Line integrals

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}} \DeclareMathOperator {\proj }{\vec {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

We accumulate vectors along a path.

In this section we introduce a new type of integrals, line integrals also known as path integrals. Let’s start with the definition of a line integral:

Let $\vec {F}:\R ^2\to \R ^2$ be a vector field, $\vec {p}:\R \to \R ^2$ be a smooth vector valued function tracing a curve $C$ exactly once as $t$ runs from $a$ to $b$ , $\begin{align*} \vec{F}(x,y) &= \vector{M(x,y), N(x,y)}\\ \vec{p}(t) &= \vector{x(t),y(t)}. \end{align*}$ A line integral is an integral of the form: $\begin{align*} \int_C \vec{F}\dotp \d \vec{p} &= \int_C \vector{M,N}\dotp\vector{\d x,\d y}\\ &= \int_C M(x,y)\d x + N(x,y)\d y \end{align*}$ Since $\d x = x'(t)\d t$ and $\d y = y'(t)\d t$ , we may write $\begin{align*} &= \int_a^b \left(M(x(t),y(t))\cdot x'(t) + N(x(t),y(t))\cdot y'(t)\right) \d t\\ &= \int_a^b \vector{M(x(t),y(t)),N(x(t),y(t))}\dotp\vector{x'(t), y'(t)} \d t \end{align*}$

If the path $C$ is closed, then sometimes people write a “circle” on the intgeral sign: $\oint _C \vec {F}\dotp \d \vec {p}$ This notation is not critical, but it can sometimes help us from making silly mistakes.

Which of the following are line integrals?

$\int _R (x^2+y^2) \d A$ $\int _C \left ( -y\d x + x\d y \right )$ $\int _3^4\int _2^3 \ln (xy) \d x \d y$ $\int _0^{2\pi } \vector {-\sin (t),\cos (t)}\dotp \vector {-\sin (t),\cos (t)}\d t$

Read on to learn the meaning of this new integral.

What do line integrals measure?

A line integral measures the flow of a vector field along a path. The basic idea is that there is some vector field given by $\vec {F}$ :

Now we add an oriented path $C$ that is parameterized by $\vec {p}(t) = \vector {x(t),y(t)}$ . This can be thought of as a path that an object takes through the field. To reason via a specific example, we’ll add a path below:

To figure out if the flow of the vector field is “with” the direction of the path, we use the dot product: $\underbrace {\vec {F}(x(t),y(t))}_{\text {direction of field}} \dotp \underbrace {\vector {x'(t) \d t,y'(t) \d t}}_{\text {direction of path}}$

When the direction of the field and the direction of the path are in alignment, the dot product is…

positive zero negative

When the direction of the field and the direction of the path are orthogonal, the dot product is…

positive zero negative

When the direction of the field and the direction of the path are in opposite direction, the dot product is…

positive zero negative

Integrating over the path sums these infinitesimal measurements: $\vec {F}(x(t),y(t))\dotp \vector {x'(t),y'(t)} \d t = \vec {F} \dotp \d \vec {p}$ Thus the line integral $\int _C \vec {F}\dotp \d \vec {p}$ measures the flow of a field along a path. In particular, if the value of the line integral is positive, then the flow is with the path; if the value is negative, then the flow is against the path.

Consider the following vector field along with a (directed) curve $C$ .

Do you expect $\int _C \vec {F}\dotp \d \vec {p}$ to be positive, zero, or negative?

positive zero negative

We can think about this better if we break the path into pieces: $C_1$ , $C_2$ , $C_3$ .

We can see that the vectors are flowing with the direction of $C_1$ . Note that the the magnitude of these vectors is large, so this contributes a large positive value to our integral.

The field vectors are orthogonal to the direction of the path $C_2$ . So this part contributes nothing to the integral.

The field vectors are flowing against the direction of $C_3$ . However, their magnitude is much less than the vectors that flowed with $C_1$ . So this contributes a small negative value to our integral.

Consider the following vector field along with a (directed) curve $C$ .

Do you expect $\int _C \vec {F}\dotp \d \vec {p}$ to be positive, zero, or negative?

positive zero negative

We can think about this better if we break the path into pieces: $C_1$ , $C_2$ , $C_3$ .

The field vectors are orthogonal to the direction of the path $C_1$ . So this part contributes nothing to the integral.

The field vectors are flowing against the direction of $C_2$ . This contributes a negative value to our integral.

The field vectors are again orthogonal to the direction of the path $C_3$ . So this part contributes nothing to the integral.

Consider the following vector field along with a (directed) curve $C$ .

Do you expect $\int _C \vec {F}\dotp \d \vec {p}$ to be positive, zero, or negative?

positive zero negative

Think about what the tangent vectors to the parameterized curve look like, and whether they point with the field or against the field.

Consider the following vector field along with a (directed) curve $C$ .

Do you expect $\int _C \vec {F}\dotp \d \vec {p}$ to be positive, zero, or negative?

positive zero negative

Think about what the tangent vectors to the parameterized curve look like, and whether they point with the field or against the field.

Let’s attempt to solve a discrete problem.

Below we have a very simple directed curve $C$ (it’s a line) along with field vectors from a vector field $\vec {F}$ .

Setting $\d x= -2$ , estimate: $\int _C \vec {F}\dotp \d \vec {p}$

If $\vec {F}(x,y) = \vector {M(x,y),N(x,y)}$ , we have that $\int _C \vec {F}\dotp \d \vec {p} = \int _C M(x,y) \d x + N(x,y)\d y.$ We know that $\d x= \answer [given]{-2}$ and $\d y = y'(x) \d x$ Since $C$ is a line of slope $\answer [given]{1/2}$ , $\d y = \answer [given]{-1}.$ Since we want to estimate $\int _C M(x,y) \d x + N(x,y)\d y.$ We will compute: $\sum \left (M(x,y) \d x + N(x,y) \d y\right )$ Write with me $\begin{align*} \answer[given]{0}\cdot (-2) &+ \answer[given]{2} \cdot (-1) \\ &+ (-1) \cdot (-2) + 2 \cdot (-1) \\ &+ (-2)\cdot (-2) + 1 \cdot (-1) \\ &+ (-3)\cdot (-2) + 0\cdot (-1) \\ &+ (-1)\cdot (-2) + (-1)\cdot (-1)\\ &+ (0)\cdot (-2) + (-1)\cdot (-1) \end{align*}$ $= \answer [given]{11}.$

Computations with line integrals

Let $\vec {F}(x,y) = \vector {-y,x}$ and let $C$ be the unit circle centered at the origin. Compute $\oint _C \vec {F}\dotp \d \vec {p}$

The path $C$ can be parameterized by $\begin{align*} x(\theta) &= \cos(\theta)\\ y(\theta) &= \sin(\theta) \end{align*}$ with $0\le \theta \le 2\pi$ . To compute the integral, write with me $\begin{align*} \oint_C \vec{F}\dotp\d\vec{p} &= \int_0^{2\pi} F(x(\theta),y(\theta))\dotp \vector{x'(\theta),y'(\theta)}\d \theta\\ &= \int_0^{2\pi} \vector{\answer[given]{-\sin(\theta)},\answer[given]{\cos(\theta)}}\dotp \vector{\answer[given]{-\sin(\theta)},\answer[given]{\cos(\theta)}}\d \theta\\ &= \int_0^{2\pi}\left(\sin^2(\theta)+\cos^2(\theta)\right)\d \theta\\ &= \int_0^{2\pi} 1\d \theta\\ &=\answer[given]{2\pi}. \end{align*}$

Any smooth path can be approximated with a polygonal path. These can be quite easy to integrate. Check out our next example.

Let $\vec {F}(x,y) = \vector {0,x}$ and let $C$ be the polygonal path below parameterized in a counterclockwise direction:

Compute $\oint _C \vec {F}\dotp \d \vec {p}$

We need to parameterize our paths in a counterclockwise direction. We’ll break it into four line segments each parameterized as $t$ runs from $0$ to $1$ .

Where: $\begin{align*} \vecl_1(t) &= \vector{4t,\answer[given]{0}}\\ \vecl_2(t) &= \vector{\answer[given]{4-t},2t}\\ \vecl_3(t) &= \vector{3-2t,\answer[given]{2}}\\ \vecl_4(t) &= \vector{\answer[given]{1-t},2-2t} \end{align*}$ and each draws the line as $t$ runs from $0$ to $1$ . Write: $\begin{align*} \oint_C \vec{F}\dotp\d\vec{p} = \int_0^1 &\vec{F}(\vecl_1(t))\dotp \vecl_1'(t) \d t \\ &+ \int_0^1 \vec{F}(\vecl_2(t))\dotp \vecl_2'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_3(t))\dotp \vecl_3'(t) \d t\\ &+ \int_0^1 \vec{F}(\vecl_4(t))\dotp \vecl_4'(t) \d t \end{align*}$ For each of the integrands above, say $\vecl (t) = \vector {x(t),y(t)}$ , we will write $\vec {F}(\vecl (t))\dotp \vecl '(t) = \vector {0,x(t)} \dotp \vector {x'(t),y'(t)}$ and combine them into a single integral. Write with me $\begin{align*} \oint_C \vec{F}\dotp\d\vec{p} &= \int_0^1 \left(\answer[given]{2(4-t) -2(1-t)} \right)\d t\\ &= \eval{\answer[given]{6t}}_0^1\\ &=\answer[given]{6}. \end{align*}$

The fundamental theorems of calculus

We will soon see that there are many “Fundamental Theorems of calculus.” What makes them similar is that they all share the following rather vague description:

To compute a certain sort of integral over a region, we may do a computation on the boundary of the region that involves one fewer integrations.

Each version of the Fundamental Theorem of calculus makes the “vague” statement above precise. For example, when working with a single variable, the Fundamental Theorem concludes: $\int _a^b f'(x) \d x = f(b) - f(a)$ In this case we are doing an integral over the “region” $[a,b]$ , and the “computation” that allows “one fewer integrations” is antidifferentiation. This is the only Fundamental Theorem you have known so far in your studies. However with additional dimensions, there come additional derivatives. When working with function $F:\R ^n\to \R$ we have the gradient as a “derivative.” This brings us to our first of several new fundamental theorems.

Fundamental Theorem for Line Integrals If $C$ is a curve that starts at $\vec {a}$ and ends at $\vec {b}$ , then: $\int _C \grad F\dotp \d \vec {p} = F(\vec {b}) - F(\vec {a})$

Like the first fundamental theorem we met in our very first calculus class, the fundamental theorem for line integrals says that if we can find a potential function for a gradient field, we can evaluate a line integral over this gradient field by evaluating at the end-points. The up-shot is that whenever you are dealing with a line integral, you should always start by checking to see if you are working with a gradient field.

Let $F(x,y) = x\cos (xy)$ . Compute: $\grad F$ $\grad F(x,y) = \vector {\answer {\cos (xy) -xy\sin (xy)},\answer { -x^2\sin (xy)}}$

Now let: $\vec {G}(x,y) = \vector {\cos (xy) -xy\sin (xy), -x^2\sin (xy)}$ Compute $\int _C \vec G \dotp \d \vec {p}$ where $C$ is shown below:

$\begin{align*} \int_C \vec{G}\dotp \d \vec{p} &= F(4,0) - F(0,2)\\ &= \answer{4}. \end{align*}$

Conservative fields

When dealing with gradient fields and closed curve something very nice happens.

Suppose that $C$ is a closed curve, one that starts and stops at the same location. Compute: $\oint _C \grad F\dotp \d \vec {p} \begin {prompt} = \answer {0} \end {prompt}$

This leads us to our definition:

A conservative field is just another name for a gradient field.

If something is special enough to be named twice, we ought to do some more examples.

Let $\vec {F}(x,y) = \vector {-y,-x}$ . Let $C$ be the polygonal path below parameterized in a counterclockwise direction:

Compute $\oint _C \vec {F}\dotp \d \vec {p}$

In light of the Fundamental Theorem for line integrals, we should check to see if our field is a gradient field using the Clairaut gradient test: $\pp {x}(-x)-\pp {y}(-y) = \answer [given]{0}.$ Since our field is a gradient field, and our curve is closed, we see that $\oint _C \vec {F}\dotp \d \vec {p} = \answer [given]{0}.$

Below we see a directed curve with some field vectors attached.

Assuming that the field vectors are constant along each “side” of our polygonal path, compute: $\int _C\vec {F}\dotp \d \vec {p}$

This problem would be easier if our field was a gradient field, and while it surely isn’t, the field is constant along two paths that together make up all of $C$ :

So now we see that $\int _C \vec {F}\dotp \d \vec {p} = \int _{C_1} \vec {F}\dotp \d \vec {p}+\int _{C_2} \vec {F}\dotp \d \vec {p}.$ so we have that $\vec {F}(x,y) = \begin {cases} \vector {-1,1} &\text {on $C_1$,}\\ \vector {1,-1} &\text {on $C_2$.} \end {cases}$ Using the Clairaut gradient test we see that $\vec {F}$ isis not a gradient field when restricted to either part of $C$ . Let $F$ be a potential function for $\vec {F}$ . One candidate for the the potential function is $F(x,y) = \begin {cases} -x+y &\text {on $C_1$,}\\ x-y &\text {on $C_2$.} \end {cases}$ By the Fundamental Theorem for Line Integrals, $\begin{align*} \int_{C_1} \vec{F}\dotp\d\vec{p} &= \eval{-x+y}_{\vector{3,3}}^{\vector{-5,3}}\\ &= \answer[given]{8}, \end{align*}$ and $\begin{align*} \int_{C_2} \vec{F}\dotp\d\vec{p} &= \eval{x-y}_{\vector{-5,3}}^{\vector{-9,-5}}\\ &= \answer[given]{4}. \end{align*}$ So $\int _{C} \vec {F}\dotp \d \vec {p} = \answer [given]{12}.$