2 Line integrals of scalar products

Integrals of the form ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ , referred to at the end of the previous sub-section, occur in applications such as the following.

Figure 2:

Consider a cyclist riding along the road from $A$ to $B$ (Figure 2). Suppose it is necessary to find the total work the cyclist has to do in overcoming a wind of velocity $\underset{̲}{v}$ .

On moving from $S$ to $T$ , the work done is given by ‘Force $\times$ distance’ $=F\times \delta rcos\theta$ where $F$ , the force, is directly proportional to $v$ , but in the opposite direction, and $\delta rcos\theta$ is the component of the distance travelled in the direction of the wind.

So, the work done travelling $\underset{̲}{\delta r}$ is $-k\underset{̲}{v}\cdot \underset{̲}{\delta r}$ . Letting $\underset{̲}{\delta r}$ become infinitesimally small, the work done becomes $-k\underset{̲}{v}\cdot \underset{̲}{dr}$ and the total work is $-k{\int \; <!--nolimits\; -->}_A^B\underset{̲}{v}\cdot \underset{̲}{dr}$ .

This is an example of the integral along a line of the scalar product of a vector field and a vector describing the line. The term scalar line integral is often used for integrals of this form. The vector $\underset{̲}{dr}$ may be considered to be $dx\underset{̲}{i}+dy\underset{̲}{j}+dx\underset{̲}{k}$ .

Multiplying out the scalar product, in three dimensions, the ’scalar line integral’ of the vector $\underset{̲}{F}$ along contour $C$ is given by ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ and equals ${\int \; <!--nolimits\; -->}_C\left\{F_{x}dx+F_{y}dy+F_{z}dz\right\}$ in three dimensions ( ${\int \; <!--nolimits\; -->}_C\left\{F_{x}dx+F_{y}dy\right\}$ in two dimensions.)

If the contour $C$ has its start and end points in the same positions i.e. it represents a closed contour, the symbol ${\oint }_C$ rather than ${\int \; <!--nolimits\; -->}_C$ is used, i.e. ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ .

As before, to evaluate the line integral, express the path and the function $\underset{̲}{F}$ in terms of either $x$ , $y$ and $z$ , or in terms of a parameter $t$ . Note that in examples $t$ often represents time.

Example 5

Find ${\int \; <!--nolimits\; -->}_C\left\{2xydx-5xdy\right\}$ where $C$ is the curve $y=x^3$ with $x$ varying from $x=0$ to $x=1$ .

[This is the integral ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ where $\underset{̲}{F}=2xy\underset{̲}{i}-5x\underset{̲}{j}$ and $\underset{̲}{dr}=dx\underset{̲}{i}+dy\underset{̲}{j}$ .]

Solution

It is possible to split this integral into two different integrals and express the first term as a function of $x$ and the second term as a function of $y$ . However, it is also possible to express everything in terms of $x$ . Note that on $C$ , $y=x^3$ so $dy=3x^{2}dx$ and the integral becomes

Example 6

Three paths from $\left(0,0\right)$ to $\left(1,2\right)$ are defined by

1. $C_1:y=2x$
2. $C_2:y=2x^2$
3. $C_3:y=0$ from $\left(0,0\right)$ to $\left(1,0\right)$ and $x=1$ from $\left(1,0\right)$ to $\left(1,2\right)$

Sketch each path and find $\int \; <!--nolimits\; -->\underset{̲}{F}\cdot \underset{̲}{dr}$ , where $\underset{̲}{F}=y^2\underset{̲}{i}+xy\underset{̲}{j}$ , along each path.

Solution

1. $\int \; <!--nolimits\; -->\underset{̲}{F}\cdot \underset{̲}{dr}=\int \; <!--nolimits\; -->\left\{y^{2}dx+xydy\right\}$ . Along $y=2x$ , $\frac{dy}{dx}=2$ so $dy=2dx$ . Then $$\begin{array}{rcll}{\int \; <!--nolimits\; -->}_{C_1}\underset{̲}{F}\cdot \underset{̲}{dr}& =& {\int \; <!--nolimits\; -->}_{x=0}^1\left\{{\left(2x\right)}^{2}dx+x\left(2x\right)\left(2dx\right)\right\}& \\ & =& {\int \; <!--nolimits\; -->}_0^1\left(4x^2+4x^2\right)dx={\int \; <!--nolimits\; -->}_0^{1}8x^{2}dx={\left[\frac{8}{3}{x}^2\right]}_0^1=\frac{8}{3}& \end{array}$$

   Figure 3(a):

   

2. $\int \; <!--nolimits\; -->\underset{̲}{F}\cdot \underset{̲}{dr}=\int \; <!--nolimits\; -->\left\{y^{2}dx+xydy\right\}$ . Along $y=2x^2$ , $\frac{dy}{dx}=4x$ so $dy=4xdx$ . Then
   
   ${\int \; <!--nolimits\; -->}_{C_2}\underset{̲}{F}\cdot \underset{̲}{dr}={\int \; <!--nolimits\; -->}_{x=0}^1\left\{{\left(2x^2\right)}^{2}dx+x\left(2x^2\right)\left(4xdx\right)\right\}={\int \; <!--nolimits\; -->}_0^{1}12x^{4}dx={\left[\frac{12}{5}{x}^5\right]}_0^1=\frac{12}{5}$
   
   

   Figure 3(b):

   

   Note that the answer is different to part (a), i.e., the line integral depends upon the path taken.

3. As the contour $C_3$ , has two distinct parts with different equations, it is necessary to break the full contour $OA$ into the two parts, namely $OB$ and $BA$ where $B$ is the point $\left(1,0\right)$ . Hence

   ${\int \; <!--nolimits\; -->}_{C_3}\underset{̲}{F}\cdot \underset{̲}{dr}={\int \; <!--nolimits\; -->}_O^B\underset{̲}{F}\cdot \underset{̲}{dr}+{\int \; <!--nolimits\; -->}_B^A\underset{̲}{F}\cdot \underset{̲}{dr}$

   Along $OB$ , $y=0$ so $dy=0$ . Then

   ${\int \; <!--nolimits\; -->}_O^B\underset{̲}{F}\cdot \underset{̲}{dr}={\int \; <!--nolimits\; -->}_{x=0}^1\left(0^{2}dx+x\times 0\times 0\right)={\int \; <!--nolimits\; -->}_0^{1}0dx=0$

   Along $AB$ , $x=1$ so $dx=0$ . Then

   ${\int \; <!--nolimits\; -->}_A^B\underset{̲}{F}\cdot \underset{̲}{dr}={\int \; <!--nolimits\; -->}_{y=0}^2\left(y^2\times 0+1\times y\times dy\right)={\int \; <!--nolimits\; -->}_0^{2}ydy={\left[\frac{1}{2}{y}^2\right]}_0^2=2.$

   Hence ${\int \; <!--nolimits\; -->}_{C_3}\underset{̲}{F}\cdot \underset{̲}{dr}=0+2=2$

   Figure 3(c):

   

   Once again, the result is path dependent.

Example 7

Find ${\int \; <!--nolimits\; -->}_A^O\underset{̲}{F}\cdot \underset{̲}{dr}$ , where $\underset{̲}{F}=y^2\underset{̲}{i}+xy\underset{̲}{j}$ (as in Example 6) and the path from $A$ to $O$ is the straight line from $\left(1,2\right)$ to $\left(0,0\right)$ , that is the reverse of $C_1$ in Example 6(a).

Deduce ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ , the integral around the closed path $C$ formed by the parabola $y=2x^2$ from $\left(0,0\right)$ to $\left(1,2\right)$ and the line $y=2x$ from $\left(1,2\right)$ to $\left(0,0\right)$ .

Solution

Reversing the path swaps the limits of integration, this results in a change of sign for the value of the integral.

${\int \; <!--nolimits\; -->}_A^O\underset{̲}{F}\cdot \underset{̲}{dr}=-{\int \; <!--nolimits\; -->}_O^A\underset{̲}{F}\cdot \underset{̲}{dr}=-\frac{8}{3}$

The integral along the parabola (calculated in (iii) above) evaluates to $\frac{12}{5}$ , then

${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}={\oint }_{C_2}\underset{̲}{F}\cdot \underset{̲}{dr}+{\oint }_{C_4}\underset{̲}{F}\cdot \underset{̲}{dr}=\frac{12}{5}-\frac{8}{3}=-\frac{4}{15}\approx -0.267$

Example 8

Consider the vector field

$\underset{̲}{F}=y^2{z}^3\underset{̲}{i}+2xy{z}^3\underset{̲}{j}+3x{y}^2{z}^2\underset{̲}{k}$

Let $C_1$ and $C_2$ be the curves from $O=\left(0,0,0\right)$ to $A=\left(1,1,1\right)$ , given by

1. Evaluate the scalar integral of the vector field along each path.
2. Find the value of ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ where $C$ is the closed path along $C_1$ from $O$ to $A$ and back along $C_2$ from $A$ to $O$ .

Solution

1. The path $C_1$ is given in terms of the parameter $t$ by $x=t$ , $y=t$ and $z=t$ . Hence

   $\frac{dx}{dt}=\frac{dy}{dt}=\frac{dz}{dt}=1\text{ and }\frac{d\underset{̲}{r}}{dt}=\frac{dx}{dt}\underset{̲}{i}+\frac{dy}{dt}\underset{̲}{j}+\frac{dz}{dt}\underset{̲}{k}=\underset{̲}{i}+\underset{̲}{j}+\underset{̲}{k}$

   Now by substituting for $x=y=z=t$ in $\underset{̲}{F}$ we have

   $\underset{̲}{F}=t^5\underset{̲}{i}+2t^5\underset{̲}{j}+3t^5\underset{̲}{k}$

   Hence $\underset{̲}{F}\cdot \frac{d\underset{̲}{r}}{dt}=t^5+2t^5+3t^5=6t^5$ . The values of $t=0$ and $t=1$ correspond to the start and end point of $C_1$ and so these are the required limits of integration. Now

   ${\int \; <!--nolimits\; -->}_{C_1}\underset{̲}{F}\cdot \underset{̲}{dr}={\int \; <!--nolimits\; -->}_0^1\underset{̲}{F}\cdot \frac{d\underset{̲}{r}}{dt}dt={\int \; <!--nolimits\; -->}_0^{1}6t^{5}dt={\left[t^6\right]}_0^1=1$

   For the path $C_2$ the parameterisation is $x=t^2$ , $y=t$ and $z=t^2$ so $\frac{d\underset{̲}{r}}{dt}=2t\underset{̲}{i}+\underset{̲}{j}+2t\underset{̲}{k}$ .
   
   Substituting $x=t^2$ , $y=t$ and $z=t^2$ in $\underset{̲}{F}$ we have

   $\underset{̲}{F}=t^8\underset{̲}{i}+2t^9\underset{̲}{j}+3t^8\underset{̲}{k}\text{ and }\underset{̲}{F}\cdot \frac{d\underset{̲}{r}}{dt}=2t^9+2t^9+6t^9=10t^9$

   ${\int \; <!--nolimits\; -->}_{C_2}\underset{̲}{F}\cdot \underset{̲}{dr}={\int \; <!--nolimits\; -->}_0^{1}10t^{9}dt={\left[t^{10}\right]}_0^1=1$

2. For the closed path $C$

   ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}={\int \; <!--nolimits\; -->}_{C_1}\underset{̲}{F}\cdot \underset{̲}{dr}-{\int \; <!--nolimits\; -->}_{C_2}\underset{̲}{F}\cdot \underset{̲}{dr}=1-1=0$

   (Note that the line integral round a closed path is not necessarily zero - see Example 7.)

Further points on Example 8

Task!

Consider the vector field

$\underset{̲}{G}=x\underset{̲}{i}+\left(4x-y\right)\underset{̲}{j}$

Let $C_1$ and $C_2$ be the curves from $O=\left(0,0,0\right)$ to $A=\left(1,1,1\right)$ , given by

1. Evaluate the scalar integral ${\int \; <!--nolimits\; -->}_C\underset{̲}{G}\cdot \underset{̲}{dr}$ of each vector field along each path.
2. Find the value of ${\oint }_C\underset{̲}{G}\cdot \underset{̲}{dr}$ where $C$ is the closed path along $C_1$ from $O$ to $A$ and back along $C_2$ from $A$ to $O$ .

Answer

Example 9

Find ${\int \; <!--nolimits\; -->}_C\left\{\underset{̲}{\nabla }\left(x^{2}y\right)\right\}\cdot \underset{̲}{dr}$ where $C$ is the contour $y=2x-x^2$ from $\left(0,0\right)$ to $\left(2,0\right)$ .

Solution

Note that $\underset{̲}{\nabla }\left(x^{2}y\right)=2xy\underset{̲}{i}+x^2\underset{̲}{j}$ so the integral is ${\int \; <!--nolimits\; -->}_C\left\{2xydx+x^{2}dy\right\}$ .

On $y=2x-x^2$ , $dy=\left(2-2x\right)dx$ so the integral becomes

Task!

Evaluate ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ , where $\underset{̲}{F}=\left(x-y\right)\underset{̲}{i}+\left(x+y\right)\underset{̲}{j}$ along each of the following paths

1. $C_1$ : from $\left(1,1\right)$ to $\left(2,4\right)$ along the straight line $y=3x-2$ :
2. $C_2$ : from $\left(1,1\right)$ to $\left(2,4\right)$ along the parabola $y=x^2$ :
3. $C_3$ : along the straight line $x=1$ from $\left(1,1\right)$ to $\left(1,4\right)$ then along the straight line $y=4$ from $\left(1,4\right)$ to $\left(2,4\right)$ .

Answer

Task!

For the function $\underset{̲}{F}$ and paths in the last Task, deduce $\oint \underset{̲}{F}\cdot \underset{̲}{dr}$ for the closed paths

1. $C_1$ followed by the reverse of $C_2$ .
2. $C_2$ followed by the reverse of $C_3$ .
3. $C_3$ followed by the reverse of $C_1$ .

Answer

Exercises

1. Consider ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ , where $\underset{̲}{F}=3x^2{y}^2\underset{̲}{i}+\left(2x^{3}y-1\right)\underset{̲}{j}$ . Find the value of the line integral along each of the paths from $\left(0,0\right)$ to $\left(1,4\right)$ .
   1. $y=4x$
   2. $y=4x^2$
   3. $y=4x^{1∕2}$
   4. $y=4x^3$
2. Consider the vector field $\underset{̲}{F}=2x\underset{̲}{i}+\left(xz-2\right)\underset{̲}{j}+xy\underset{̲}{k}$ and the two curves between $\left(0,0,0\right)$ and $\left(1,-1,2\right)$ defined by
   
   $C_1$ : $x=t^2$ , $y=-t$ , $z=2t$ for $0\le t\le 1$ .
   
   $C_2$ : $x=t-1$ , $y=1-t$ , $z=2t-2$ for $1\le t\le 2$ .
   
   
   1. Find ${\int \; <!--nolimits\; -->}_{C_1}\underset{̲}{F}\cdot \underset{̲}{dr}$ , ${\int \; <!--nolimits\; -->}_{C_2}\underset{̲}{F}\cdot \underset{̲}{dr}$
   2. Find ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ where $C$ is the closed path from $\left(0,0,0\right)$ to $\left(1,-1,2\right)$ along $C_1$ and back to $\left(0,0,0\right)$ along $C_2$ .
3. Consider the vector field $\underset{̲}{G}=x^{2}z\underset{̲}{i}+y^{2}z\underset{̲}{j}+\frac{1}{3}\left(x^3+y^3\right)\underset{̲}{k}$ and the two curves between $\left(0,0,0\right)$ and $\left(1,-1,2\right)$ defined by
   
   $C_1$ : $x=t^2$ , $y=-t$ , $z=2t$ for $0\le t\le 1$ .
   
   $C_2$ : $x=t-1$ , $y=1-t$ , $z=2t-2$ for $1\le t\le 2$ .
   
   
   1. Find ${\int \; <!--nolimits\; -->}_{C_1}\underset{̲}{G}\cdot \underset{̲}{dr}$ , ${\int \; <!--nolimits\; -->}_{C_2}\underset{̲}{G}\cdot \underset{̲}{dr}$
   2. Find ${\oint }_C\underset{̲}{G}\cdot \underset{̲}{dr}$ where $C$ is the closed path from $\left(0,0,0\right)$ to $\left(1,-1,2\right)$ along $C_1$ and back to $\left(0,0,0\right)$ along $C_2$ .
4. Find ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ along $y=2x$ from $\left(0,0\right)$ to $\left(2,4\right)$ for
   1. $\underset{̲}{F}=\underset{̲}{\nabla }\left(x^{2}y\right)$
   2. $\underset{̲}{F}=\underset{̲}{\nabla }\times \left(\frac{1}{2}{x}^2{y}^2\underset{̲}{k}\right)$

Answer