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

${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)dw$

There are three main features determining this integral:

$F\left(x,y\right)$ This is the scalar function to be integrated   e.g.   $F\left(x,y\right)=x^2+4y^2.$

$C$  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 $t$ ).

$dw$ 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 $F\left(x,y\right)dw$ has the form: $F_{1}dx+F_{2}dy$ . This is a combination

   of the cases $dx$ and $dy$ .

The integral ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)ds$ represents the area beneath the surface $z=F\left(x,y\right)$ but above the line $C$ .

The integrals ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)dx$ and ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)dy$ represent the projections of this area onto the $xz$ and $yz$ planes respectively.

A particular case of the integral ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)ds$ is the integral ${\int \; <!--nolimits\; -->}_{C}1ds$ . 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   ${\int \; <!--nolimits\; -->}_{c}x\left(1+4y\right)dx$  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

Example 2

Find   ${\int \; <!--nolimits\; -->}_{c}x\left(1+4y\right)dy$  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

Example 3

Find   ${\int \; <!--nolimits\; -->}_{c}x\left(1+4y\right)ds$  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 $y=x^2$ , $ds=\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx=\sqrt{1+{\left(2x\right)}^2}dx=\sqrt{1+4x^2}dx$

So, the integral is

${\int \; <!--nolimits\; -->}_{c}x\left(1+4y\right)ds={\int \; <!--nolimits\; -->}_{x=0}^{1}x\left(1+4x^2\right)\sqrt{1+4x^2}dx={\int \; <!--nolimits\; -->}_{x=0}^{1}x{\left(1+4x^2\right)}^{3∕2}dx$

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

The integral therefore equals

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 ${\int \; <!--nolimits\; -->}_{C}xydx$ where, on $C$ , $x$ and $y$ are given by $x=3t^2$ , $y=t^3-1$ for $t$ starting at $t=0$ and progressing to $t=1$ .

Solution

Everything can be expressed in terms of $t$ , the parameter. Here $x=3t^2$ so $dx=6tdt$ . The limits on $t$ are $t=0$ and $t=1$ . The integral becomes

Task!

For $F\left(x,y\right)=2x+y^2$ , find ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)dx$ , ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)dy$ and ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)ds$ 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:

Answer

Task!

Find ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)dx$ , ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)dy$ and ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)ds$ 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 $\left(2,2-\frac{1}{4}ln2\right)$ .

Answer

Task!

Find ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)dx$ , ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)dy$ and ${\int \; <!--nolimits\; -->}_{C}F\left(x,y\right)ds$ where $F\left(x,y\right)=sin2x$ and $C$ is the curve $y=sinx$ from $\left(0,0\right)$ to $\left(\frac{\pi }{2},1\right)$ .

Answer