### 3 Conservative vector fields

For some line integrals in the previous section, the integral depended only on the vector field $\underset{̲}{F}$ and the start and end points of the line but not on the actual path of the line between the start and end points. However, for other line integrals, the result depended on the actual details of the path of the line.

Vector fields are classified according to whether the line integrals are path dependent or path independent. Those vector fields for which all line integrals between all pairs of points are path independent are called conservative vector fields .

There are five properties of a conservative vector field (P1 to P5). It is impossible to check the value of every line integral over every path, but instead it is possible to use any one of these five properties (and in particular property P3 below) to determine whether a vector field is conservative. They are also used to simplify calculations with conservative vector fields.

P1.
The line integral ${\int }_{A}^{B}\underset{̲}{F}\cdot \underset{̲}{dr}$ depends only on the end points $A$ and $B$ and is independent of the actual path taken.
P2.
The line integral around any closed curve is zero. That is ${\oint }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}=0$ for all $C$ .
P3.
The curl of a conservative vector field $\underset{̲}{F}$ is zero i.e. $\underset{̲}{\nabla }×\underset{̲}{F}=\underset{̲}{0}$ .
P4.
For any conservative vector field $\underset{̲}{F}$ , it is possible to find a scalar field $\varphi$ such that $\underset{̲}{\nabla }\varphi =\underset{̲}{F}$ . Then, ${\oint }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}=\varphi \left(B\right)-\varphi \left(A\right)$ where $A$ and $B$ are the start and end points of contour $C$ .

[This is sometimes called the Fundamental Theorem of Line Integrals and is comparable with the Fundamental Theorems of Calculus.]
P5.
All gradient fields are conservative. That is, $\underset{̲}{F}=\underset{̲}{\nabla }\varphi$ is a conservative vector field for any scalar field $\varphi$ .
##### Example 10

The following vector fields were considered in the Examples of the previous subsection.

1. ${\underset{̲}{F}}_{1}={y}^{2}\underset{̲}{i}+xy\underset{̲}{j}$ (Example 6)
2. ${\underset{̲}{F}}_{2}=2x\underset{̲}{i}+2y\underset{̲}{j}$ (Example 7)
3. ${\underset{̲}{F}}_{3}={y}^{2}{z}^{3}\underset{̲}{i}+2xy{z}^{3}\underset{̲}{j}+3x{y}^{2}{z}^{2}\underset{̲}{k}$ (Example 8)
4. ${\underset{̲}{F}}_{4}=x\underset{̲}{i}+\left(4x-y\right)\underset{̲}{j}$ (Task on page 13)

Determine which of these vector fields are conservative e.g. by referring to the answers given in the solution. For those that are conservative find a scalar field $\varphi$ such that $\underset{̲}{F}=\underset{̲}{\nabla }\varphi$ and use property P4 to verify the line integrals found.

##### Solution
1. Two different values were obtained for line integrals over the paths ${C}_{1}$ and ${C}_{2}$ . Hence, by P1, ${\underset{̲}{F}}_{1}$ is not conservative. [It is also possible to reach this conclusion from P3 by finding that $\underset{̲}{\nabla }×\underset{̲}{F}=-y\underset{̲}{k}\ne \underset{̲}{0}$ .]
2. Both line integrals from $\left(0,0\right)$ to $\left(4,2\right)$ had the same value i.e. $20$ and for the closed path the line integral was $0$ . This alone does not mean that ${\underset{̲}{F}}_{2}$ is conservative as there could be other untried paths giving different values. So by using P3

$\phantom{\rule{2em}{0ex}}\underset{̲}{\nabla }×{\underset{̲}{F}}_{2}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill \\ \hfill 2x\hfill & \hfill 2y\hfill & \hfill 0\hfill \end{array}\right|=\underset{̲}{i}\left(0-0\right)-\underset{̲}{j}\left(0-0\right)+\underset{̲}{k}\left(0-0\right)=\underset{̲}{0}$

As $\underset{̲}{\nabla }×{\underset{̲}{F}}_{2}=\underset{̲}{0}$ , P3 gives that ${\underset{̲}{F}}_{2}$ is a conservative vector field.

Now, find a $\varphi$ such that ${\underset{̲}{F}}_{2}=\underset{̲}{\nabla }\varphi$ . Then $\frac{\partial \varphi }{\partial x}\underset{̲}{i}+\frac{\partial \varphi }{\partial y}\underset{̲}{j}=2x\underset{̲}{i}+2y\underset{̲}{j}.$

Thus

Using P4: ${\int }_{\left(0,0\right)}^{\left(4,2\right)}{\underset{̲}{F}}_{2}\cdot \underset{̲}{dr}={\int }_{\left(0,0\right)}^{\left(4,2\right)}\left(\underset{̲}{\nabla }\varphi \right)\cdot \underset{̲}{dr}=\varphi \left(4,2\right)-\varphi \left(0,0\right)=\left({4}^{2}+{2}^{2}\right)-\left({0}^{2}+{0}^{2}\right)=20$ .

3. The fact that line integrals along two different paths between the same start and end points is consistent with ${\underset{̲}{F}}_{3}$ being a conservative field according to P1. So too is the fact that the integral around a closed path is zero according to P2. However, neither fact can be used to conclude that ${\underset{̲}{F}}_{3}$ is a conservative field. This can be done by showing that $\underset{̲}{\nabla }×{\underset{̲}{F}}_{3}=\underset{̲}{0}$ .

Now, $\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill {y}^{2}{z}^{3}\hfill & \hfill 2xy{z}^{3}\hfill & \hfill 3x{y}^{2}{z}^{2}\hfill \end{array}\right|=\left(6xy{z}^{2}-6xy{z}^{2}\right)\underset{̲}{i}-\left(3{y}^{2}{z}^{2}-3{y}^{2}{z}^{2}\right)\underset{̲}{j}+\left(2y{z}^{3}-2y{z}^{3}\right)\underset{̲}{k}=\underset{̲}{0}$ .

As $\underset{̲}{\nabla }×{\underset{̲}{F}}_{3}=\underset{̲}{0}$ , P3 gives that ${\underset{̲}{F}}_{3}$ is a conservative field.

To find $\varphi$ that satisfies $\underset{̲}{\nabla }\varphi ={\underset{̲}{F}}_{3}$ , it is necessary to satisfy

$\phantom{\rule{2em}{0ex}}\begin{array}{ccc}\frac{\partial \varphi }{\partial x}={y}^{2}{z}^{3}\hfill & \hfill \to \hfill & \hfill \varphi =x{y}^{2}{z}^{3}+f\left(y,z\right)\hfill \\ \hfill \\ \frac{\partial \varphi }{\partial y}=2xy{z}^{3}\hfill & \hfill \to \hfill & \hfill \varphi =x{y}^{2}{z}^{3}+g\left(x,z\right)\hfill \\ \hfill \\ \frac{\partial \varphi }{\partial z}=3x{y}^{2}{z}^{2}\hfill & \hfill \to \hfill & \hfill \varphi =x{y}^{2}{z}^{3}+h\left(x,y\right)\hfill \end{array}}\to \varphi =x{y}^{2}{z}^{3}$

Using P4: ${\int }_{\left(0,0,0\right)}^{\left(1,1,1\right)}{\underset{̲}{F}}_{3}\cdot \underset{̲}{dr}=\varphi \left(1,1,1\right)-\varphi \left(0,0,0\right)=1-0=1$ .

4. As the integral along ${C}_{1}$ is $2$ and the integral along ${C}_{2}$ (same start and end points but different intermediate points) is $\frac{4}{3}$ , ${F}_{4}$ is not a conservative field using P1.

Note that $\underset{̲}{\nabla }×{\underset{̲}{F}}_{4}=4\underset{̲}{k}\ne \underset{̲}{0}$ so, using P3, this is an independent conclusion that ${\underset{̲}{F}}_{4}$ is not conservative.