### 2 Line integrals of scalar products

Integrals of the form ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ , 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 $×$ distance’ $=F×\delta rcos\theta$ 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 $-k{\int }_{A}^{B}\underset{̲}{v}\cdot \underset{̲}{dr}$ .

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 ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ and equals ${\int }_{C}\left\{{F}_{x}\phantom{\rule{0.3em}{0ex}}dx+{F}_{y}\phantom{\rule{0.3em}{0ex}}dy+{F}_{z}\phantom{\rule{0.3em}{0ex}}dz\right\}$ in three dimensions ( ${\int }_{C}\left\{{F}_{x}\phantom{\rule{0.3em}{0ex}}dx+{F}_{y}\phantom{\rule{0.3em}{0ex}}dy\right\}$ 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 ${\int }_{C}$ 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 ${\int }_{C}\left\{2xy\phantom{\rule{0.3em}{0ex}}dx-5x\phantom{\rule{0.3em}{0ex}}dy\right\}$ where $C$ is the curve $y={x}^{3}$ with $x$ varying from $x=0$ to $x=1$ .

[This is the integral ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ 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 $dy=3{x}^{2}\phantom{\rule{0.3em}{0ex}}dx$ and the integral becomes

$\begin{array}{rcll}{\int }_{C}\left\{2xy\phantom{\rule{0.3em}{0ex}}dx-5x\phantom{\rule{0.3em}{0ex}}dy\right\}& =& {\int }_{x=0}^{1}\left(2x\phantom{\rule{0.3em}{0ex}}{x}^{3}\phantom{\rule{0.3em}{0ex}}dx-5x\phantom{\rule{0.3em}{0ex}}3{x}^{2}\phantom{\rule{0.3em}{0ex}}dx\right)={\int }_{0}^{1}\left(2{x}^{4}-15{x}^{3}\right)\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\left[\frac{2}{5}{x}^{5}-\frac{15}{4}{x}^{4}\right]}_{0}^{1}=\frac{2}{5}-\frac{15}{4}-0=-\frac{67}{20}& \text{}\end{array}$
##### Key Point 2

An integral of the form ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ may be expressed as ${\int }_{C}\left\{{F}_{x}\phantom{\rule{0.3em}{0ex}}dx+{F}_{y}\phantom{\rule{0.3em}{0ex}}dy+{F}_{z}\phantom{\rule{0.3em}{0ex}}dz\right\}$ . Knowing the expression for the path $C$ , every term in the integral can be further expressed in terms of one of the variables $x$ , $y$ or $z$ or in terms of a parameter $t$ and hence integrated.

If an integral is two-dimensional there are no terms involving $z$ .

The integral ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ evaluates to a scalar.

##### Example 6

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

1. ${C}_{1}:y=2x$
2. ${C}_{2}:y=2{x}^{2}$
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 $\int \underset{̲}{F}\cdot \underset{̲}{dr}$ , where $\underset{̲}{F}={y}^{2}\underset{̲}{i}+xy\underset{̲}{j}$ , along each path.

##### Solution
1. $\int \underset{̲}{F}\cdot \underset{̲}{dr}=\int \left\{{y}^{2}dx+xydy\right\}$ . Along $y=2x$ , $\frac{dy}{dx}=2$ so $dy=2dx$ . Then $\begin{array}{rcll}{\int }_{{C}_{1}}\underset{̲}{F}\cdot \underset{̲}{dr}& =& {\int }_{x=0}^{1}\left\{{\left(2x\right)}^{2}dx+x\left(2x\right)\left(2dx\right)\right\}& \text{}\\ & =& {\int }_{0}^{1}\left(4{x}^{2}+4{x}^{2}\right)dx={\int }_{0}^{1}8{x}^{2}dx={\left[\frac{8}{3}{x}^{2}\right]}_{0}^{1}=\frac{8}{3}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Figure 3(a):

2. $\phantom{\rule{1em}{0ex}}\int \underset{̲}{F}\cdot \underset{̲}{dr}=\int \left\{{y}^{2}dx+xydy\right\}$ . Along $y=2{x}^{2}$ , $\frac{dy}{dx}=4x$ so $dy=4xdx$ . Then

${\int }_{{C}_{2}}\underset{̲}{F}\cdot \underset{̲}{dr}={\int }_{x=0}^{1}\left\{{\left(2{x}^{2}\right)}^{2}dx+x\left(2{x}^{2}\right)\left(4xdx\right)\right\}={\int }_{0}^{1}12{x}^{4}dx={\left[\frac{12}{5}{x}^{5}\right]}_{0}^{1}=\frac{12}{5}$

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

$\phantom{\rule{2em}{0ex}}{\int }_{{C}_{3}}\underset{̲}{F}\cdot \underset{̲}{dr}={\int }_{O}^{B}\underset{̲}{F}\cdot \underset{̲}{dr}+{\int }_{B}^{A}\underset{̲}{F}\cdot \underset{̲}{dr}$

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

$\phantom{\rule{2em}{0ex}}{\int }_{O}^{B}\underset{̲}{F}\cdot \underset{̲}{dr}={\int }_{x=0}^{1}\left({0}^{2}dx+x×0×0\right)={\int }_{0}^{1}0dx=0$

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

$\phantom{\rule{2em}{0ex}}{\int }_{A}^{B}\underset{̲}{F}\cdot \underset{̲}{dr}={\int }_{y=0}^{2}\left({y}^{2}×0+1×y×dy\right)={\int }_{0}^{2}ydy={\left[\frac{1}{2}{y}^{2}\right]}_{0}^{2}=2.$

Hence ${\int }_{{C}_{3}}\underset{̲}{F}\cdot \underset{̲}{dr}=0+2=2$

Figure 3(c):

Once again, the result is path dependent.

##### Key Point 3

In general, the value of the line integral depends on the path of integration as well as the end points.

##### Example 7

Find ${\int }_{A}^{O}\underset{̲}{F}\cdot \underset{̲}{dr}$ , 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 $y=2{x}^{2}$ 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.

$\phantom{\rule{2em}{0ex}}{\int }_{A}^{O}\underset{̲}{F}\cdot \underset{̲}{dr}=-{\int }_{O}^{A}\underset{̲}{F}\cdot \underset{̲}{dr}=-\frac{8}{3}$

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

$\phantom{\rule{2em}{0ex}}{\oint }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}={\oint }_{{C}_{2}}\underset{̲}{F}\cdot \underset{̲}{dr}+{\oint }_{{C}_{4}}\underset{̲}{F}\cdot \underset{̲}{dr}=\frac{12}{5}-\frac{8}{3}=-\frac{4}{15}\approx -0.267$

##### Example 8

Consider the vector field

$\phantom{\rule{2em}{0ex}}\underset{̲}{F}={y}^{2}{z}^{3}\underset{̲}{i}+2xy{z}^{3}\underset{̲}{j}+3x{y}^{2}{z}^{2}\underset{̲}{k}$

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

$\begin{array}{rcll}{C}_{1}& :& x=t,\phantom{\rule{2em}{0ex}}y=t,\phantom{\rule{2em}{0ex}}z=t\phantom{\rule{2em}{0ex}}\left(0\le t\le 1\right)& \text{}\\ {C}_{2}& :& x={t}^{2},\phantom{\rule{2em}{0ex}}y=t,\phantom{\rule{2em}{0ex}}z={t}^{2}\phantom{\rule{2em}{0ex}}\left(0\le t\le 1\right)& \text{}\end{array}$
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

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

$\phantom{\rule{2em}{0ex}}\underset{̲}{F}={t}^{5}\underset{̲}{i}+2{t}^{5}\underset{̲}{j}+3{t}^{5}\underset{̲}{k}$

Hence $\underset{̲}{F}\cdot \frac{d\underset{̲}{r}}{dt}={t}^{5}+2{t}^{5}+3{t}^{5}=6{t}^{5}$ . 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

$\phantom{\rule{2em}{0ex}}{\int }_{{C}_{1}}\underset{̲}{F}\cdot \underset{̲}{dr}={\int }_{0}^{1}\underset{̲}{F}\cdot \frac{d\underset{̲}{r}}{dt}dt={\int }_{0}^{1}6{t}^{5}dt={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{t}^{6}\right]}_{0}^{1}=1$

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

$\phantom{\rule{2em}{0ex}}{\int }_{{C}_{2}}\underset{̲}{F}\cdot \underset{̲}{dr}={\int }_{0}^{1}10{t}^{9}dt={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{t}^{10}\right]}_{0}^{1}=1$

2. For the closed path $C$

$\phantom{\rule{2em}{0ex}}{\oint }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}={\int }_{{C}_{1}}\underset{̲}{F}\cdot \underset{̲}{dr}-{\int }_{{C}_{2}}\underset{̲}{F}\cdot \underset{̲}{dr}=1-1=0$

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

Further points on Example 8

Note that the line integral of $\underset{̲}{F}$ is $1$ for both paths ${C}_{1}$ and ${C}_{2}$ . In fact, this would hold for any path from $\left(0,0,0\right)$ to $\left(1,1,1\right)$ .

The field $\underset{̲}{F}$ is an example of a conservative vector field ; these are discussed in detail in the next subsection.

In ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ , the vector field $\underset{̲}{F}$ may be the gradient of a scalar field or the curl of a vector field.

Consider the vector field

$\phantom{\rule{2em}{0ex}}\underset{̲}{G}=x\underset{̲}{i}+\left(4x-y\right)\underset{̲}{j}$

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

$\begin{array}{rcll}{C}_{1}& :& x=t,\phantom{\rule{2em}{0ex}}y=t,\phantom{\rule{2em}{0ex}}z=t\phantom{\rule{2em}{0ex}}\left(0\le t\le 1\right)& \text{}\\ {C}_{2}& :& x={t}^{2},\phantom{\rule{2em}{0ex}}y=t,\phantom{\rule{2em}{0ex}}z={t}^{2}\phantom{\rule{2em}{0ex}}\left(0\le t\le 1\right)& \text{}\end{array}$
1. Evaluate the scalar integral ${\int }_{C}\underset{̲}{G}\cdot \underset{̲}{dr}$ 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 $O$ .
1. The path ${C}_{1}$ is given in terms of the parameter $t$ by $x=t$ , $y=t$ and $z=t$ . Hence

Substituting for $x=y=z=t$ in $\underset{̲}{G}$ we have

The limits of integration are $t=0$ and $t=1$ , then

$\phantom{\rule{2em}{0ex}}{\int }_{{C}_{1}}\underset{̲}{G}\cdot \underset{̲}{dr}={\int }_{0}^{1}\underset{̲}{G}\cdot \frac{d\underset{̲}{r}}{dt}dt={\int }_{0}^{1}4tdt={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}2{t}^{2}\right]}_{0}^{1}=2$

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 $\underset{̲}{G}$ we have

$\phantom{\rule{2em}{0ex}}{\int }_{{C}_{2}}\underset{̲}{G}\cdot \underset{̲}{dr}={\int }_{0}^{1}\left(2{t}^{3}+4{t}^{2}-t\right)dt={\left[\frac{1}{2}{t}^{4}+\frac{4}{3}{t}^{3}-\frac{1}{2}{t}^{2}\right]}_{0}^{1}=\frac{4}{3}$

2. For the closed path $C$ $\phantom{\rule{2em}{0ex}}{\oint }_{C}\underset{̲}{G}\cdot \underset{̲}{dr}={\int }_{{C}_{1}}\underset{̲}{G}\cdot \underset{̲}{dr}-{\int }_{{C}_{2}}\underset{̲}{G}\cdot \underset{̲}{dr}=2-\frac{4}{3}=\frac{2}{3}$

(Note that the integral around the closed path is non-zero, unlike Example 8.)

##### Example 9

Find ${\int }_{C}\left\{\underset{̲}{\nabla }\left({x}^{2}y\right)\right\}\cdot \underset{̲}{dr}$ 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 ${\int }_{C}\left\{2xy\phantom{\rule{0.3em}{0ex}}dx+{x}^{2}\phantom{\rule{0.3em}{0ex}}dy\right\}$ .

On $y=2x-{x}^{2}$ , $dy=\left(2-2x\right)\phantom{\rule{0.3em}{0ex}}dx$ so the integral becomes

$\begin{array}{rcll}{\int }_{C}\left\{2xy\phantom{\rule{0.3em}{0ex}}dx+{x}^{2}\phantom{\rule{0.3em}{0ex}}dy\right\}& =& {\int }_{x=0}^{2}\left\{2x\left(2x-{x}^{2}\right)\phantom{\rule{0.3em}{0ex}}dx+{x}^{2}\left(2-2x\right)\phantom{\rule{0.3em}{0ex}}dx\right\}& \text{}\\ & =& {\int }_{0}^{2}\left(6{x}^{2}-4{x}^{3}\right)\phantom{\rule{0.3em}{0ex}}dx={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}2{x}^{3}-{x}^{4}\right]}_{0}^{2}=0\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Evaluate ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ , 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 $\left(2,4\right)$ .
1. ${\int }_{1}^{2}\left(10x-4\right)\phantom{\rule{0.3em}{0ex}}dx=11$ ,
2. ${\int }_{1}^{2}\left(x+{x}^{2}+2{x}^{3}\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{35}{3}$ , (this differs from 1. showing path dependence)
3. ${\int }_{1}^{4}\left(1+y\right)\phantom{\rule{0.3em}{0ex}}dy+{\int }_{1}^{2}\left(x-4\right)\phantom{\rule{0.3em}{0ex}}dx=8$

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 ${C}_{1}$ .
1. $-\frac{1}{3}$ ,
2. $\frac{10}{3}$ ,
3. $-3$ . (note that all these are non-zero.)
##### Exercises
1. Consider ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ , where $\underset{̲}{F}=3{x}^{2}{y}^{2}\underset{̲}{i}+\left(2{x}^{3}y-1\right)\underset{̲}{j}$ . 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. $y=4{x}^{2}$
3. $y=4{x}^{1∕2}$
4. $y=4{x}^{3}$
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 ${\int }_{{C}_{1}}\underset{̲}{F}\cdot \underset{̲}{dr}$ , ${\int }_{{C}_{2}}\underset{̲}{F}\cdot \underset{̲}{dr}$
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 ${\int }_{{C}_{1}}\underset{̲}{G}\cdot \underset{̲}{dr}$ , ${\int }_{{C}_{2}}\underset{̲}{G}\cdot \underset{̲}{dr}$
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 ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ 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. $\underset{̲}{F}=\underset{̲}{\nabla }×\left(\frac{1}{2}{x}^{2}{y}^{2}\underset{̲}{k}\right)$
1. All are $12$ , and in fact the integral would be 12 for any path from (0,0) to (1,4).
1. $2$ , $\frac{5}{3}$ ,
2. $0$ .
1. $0$ , $\frac{1}{3}$ ,
2. $0$ .
2. $16$ , $-16$ .