### 5 Vector line integrals

It is also possible to form the less commonly used integrals ${\int }_{C}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}\underset{̲}{dr}$ and ${\int }_{C}\underset{̲}{F}\left(x,y,z\right)×\underset{̲}{dr}$ . Each of these integrals evaluates to a vector.

Remembering that $\underset{̲}{dr}=dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+dy\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}+dz\phantom{\rule{0.3em}{0ex}}\underset{̲}{k}$ , an integral of the form ${\int }_{C}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}\underset{̲}{dr}$ becomes

${\int }_{C}f\left(x,y,z\right)dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+{\int }_{C}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}+{\int }_{C}f\left(x,y,z\right)dz\phantom{\rule{0.3em}{0ex}}\underset{̲}{k}$ . The first term can be evaluated by expressing $y$ and $z$ in terms of $x$ . Similarly the second and third terms can be evaluated by expressing all terms as functions of $y$ and $z$ respectively. Alternatively, all variables can be expressed in terms of a parameter $t$ . If an integral is two-dimensional, the term in $z$ will be absent.

##### Example 12

Evaluate the integral ${\int }_{C}x{y}^{2}\underset{̲}{dr}$ where $C$ represents the contour $y={x}^{2}$ from $\left(0,0\right)$ to $\left(1,1\right)$ .

##### Solution

This is a two-dimensional integral so the term in $z$ will be absent.

$\begin{array}{rcll}I& =& {\int }_{C}x{y}^{2}\underset{̲}{dr}& \text{}\\ & =& {\int }_{C}x{y}^{2}\left(dx\underset{̲}{i}+dy\underset{̲}{j}\right)& \text{}\\ & =& {\int }_{C}x{y}^{2}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+{\int }_{C}x{y}^{2}\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}& \text{}\\ & =& {\int }_{x=0}^{1}x{\left({x}^{2}\right)}^{2}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+{\int }_{y=0}^{1}{y}^{1∕2}{y}^{2}\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}& \text{}\\ & =& {\int }_{0}^{1}{x}^{5}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+{\int }_{0}^{1}{y}^{5∕2}\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {\left[\frac{1}{6}{x}^{6}\right]}_{0}^{1}\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+{\left[\frac{2}{7}{x}^{7∕2}\right]}_{0}^{1}\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}& \text{}\\ & =& \frac{1}{6}\underset{̲}{i}+\frac{2}{7}\underset{̲}{j}& \text{}\end{array}$
##### Example 13

Find $I={\int }_{C}x\underset{̲}{dr}$ for the contour $C$ given parametrically by $x=cost$ , $y=sint$ , $z=t-\pi$ starting at $t=0$ and going to $t=2\pi$ , i.e. the contour starts at $\left(1,0,-\pi \right)$ and finishes at $\left(1,0,\pi \right)$ .

##### Solution

The integral becomes ${\int }_{C}x\left(dx\underset{̲}{i}+dy\underset{̲}{j}+dz\underset{̲}{k}\right).$

Now, $x=cost$ , $y=sint$ , $z=t-\pi$ so $dx=-sint\phantom{\rule{0.3em}{0ex}}dt$ , $dy=cost\phantom{\rule{0.3em}{0ex}}dt$ and $dz=dt$ . So

$\begin{array}{rcll}I& =& {\int }_{0}^{2\pi }cost\left(-sint\phantom{\rule{0.3em}{0ex}}dt\underset{̲}{i}+cost\phantom{\rule{0.3em}{0ex}}dt\underset{̲}{j}+dt\underset{̲}{k}\right)& \text{}\\ & =& -{\int }_{0}^{2\pi }costsint\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+{\int }_{0}^{2\pi }{cos}^{2}t\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}+{\int }_{0}^{2\pi }cost\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{0.3em}{0ex}}\underset{̲}{k}& \text{}\\ & =& -\frac{1}{2}{\int }_{0}^{2\pi }sin2t\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+\frac{1}{2}{\int }_{0}^{2\pi }\left(1+cos2t\right)\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}+{\left[sint\right]}_{0}^{2\pi }\phantom{\rule{0.3em}{0ex}}\underset{̲}{k}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& \frac{1}{4}{\left[cos2t\right]}_{0}^{2\pi }\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}\phantom{\rule{1em}{0ex}}+\phantom{\rule{1em}{0ex}}\frac{1}{2}{\left[t+\frac{1}{2}sin2t\right]}_{0}^{2\pi }\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}\phantom{\rule{1em}{0ex}}+\phantom{\rule{1em}{0ex}}0\underset{̲}{k}& \text{}\\ & =& 0\underset{̲}{i}+\pi \phantom{\rule{0.3em}{0ex}}\underset{̲}{j}=\pi \underset{̲}{j}& \text{}\end{array}$

Integrals of the form ${\int }_{C}\underset{̲}{F}×\underset{̲}{dr}$ can be evaluated as follows. The vector field $\underset{̲}{F}={F}_{1}\underset{̲}{i}+{F}_{2}\underset{̲}{j}+{F}_{3}\underset{̲}{k}$ and $\underset{̲}{dr}=dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+dy\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}+dz\phantom{\rule{0.3em}{0ex}}\underset{̲}{k}$ so

$\begin{array}{rcll}\underset{̲}{F}×\underset{̲}{dr}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill {F}_{1}\hfill & \hfill {F}_{2}\hfill & \hfill {F}_{3}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill dx\hfill & \hfill dy\hfill & \hfill dz\hfill \end{array}\right|& =& \left({F}_{2}\phantom{\rule{0.3em}{0ex}}dz-{F}_{3}\phantom{\rule{0.3em}{0ex}}dy\right)\underset{̲}{i}+\left({F}_{3}\phantom{\rule{0.3em}{0ex}}dx-{F}_{1}\phantom{\rule{0.3em}{0ex}}dz\right)\underset{̲}{j}+\left({F}_{1}\phantom{\rule{0.3em}{0ex}}dy-{F}_{2}\phantom{\rule{0.3em}{0ex}}dx\right)\underset{̲}{k}& \text{}\\ & =& \left({F}_{3}\underset{̲}{j}-{F}_{2}\underset{̲}{k}\right)dx+\left({F}_{1}\underset{̲}{k}-{F}_{3}\underset{̲}{i}\right)dy+\left({F}_{2}\underset{̲}{i}-{F}_{1}\underset{̲}{j}\right)dz& \text{}\end{array}$

There are a maximum of six terms involved in one such integral; the exact details may dictate which form to use.

##### Example 14

Evaluate the integral ${\int }_{C}\left({x}^{2}\underset{̲}{i}+3xy\underset{̲}{j}\right)×\underset{̲}{dr}$ where $C$ represents the curve $y=2{x}^{2}$ from $\left(0,0\right)$ to $\left(1,2\right)$ .

##### Solution

Note that the $z$ component of $\underset{̲}{F}$ and $\underset{̲}{dr}$ are both zero.

So $\underset{̲}{F}×\underset{̲}{dr}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill {x}^{2}\hfill & \hfill 3xy\hfill & \hfill 0\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill dx\hfill & \hfill dy\hfill & \hfill 0\hfill \end{array}\right|=\left({x}^{2}dy-3xydx\right)\underset{̲}{k}$

and ${\int }_{C}\left({x}^{2}\underset{̲}{i}+3xy\underset{̲}{j}\right)×\underset{̲}{dr}={\int }_{C}\left({x}^{2}dy-3xydx\right)\underset{̲}{k}$

Now, on $C$ , $y=2{x}^{2}$ so $dy=4xdx$ and

$\begin{array}{rcll}{\int }_{C}\left({x}^{2}\underset{̲}{i}+3xy\underset{̲}{j}\right)×\underset{̲}{dr}& \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}& {\int }_{C}\left\{{x}^{2}dy-3xydx\right\}\underset{̲}{k}& \text{}\\ & \phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}& {\int }_{x=0}^{1}\left\{{x}^{2}×4xdx-3x×2{x}^{2}dx\right\}\underset{̲}{k}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {\int }_{0}^{1}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}-2{x}^{3}dx\underset{̲}{k}& \text{}\\ & =& -{\left[\frac{1}{2}{x}^{4}\right]}_{0}^{1}\underset{̲}{k}& \text{}\\ & =& -\frac{1}{2}\underset{̲}{k}& \text{}\end{array}$