y}dx$$

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Curl as measure of circulation

Curl as measure of circulation

The scalar curl field can be explained as a measure of circulation of the vector field around each point in space.

It is good to understand how this works. Below we calculate the circulation of $\mathbf{F}$ around a small box. In a Problem, you calculate the circulation of $\mathbf{F}$ around a small circular loop.

Consider a small box $\mathcal{C}$ with bottom left corner at a point $\mathbf{p}=(a,b)$ in the plane, and sides of length $\Delta x=h$ and $\Delta y=k$. Label the sides with subscripts for left, right, top, bottom as in the picture:

The ‘circulation’ we wish to calculate is the aligned vector line integral around this box:

$${\oint }_{\mathcal{C}}\mathbf{F}\cdot d\mathbf{s}.$$

Break this into four integrals, one over each side of the box. Define parametric curves traversing each side, with $t\in [0,1]$ for each case:

• For side $C_B$, take ${\mathbf{r}}_B(t)=\mathbf{p}+th{\mathbf{e}}_{\mathbf{i}}$.
  • Observe that ${\mathbf{r}}_B^{\prime }(t)=h{\mathbf{e}}_{\mathbf{i}}$ and $d\mathbf{s}=(h,0)dt$.
• For side $C_R$, take ${\mathbf{r}}_R(t)=\mathbf{p}+h{\mathbf{e}}_{\mathbf{i}}+tk{\mathbf{e}}_{\mathbf{j}}$.
  • Observe that ${\mathbf{r}}_R^{\prime }(t)=k{\mathbf{e}}_{\mathbf{j}}$ and $d\mathbf{s}=(0,k)dt$.
• For side $C_T$, take ${\mathbf{r}}_T(t)=\mathbf{p}+(1-t)h{\mathbf{e}}_{\mathbf{i}}+k{\mathbf{e}}_{\mathbf{j}}$.
  • Observe that ${\mathbf{r}}_T^{\prime }(t)=-h{\mathbf{e}}_{\mathbf{i}}$ and $d\mathbf{s}=(-h,0)dt$.
• For side $C_L$, take ${\mathbf{r}}_L(t)=\mathbf{p}+k(1-t){\mathbf{e}}_{\mathbf{j}}$.
  • Observe that ${\mathbf{r}}_L^{\prime }(t)=-k{\mathbf{e}}_{\mathbf{j}}$ and $d\mathbf{s}=(0,-k)dt$.

Notice that $\mathbf{F}\cdot d\mathbf{s}=F_1({\mathbf{r}}_B(t))hdt$ on $C_B$, and the others simplify in a similar way because of the zeros in the $d\mathbf{s}$ vectors.

We are considering small boxes $\mathcal{C}$, and we will in fact take a limit as the sides $h,k$ go to zero. It is appropriate to use the linear approximation for each component $F_1$ and $F_2$ on the box, with differentials from the point $\mathbf{p}$. This lets us write the circulation integral in terms of $\frac{\partial F_i}{\partial x}$ and $\frac{\partial F_i}{\partial y}$, evaluated at the point $\mathbf{p}$; these appear in the curl formula we are trying to explain. Write $(q_1,q_2)$ for the components of $\mathbf{F}(\mathbf{p})$.

• For side $C_B$, we have $F_1({\mathbf{r}}_B(t))\approx q_1+\frac{\partial F_1}{\partial x}th$.
• For side $C_R$, we have $F_2({\mathbf{r}}_R(t))\approx q_2+\frac{\partial F_2}{\partial x}h+\frac{\partial F_2}{\partial y}tk$.
• For side $C_T$, we have $F_1({\mathbf{r}}_T(t))\approx q_1+\frac{\partial F_1}{\partial x}(1-t)h+\frac{\partial F_1}{\partial y}k$.
• For side $C_L$, we have $F_2({\mathbf{r}}_L(t))\approx q_2+\frac{\partial F_2}{\partial y}(1-t)k$.

Now plug everything in and compute four integrals in $t$:

$$\begin{array}{rl}{\oint }_{\mathcal{C}}\mathbf{F}\cdot d\mathbf{s}& \approx {\int }_0^1\left(q_1+\frac{\partial F_1}{\partial x}th\right)hdt\\ & +{\int }_0^1\left(q_2+\frac{\partial F_2}{\partial x}h+\frac{\partial F_2}{\partial y}tk\right)kdt\\ & +{\int }_0^1\left(q_1+\frac{\partial F_1}{\partial x}(1-t)h+\frac{\partial F_1}{\partial y}k\right)(-h)dt\\ & +{\int }_0^1\left(q_2+\frac{\partial F_2}{\partial y}(1-t)k\right)(-k)dt.\end{array}$$

Simplify this by cancelling some terms and combining some others:

$$\begin{array}{rl}{\oint }_{\mathcal{C}}\mathbf{F}\cdot d\mathbf{s}& \approx {\int }_0^1{h}^2\frac{\partial F_1}{\partial x}(2t-1)+k^2\frac{\partial F_2}{\partial y}(2t-1)+hk\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dt\\ & =hk\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)=hk(\nabla {\times }_z\mathbf{F}).\end{array}$$

Notice that $hk=\Delta x\Delta y$ is the area of the box. The bigger the box, the greater the effect of circulation.

The scalar curl is the limit of the ratio of circulation to the enclosed area, for boxes shrinking to zero:

$$\nabla {\times }_z\mathbf{F}=\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=\underset{h,k\to 0}{\lim }\frac{1}{\text{Area}(\mathcal{C})}{\oint }_{\mathcal{C}}\mathbf{F}\cdot d\mathbf{s}.$$Link to original

Exercise

05

Circulation calculated around circles

Repeat the reasoning above but going around a small circle instead of a box.
You can use the single parametrization $\mathbf{r}(t)=(a+r\cos (t),b+r\sin (t))$ for $r$ a small constant, and $(a,b)$ the center point of the circle.

The derivation is faster with this method, but circles do not tile the plane the way boxes do, and we need a tiling of the plane to prove Green’s Theorem.

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Green’s Theorem

Introduction

The main theorem of 2D calculus is called Green’s Theorem, but it could be called more expressively “The Fundamental Theorem of 2D Calculus”. It says that the integral of curl over a region is equal to the total circulation on the boundary:

Green’s Theorem

When $\mathcal{D}$ is a region of the $xy$-plane with boundary $\partial \mathcal{D}$, then:

$${\iint }_{\mathcal{D}}\nabla {\times }_z\mathbf{F}dA={\oint }_{\partial \mathcal{D}}\mathbf{F}\cdot d\mathbf{s}$$

Writing this in terms of component functions, we have:

$${\iint }_{\mathcal{D}}\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dxdy={\oint }_{\partial \mathcal{D}}{F}_{1}dx+F_{2}dy.$$ Link to original

Example

Area of a region calculated from the boundary

Area of a region calculated from the boundary

Set $\mathbf{F}=(0,x)$. Then $\frac{\partial F_2}{\partial x}=1$ while $\frac{\partial F_1}{\partial y}=0$, so the LHS of Green’s Theorem reduces to the enclosed area. We then have:

$$\text{Area}(\mathcal{D})={\oint }_{\mathcal{C}}xdy.$$

For example, an ellipse is given by $\mathbf{r}(t)=(a\cos (t),b\sin (t))$. Note $dy=b\cos (t)dt$. The area of an ellipse is therefore:

$$\text{Area}(\text{ellipse})={\int }_0^{2\pi }a\cos (t)b\cos (t)dt=\pi ab.$$

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Example

Faster calculation of line integral using Green’s Theorem

Faster calculation of line integral using Green’s Theorem

Problem: Find ${\oint }_{\mathcal{C}}{y}^{2}dx+3xydy$, where $\mathcal{C}$ is given as in the picture:

Solution: The interior of the region is covered by one double integral in polar coordinates, and that is easier to compute than four line integrals, so we use Green’s Theorem to convert to a double integral. We choose the vector field $\mathbf{F}=(y^2,3xy)$ so that $\mathbf{F}\cdot d\mathbf{s}$ gives the line element. Then $\nabla {\times }_z\mathbf{F}=\frac{\partial }{\partial x}(3xy)-\frac{\partial }{\partial y}{y}^2=y$, and we compute:

$$\begin{array}{rl}{\oint }_{\mathcal{C}}{y}^{2}dx+3xydy& ={\iint }_{\mathcal{D}}\nabla {\times }_z\mathbf{F}d\mathbf{s}={\iint }_{\mathcal{D}}ydA\\ & ={\int }_0^{\pi }{\int }_1^{2}r\sin (\theta )rdrd\theta =\frac{14}{3}.\end{array}$$

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Exercise

01

Faster calculation of line integral using Green’s Theorem

Compute ${\oint }_{\mathcal{T}}{e}^{2x+y}dx+e^{-y}dy$, where $\mathcal{T}$ is the triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$ (traversed in order).

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Understanding Green’s Theorem

Understanding Green’s Theorem

To understand the reasoning behind Green’s Theorem, it is important to remember why the 1D version of the Fundamental Theorem of Calculus is true.

FTC derivation reminder: telescoping summation

Let $F^{\prime }(x)=f(x)$. We would like to see why ${\int }_a^{b}f(x)dx=F(b)-F(a)$.

If $a$ and $b$ are very close, we could write $F(b)=F(a)+f(a)\Delta x+\Delta x\epsilon (x)$. Ignoring the error term, we see that $F(b)-F(a)\approx f(a)\Delta x$, and the RHS is the area of a box with base $[a,b]$ and height $f(a)$. This is $\approx {\int }_a^{b}fdx$ when $\Delta x$ is very small.

If $a$ and $b$ are farther apart, divide $F(b)-F(a)$ into intervals:

$$\begin{array}{rl}F(b)-F(a)& =F(b)-F(x_{n-1})+F(x_{n-1})-F(x_{n-2})+\cdots +F(x_2)-F(a)\\ & =\sum _{i=1}^{n}F(x_{i+1})-F(x_i).\end{array}$$

Assume for simplicity that $x_{i+1}-x_i=\Delta x$ for every $i$. Now analyze each small difference using a linear approximation:

$$F(x_{i+1})=F(x_i)+f(x_i)\Delta x+\Delta x{\epsilon }_i(x).$$

Subtracting $F(x_i)$ from both sides and plugging into the summation, obtain:

$$F(b)-F(a)=\sum _{i=1}^{n}f(x_i)\Delta x+\sum _{i=1}^n\Delta x{\epsilon }_i(x).$$

If we ignore the error term and take the limit as $n\to \infty$ with $\Delta x\to 0$, then the first summation (a Riemann sum) becomes ${\int }_a^{b}fdx$.

To account for the error term properly, take $\Delta x$ small enough that $|{\epsilon }_i(x)|<\Delta xM$ for some fixed positive number $M$. Then as $n\to \infty$ the error summation limits to zero:

$$\left|\sum _{i=1}^n\Delta x{\epsilon }_i(x)\right|\le \sum _{i=1}^n\Delta x^{2}M=n\Delta x^{2}M=(b-a)\Delta xM\to 0.$$

To summarize, the key idea is that we add up all the small differences $F(x_{i+1})-F(x_i)$. In the sum, internal cancellation occurs, leaving just the boundary difference $F(b)-F(a)$. On the other hand, each small difference is approximated by $f(x_i)\Delta x$. (The linear approximation of $F$ using $F^{\prime }=f$.) The sum of these terms $f(x_i)\Delta x$ is exactly the Riemann sum for the integral ${\int }_a^{b}fdx$.

Even more briefly, the integral adds up many microscopic differences in $F$ which are written in the form $f(x_i)\Delta x$.

Green’s Theorem can be explained with exactly the same process, except that instead of differences, we have circulations, and instead of $f(x_i)\Delta x$ that approximates a small difference, we have $\nabla {\times }_z\mathbf{F}dA$ that approximates a small circulation. To add up circulations, we collect 2D squares, but there is still internal cancellation leaving only the terms on the boundary curve of the region.

Derivation of Green’s Theorem

We need to divide the region $\mathcal{D}$ into microscopic squares. Let each square be identified by its bottom-left corner at point $(x_i,y_j)$, where $i,j\in 1,\dots ,n$. Each square has width and height $\Delta x=\Delta y=h$. Let ${\mathcal{I}}_n$ denote the set of squares fully contained inside $\mathcal{D}$ (specifically, the set of ordered pairs $(i,j)$ identifying the bottom-left corner of a square contained in $\mathcal{D}$).

The circulation around the square at $(x_i,y_j)$ is approximated by $h^2(\nabla {\times }_z\mathbf{F})(x_i,y_j)$. The error terms need to be collected as well; the exact circulation differs by an error term with magnitude less than $h^2\epsilon (h)$, where $\epsilon (h)$ is a positive function of $h$ satisfying $\epsilon (h)\le Mh$, for some $M$, as $h\to 0$. (Notice that this $\epsilon (h)$ can be chosen so it doesn’t depend on $(i,j)$ by taking the biggest one of those used for any of the squares in ${\mathcal{I}}_n$.)

Now we add up all the circulations. The internal sides of boxes, which are shared between two boxes, will have a component of the integral traversed in each direction. These traversals contribute the same number with opposite signs, and therefore cancel. So, the total sum of circulations is equal to the circulation around the outer edges of the shape made by all squares of width $h$ which fit entirely in the domain. This is a 2D version of the cancellation in 1D that occurs between $F(x_{i+1})-F(x_i)$ and $F(x_i)-F(x_{i-1})$, for example.

Considering now the sum of approximations, we have the sum of the values $h^2(\nabla {\times }_z\mathbf{F})(x_i,y_j)$ multiplied by the area of the squares, $h^2$. This is a Riemann sum that limits to ${\int }_{\mathcal{D}}\nabla {\times }_z\mathbf{F}dA$ as $h\to 0$ and the number of boxes $n^2$ goes to $\infty$. So the circulations add up to the boundary circulation around the squares contained in $\mathcal{D}$, and the sum of approximations using $\nabla {\times }_z\mathbf{F}$ adds up to the integral of $\nabla {\times }_z\mathbf{F}$ over the region.

Now let’s address the error term. Each small square has an error less than $h^2\epsilon (h)$, which is less than $M{h}^3$ for some positive number $M$ based on the guarantees on the behavior of $\epsilon$. The number of boxes is $n^2$. Let’s say the total area of the region is $A$. The area covered by the boxes is $n^2{h}^2$ so $n^2{h}^2\le A$, and thus $n^2\le \frac{A}{h^2}$. The total error with all $n^2$ boxes is less than $M{h}^3{n}^2\le MhA$. But $MhA\to 0$ as $h\to 0$. So the total error goes to zero.

Going deeper

One remaining issue should be addressed. Does the boundary converge to the boundary $\partial \mathcal{D}=\mathcal{C}$ of $\mathcal{D}$? Or, at least, does the circulation on the boundary of the collection of squares converge to the circulation on $\mathcal{C}$?

This question is difficult to address using the concepts in this course. Instead of addressing it fully, we just explain how the circulation on a squared boundary can approximate the correct circulation on an angled boundary. Strictly speaking, this kind of explanation will then work for polygons, but by considering linear approximations of boundaries, the explanation is also good for smooth boundaries that have first derivatives.

Consider an angled line segment $\mathcal{C}$ parametrized by $\mathbf{r}(t)=(t,2t)$ for $t\in [0,1]$. Approximate this by rectangular stair steps, two vertical for every one horizontal, of width $h$. Consider the bottom step $C_1$, with horizontal piece $C_{H_1}$ and vertical piece $C_{V_1}$. The line integral over this step, horizontal and vertical, is:

$$\begin{array}{rl}{\int }_{C_1}{F}_{1}dx+F_{2}dy& ={\int }_{C_{H_1}}{F}_{1}dx+F_{2}dy+{\int }_{C_{V_1}}{F}_{1}dx+F_{2}dy\\ & ={\int }_0^h{F}_1(t,0)dt+{\int }_0^h{F}_2(h,2t)2dt\\ & ={\int }_0^h{F}_1(t,0)+2F_2(h,2t)dt,\end{array}$$

whereas the line integral over the corresponding segment of $\mathcal{C}$, written ${\mathcal{C}}_1$, is:

$${\int }_{{\mathcal{C}}_1}{F}_{1}dx+F_{2}dy={\int }_0^h{F}_1(t,2t)+2F_2(t,2t)dt.$$

If we assume $h$ is small enough that $\mathbf{F}=(F_1,F_2)$ is (nearly) constant on the square containing ${\mathcal{C}}_h$ and $C_1$, then these two values are nearly equal. One can then take the sums over all $C_i=C_{H_1}\cup C_{V_1}$ on one hand, and ${\mathcal{C}}_i$ on the other hand, and track errors as before, and find that as $h\to 0$ they converge to the same thing.

Interior boundary curves

Some shapes have boundary lines on the inside. Do not forget to include these lines as part of the total boundary in Green’s Theorem! The orientation of these lines should be consistent with counter-clockwise circulation around infinitesimal boxes near the boundary. In the figure below, ${\mathcal{C}}_3$ and ${\mathcal{C}}_4$ are oriented correctly, while ${\mathcal{C}}_5$ has the wrong orientation. The boundary $\partial {\mathcal{D}}_2$ is therefore ${\mathcal{C}}_3+{\mathcal{C}}_4-{\mathcal{C}}_5$.

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Path dependence

Introduction

Green’s Theorem expresses a relationship between circulation around the boundary and infinitesimal circulation in the interior. (Infinitesimal circulation is the curl times $dA$). It says that adding up all the infinitesimal circulations gives the boundary circulation.

This relationship, considered from another angle, is connected to the idea of path independence. Consider two particle paths travelling between the same start and end points. Write $C_1$ for the first path, parametrized by ${\mathbf{r}}_1(t)$, and $C_2$ for the second path, parametrized by ${\mathbf{r}}_2$. Write $P_1={\mathbf{r}}_1(0)={\mathbf{r}}_2(0)$ for the starting point, and $P_2={\mathbf{r}}_1(1)={\mathbf{r}}_2(1)$ for the ending point.

Now consider the question: do the line integrals ${\int }_{C_1}\mathbf{F}\cdot d\mathbf{s}$ and ${\int }_{C_2}\mathbf{F}\cdot d\mathbf{s}$ have the same value for these two paths?

The vector field $\mathbf{F}$ is said to be conservative or path independent if the answer is yes, and path dependent or non-conservative if the answer is no.

Consider the question more quantitatively. What is the value of the difference ${\int }_{C_2}\mathbf{F}\cdot d\mathbf{s}-{\int }_{C_1}\mathbf{F}\cdot d\mathbf{s}$? In fact, this difference is the same as the sum of line integrals around a closed loop starting and ending at $P_1$. The complete path is written as $C_2\cup -C_1$:

(Here the path $-C_1$ is the same path as $C_1$ but traversed in the opposite direction.)

By Green’s Theorem, the integral of $\nabla {\times }_z\mathbf{F}$ over the region between the two paths gives the difference ${\int }_{C_2}\mathbf{F}\cdot d\mathbf{s}-{\int }_{C_1}\mathbf{F}\cdot d\mathbf{s}$. In particular, if $\nabla {\times }_z\mathbf{F}=0$ (in some region), then by Green’s Theorem the line integral from $P_1$ to $P_2$ does not depend at all on the path taken! (Provided the path remains in the region where $\nabla {\times }_z\mathbf{F}=0$.)

Path independence is also connected to the gradient:

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Exercise

01

Gradient fields are conservative

Suppose $\mathbf{F}=\nabla f$ for some scalar field $f$. Show that $\nabla {\times }_z\mathbf{F}=0$.

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Challenge problem

02

(not for presentation)

Conservative fields are gradients

Suppose that $\nabla {\times }_z\mathbf{F}=0$. Find a scalar field $f$ such that $\mathbf{F}=\nabla f$. (Hint: define it using line integrals along paths!)

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Introduction (part 2)

Simply connected regions

Path independence is equivalent to $\nabla {\times }_z\mathbf{F}=0$ only for simply connected domains. These are domains without any holes, meaning that any loop can be tightened down to a point in the domain.

If there are holes, then it is possible to construct a field $\mathbf{F}$ in the part of the domain outside the hole, such that $\nabla {\times }_z\mathbf{F}=0$ there, but path independence is no longer strictly valid. Paths that wrap around the missing hole can have different line integrals. In fact, the line integral measures the number of times a path wraps around the hole. Two paths that wrap the same number of times differ by some enclosed region outside the hole where $\nabla {\times }_z\mathbf{F}=0$, so the line integrals on these paths must agree!

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Exercise

03

Gradient of angle

Let $f(x,y)=\theta ={\tan }^{-1}(\frac{y}{x})$. Show that:

$$\mathbf{F}=\nabla f=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right).$$

We saw above that $\nabla {\times }_z\mathbf{F}=0$ for $(x,y)\ne (0,0)$. Compute the integral ${\int }_{{\mathcal{C}}_n}\mathbf{F}\cdot d\mathbf{s}$, where ${\mathcal{C}}_n$ is a path that winds $n$-times around a circle of radius $R$. (For example, $\mathbf{r}(t)=(R\cos (t),R\sin (t))$ with $t\in [0,2\pi n]$.)

Notice that $\mathbf{F}$ is not defined at $(x,y)$. So the applicable domain for Green’s Theorem has a ‘hole’ containing $(0,0)$.

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Fundamental Theorem of Line Integrals

Introduction

The following conditions for a vector field $\mathbf{F}(x,y)$ are equivalent:

• $\int \mathbf{F}\cdot d\mathbf{s}$ is path independent,
• $\nabla {\times }_z\mathbf{F}=0$ in the region (which is simply connected, i.e. no holes),
• $\mathbf{F}=\nabla f$ for some scalar field $f(x,y)$.

When these conditions hold, $\mathbf{F}$ is called conservative. Then ${\int }_{C_1}\mathbf{F}\cdot d\mathbf{s}={\int }_{C_2}\mathbf{F}\cdot d\mathbf{s}$ for two paths $C_1$ and $C_2$ from $\mathbf{p}$ to $\mathbf{q}$, and $\mathbf{F}=\nabla f$ for a scalar field $f(x,y)$.

(Note: when the domain is not simply connected, then the curl condition is not strong enough, but the first and last conditions are still equivalent to each other and the word ‘conservative’ applies when these two conditions hold.)

The Fundamental Theorem of Line Integrals gives the value that is common to these line integrals over various paths:

FT of Line Integrals

For $\mathcal{C}$ any path from point $\mathbf{p}$ to point $\mathbf{q}$, we have:

$${\int }_{\mathcal{C}}\nabla f\cdot d\mathbf{s}=f(\mathbf{q})-f(\mathbf{p}).$$

If $f(x,y)$ gives the elevation on a mountain at point $(x,y)$, then $\nabla f$ gives the rate of incline in the direction of steepest ascent. The line integral of $f$ along any hiking path $\mathcal{C}$ then gives the total change in elevation.

Derivation of FT of Line Integrals

Take any parametrization $\mathbf{r}(t)$ of $\mathcal{C}$ given for $t\in [a,b]$. Now $\nabla f\cdot d\mathbf{s}=\nabla f\cdot {\mathbf{r}}^{\prime }(t)dt$. By the chain rule for paths (in reverse), this is equal to $\frac{d}{dt}f(\mathbf{r}(t))$. So can apply the ordinary FTC:

$$\begin{array}{rl}{\int }_{\mathcal{C}}\nabla f\cdot d\mathbf{s}& ={\int }_a^b(\nabla f)(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)dt\\ & ={\int }_a^b\frac{d}{dt}f(\mathbf{r}(t))dt\\ & =f(\mathbf{r}(b))-f(\mathbf{r}(a))\\ & =f(\mathbf{q})-f(\mathbf{p}).\end{array}$$

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Exercise

01

FT of Line Integrals: elevation gain

Consider the vector field $\mathbf{F}=(2x+y,x)$. Choose any path between $(0,0)$ and $(2,3)$. Verify the FT for line integrals by computing the elevation gain on your chosen path in two ways (computing both sides of the FT). For the right side, you will need to first find a field $f(x,y)$ such that $\nabla f=\mathbf{F}$.

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