### 1 Line integrals

HELM booklet  28 was concerned with evaluating an integral over all points within a rectangle or other shape (or over a cuboid or other volume). In a related manner, an integral can take place over a line or curve running through a two-dimensional (or three-dimensional) shape. Line integrals may involve scalar or vector fields. Those involving scalar fields are dealt with first.

#### 1.1 Line integrals in two dimensions

A line integral in two dimensions may be written as

$\phantom{\rule{2em}{0ex}}{\int }_{C}F\left(x,y\right)dw$

There are three main features determining this integral:

$\phantom{\rule{2em}{0ex}}F\left(x,y\right)$ This is the scalar function to be integrated   e.g.   $F\left(x,y\right)={x}^{2}+4{y}^{2}.$

$\phantom{\rule{2em}{0ex}}C$  This is the curve along which integration takes place.  e.g.  $y={x}^{2}$  or $x=siny$

or $x=t-1$ ;   $y={t}^{2}$ (where $x$ and $y$ are expressed in terms of a parameter $t$ ).

$\phantom{\rule{2em}{0ex}}dw$ This states the variable of the integration. Three main cases are $dx$ , $dy$ and $ds$ .

Here ‘ $s$ ’ is arc length and so indicates position along the curve $C$ .

$ds$   may be written as $ds=\sqrt{{\left(dx\right)}^{2}+{\left(dy\right)}^{2}}$ or   $ds=\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$ .

A fourth case is when $F\left(x,y\right)\phantom{\rule{1em}{0ex}}dw$ has the form: ${F}_{1}dx+{F}_{2}dy$ . This is a combination

of the cases $dx$ and $dy$ .

The integral ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}ds$ represents the area beneath the surface $z=F\left(x,y\right)$ but above the line $C$ .

The integrals ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dx$ and ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dy$ represent the projections of this area onto the $xz$ and $yz$ planes respectively.

A particular case of the integral ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}ds$ is the integral ${\int }_{C}1\phantom{\rule{0.3em}{0ex}}ds$ . This is a means of calculating the length along a curve i.e. an arc length.

Figure 1:

The technique with a line integral is to express all quantities in an integral in terms of a single variable.  Often,  if  the  integral  is  with  respect  to  ’ $x$ ’  or  ’ $y$ ’,  the  curve  ’ $C$ ’  and  the  function  ’ $F$ ’  may  be  expressed  in  terms  of  the  relevant  variable. If the integral is carried out with respect to $ds$ , normally everything is expressed in terms of $x$ . If $x$ and $y$ are given in terms of a parameter $t$ , normally everything is expressed in terms of $t$ .

##### Example 1

Find   ${\int }_{c}x\left(1+4y\right)dx$  where   $C$  is  the  curve   $y={x}^{2}$ ,  starting  from   $x=0,y=0$  and  ending  at   $x=1,y=1$ .

##### Solution

As this integral concerns only points along $C$ and the integration is carried out with respect to $x$ , $y$ may be replaced by ${x}^{2}$ . The limits on   $x$  will  be  0  to  1.  So the  integral  becomes

$\begin{array}{rcll}{\int }_{C}x\left(1+4y\right)\phantom{\rule{0.3em}{0ex}}dx& =& {\int }_{x=0}^{1}x\left(1+4{x}^{2}\right)dx={\int }_{x=0}^{1}\left(x+4{x}^{3}\right)dx\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {\left[\frac{{x}^{2}}{2}+{x}^{4}\right]}_{0}^{1}=\left(\frac{1}{2}+1\right)-\left(0\right)=\frac{3}{2}& \text{}\end{array}$
##### Example 2

Find   ${\int }_{c}x\left(1+4y\right)\phantom{\rule{0.3em}{0ex}}dy$  where   $C$  is  the  curve   $y={x}^{2}$ ,  starting  from   $x=0,y=0$  and  ending  at   $x=1,y=1$ . This is the same as Example 1 other than $dx$ being replaced by $dy$ .

##### Solution

As this integral concerns only points along $C$ and the integration is carried out with respect to $y$ , everything may be expressed in terms of $y$ , i.e. $x$ may be replaced by ${y}^{1∕2}$ . The limits on   $y$  will  be  0  to  1.  So the  integral  becomes

$\begin{array}{rcll}{\int }_{C}x\left(1+4y\right)\phantom{\rule{0.3em}{0ex}}dy& =& {\int }_{y=0}^{1}{y}^{1∕2}\left(1+4y\right)dx={\int }_{y=0}^{1}\left({y}^{1∕2}+4{y}^{3∕2}\right)dx\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {\left[\frac{2}{3}{y}^{3∕2}+\frac{8}{5}{x}^{5∕2}\right]}_{0}^{1}=\left(\frac{2}{3}+\frac{8}{5}\right)-\left(0\right)=\frac{34}{15}& \text{}\end{array}$
##### Example 3

Find   ${\int }_{c}x\left(1+4y\right)ds$  where   $C$  is  the  curve   $y={x}^{2}$ ,  starting  from   $x=0,y=0$  and  ending  at   $x=1,y=1$ . Once again, this is the same as the previous two examples other than the integration being carried out with respect to $s$ , the coordinate along the curve $C$ .

##### Solution

As this integral is with respect to $x$ , all parts of the integral can be expressed in terms of $x$ , Along $y={x}^{2}$ , $ds=\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx=\sqrt{1+{\left(2x\right)}^{2}}dx=\sqrt{1+4{x}^{2}}dx$

So, the integral is

$\phantom{\rule{2em}{0ex}}{\int }_{c}x\left(1+4y\right)ds={\int }_{x=0}^{1}x\left(1+4{x}^{2}\right)\sqrt{1+4{x}^{2}}\phantom{\rule{1em}{0ex}}dx={\int }_{x=0}^{1}x{\left(1+4{x}^{2}\right)}^{3∕2}dx$

This can be evaluated using the transformation $U=1+4{x}^{2}$  so   $dU=8xdx$  i.e.   $x\phantom{\rule{0.3em}{0ex}}dx=\frac{dU}{8}$ .  When   $x=0$ ,   $U=1$  and  when   $x=1$ ,   $U=5$ .

The integral therefore equals

$\begin{array}{rcll}{\int }_{x=0}^{1}x{\left(1+4{x}^{2}\right)}^{3∕2}dx& =& \frac{1}{8}{\int }_{U=1}^{5}{U}^{3∕2}dU& \text{}\\ & =& \frac{1}{8}×\frac{2}{5}{\left[{U}^{5∕2}\right]}_{1}^{5}=\frac{1}{20}\left[{5}^{5∕2}-1\right]\approx 2.745\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Note that the results for Examples 1,2 and 3 are all different: Example 3 is the area between a curve and a surface above; Examples 1 and 2 give projections of this area onto other planes.

##### Example 4

Find ${\int }_{C}xy\phantom{\rule{0.3em}{0ex}}dx$ where, on $C$ , $x$ and $y$ are given by $x=3{t}^{2}$ , $y={t}^{3}-1$ for $t$ starting at $t=0$ and progressing to $t=1$ .

##### Solution

Everything can be expressed in terms of $t$ , the parameter. Here $x=3{t}^{2}$ so $dx=6t\phantom{\rule{0.3em}{0ex}}dt$ . The limits on $t$ are $t=0$ and $t=1$ . The integral becomes

$\begin{array}{rcll}{\int }_{C}xy\phantom{\rule{0.3em}{0ex}}dx& =& {\int }_{t=0}^{1}3{t}^{2}\phantom{\rule{0.3em}{0ex}}\left({t}^{3}-1\right)\phantom{\rule{0.3em}{0ex}}6t\phantom{\rule{0.3em}{0ex}}dt={\int }_{t=0}^{1}\left(18{t}^{6}-18{t}^{3}\right)\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {\left[\frac{18}{7}{t}^{7}-\frac{18}{4}{t}^{4}\right]}_{0}^{1}=\frac{18}{7}-\frac{9}{2}-0=-\frac{27}{14}& \text{}\end{array}$

##### Key Point 1

A line integral is normally evaluated by expressing all variables in terms of one variable.

In general

${\int }_{C}f\left(x,y\right)\phantom{\rule{1em}{0ex}}ds\ne {\int }_{C}f\left(x,y\right)\phantom{\rule{1em}{0ex}}dy\ne {\int }_{C}f\left(x,y\right)\phantom{\rule{1em}{0ex}}dx$

For $F\left(x,y\right)=2x+{y}^{2}$ , find ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dx$ , ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dy$ and ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}ds$ where $C$ is the line $y=2x$ from $\left(0,0\right)$ to $\left(1,2\right)$ .

Express each integral as a simple integral with respect to a single variable and hence evaluate each integral:

${\int }_{x=0}^{1}\left(2x+4{x}^{2}\right)\phantom{\rule{1em}{0ex}}dx,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{7}{3}$ , ${\int }_{y=0}^{2}\left(y+{y}^{2}\right)\phantom{\rule{1em}{0ex}}dy,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{14}{3}$ , ${\int }_{x=0}^{1}\left(2x+4{x}^{2}\right)\sqrt{5}\phantom{\rule{1em}{0ex}}dx,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{7}{3}\sqrt{5}$

Find ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dx$ , ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dy$ and ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}ds$ where $F\left(x,y\right)=1$ and $C$ is the curve $y=\frac{1}{2}{x}^{2}-\frac{1}{4}lnx$ from $\left(1,\frac{1}{2}\right)$ to $\left(2,2-\frac{1}{4}ln2\right)$ .

${\int }_{1}^{2}1\phantom{\rule{0.3em}{0ex}}dx=1$ , ${\int }_{1∕2}^{2-1∕4ln2}1\phantom{\rule{0.3em}{0ex}}dy=\frac{3}{2}-\frac{1}{4}ln2$ ,

${\int }_{1}^{2}\left(x+\frac{1}{4x}\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{3}{2}+\frac{1}{4}ln2$ .

Find ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dx$ , ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dy$ and ${\int }_{C}F\left(x,y\right)\phantom{\rule{0.3em}{0ex}}ds$ where $F\left(x,y\right)=sin2x$ and $C$ is the curve $y=sinx$ from $\left(0,0\right)$ to $\left(\frac{\pi }{2},1\right)$ .

${\int }_{0}^{\pi ∕2}sin2x\phantom{\rule{0.3em}{0ex}}dx=1$ , ${\int }_{0}^{\pi ∕2}2sinx{cos}^{2}x\phantom{\rule{0.3em}{0ex}}dx=\frac{2}{3}$

${\int }_{0}^{\pi ∕2}sin2x\sqrt{1+{cos}^{2}x}\phantom{\rule{0.3em}{0ex}}dx$   Using the substitution $u=1+{cos}^{2}x$ gives  $\frac{2}{3}\left(2\sqrt{2}-1\right)$ .