5 Vector line integrals

It is also possible to form the less commonly used integrals ${\int \; <!--nolimits\; -->}_{C}f\left(x,y,z\right)\underset{̲}{dr}$ and ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\left(x,y,z\right)\times \underset{̲}{dr}$ . Each of these integrals evaluates to a vector.

Remembering that $\underset{̲}{dr}=dx\underset{̲}{i}+dy\underset{̲}{j}+dz\underset{̲}{k}$ , an integral of the form ${\int \; <!--nolimits\; -->}_{C}f\left(x,y,z\right)\underset{̲}{dr}$ becomes

${\int \; <!--nolimits\; -->}_{C}f\left(x,y,z\right)dx\underset{̲}{i}+{\int \; <!--nolimits\; -->}_{C}f\left(x,y,z\right)dy\underset{̲}{j}+{\int \; <!--nolimits\; -->}_{C}f\left(x,y,z\right)dz\underset{̲}{k}$ . 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 ${\int \; <!--nolimits\; -->}_{C}x{y}^2\underset{̲}{dr}$ 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.

Example 13

Find $I={\int \; <!--nolimits\; -->}_{C}x\underset{̲}{dr}$ 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 ${\int \; <!--nolimits\; -->}_{C}x\left(dx\underset{̲}{i}+dy\underset{̲}{j}+dz\underset{̲}{k}\right).$

Now, $x=cost$ , $y=sint$ , $z=t-\pi$ so $dx=-sintdt$ , $dy=costdt$ and $dz=dt$ . So

Integrals of the form ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\times \underset{̲}{dr}$ can be evaluated as follows. The vector field $\underset{̲}{F}=F_1\underset{̲}{i}+F_2\underset{̲}{j}+F_3\underset{̲}{k}$ and $\underset{̲}{dr}=dx\underset{̲}{i}+dy\underset{̲}{j}+dz\underset{̲}{k}$ so

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 ${\int \; <!--nolimits\; -->}_C\left(x^2\underset{̲}{i}+3xy\underset{̲}{j}\right)\times \underset{̲}{dr}$ where $C$ represents the curve $y=2x^2$ 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 $\underset{̲}{F}\times \underset{̲}{dr}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill x^2\hfill & \hfill 3xy\hfill & \hfill 0\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill dx\hfill & \hfill dy\hfill & \hfill 0\hfill \end{array}\right|=\left(x^{2}dy-3xydx\right)\underset{̲}{k}$

and ${\int \; <!--nolimits\; -->}_C\left(x^2\underset{̲}{i}+3xy\underset{̲}{j}\right)\times \underset{̲}{dr}={\int \; <!--nolimits\; -->}_C\left(x^{2}dy-3xydx\right)\underset{̲}{k}$

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