# Path Independence of Line Integrals

## Solved Problems

### Example 3.

Determine if the vector field $\mathbf{F} = \left( {yz,xz,xy} \right)$ is conservative?

Solution.

As $$P = yz,$$ $$Q = xz$$ and $$R = xy,$$ the curl of the vector field is

$\text{rot}\,\mathbf{F} = \nabla \times \mathbf{F} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {yz}&{xz}&{xy} \end{array}} \right| = \left( {\frac{{\partial \left( {xy} \right)}}{{\partial y}} - \frac{{\partial \left( {xz} \right)}}{{\partial z}}} \right)\mathbf{i} + \left( {\frac{{\partial \left( {yz} \right)}}{{\partial z}} - \frac{{\partial \left( {xy} \right)}}{{\partial x}}} \right)\mathbf{j} + \left( {\frac{{\partial \left( {xz} \right)}}{{\partial x}} - \frac{{\partial \left( {yz} \right)}}{{\partial y}}} \right)\mathbf{k} = \left( {x - x} \right)\mathbf{i} + \left( {y - y} \right)\mathbf{j} + \left( {z - z} \right)\mathbf{k} = \mathbf{0}.$

Hence, the vector field $$\mathbf{F}$$ is conservative.

### Example 4.

Determine if the vector field $\mathbf{F}\left( {x,y} \right) = \left( {x + y,x - y} \right)$ is conservative? If it is, find its potential.

Solution.

The components of the vector field are $$P\left( {x,y} \right) = x + y,$$ $$Q\left( {x,y} \right) = x - y.$$ It is easy to see that

$\frac{{\partial P}}{{\partial y}} = 1,\;\; \frac{{\partial Q}}{{\partial x}} = 1,\;\; \Rightarrow \frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}.$

Thus, the given vector field is conservative.

To find its potential, we first integrate $$P\left( {x,y} \right) = x + y$$ with respect to $$x.$$

$u\left( {x,y} \right) = \int {P\left( {x,y} \right)dx} + C\left( y \right) = \int {\left( {x + y} \right)dx} + C\left( y \right) = \frac{{{x^2}}}{2} + yx + C\left( y \right).$

Now we determine $$C\left( y \right)$$ by setting the partial derivative $$\frac{{\partial u}}{{\partial y}}$$ equal to $$Q\left( {x,y} \right).$$

$\frac{{\partial u}}{{\partial y}} = \frac{{\partial \left( {\frac{{{x^2}}}{2} + yx + C\left( y \right)} \right)}}{{\partial y}} = x + C'\left( y \right) = x - y.$

Hence, $$C'\left( y \right) = - y.$$ Then

$C\left( y \right) = \int {\left( { - y} \right)dy} = - \frac{{{y^2}}}{2} + {C_1},$

where $${C_1}$$ is a constant, so the scalar potential of the field is

$u\left( {x,y} \right) = \frac{{{x^2}}}{2} + yx - \frac{{{y^2}}}{2} + {C_1}.$

### Example 5.

Determine if the vector field

$\mathbf{F}\left( {x,y,z} \right) = \left( {yz,xz + 2y,xy + 1} \right)$

is conservative. If it is, find its potential.

Solution.

Here $$P = yz,$$ $$Q = xz + 2y,$$ $$R = xy + 1.$$ Calculate the curl of the vector field.

$\text{rot}\,\mathbf{F} = \nabla \times \mathbf{F} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {yz} & {xz + 2y} & {xy + 1} \end{array}} \right| = \left( {\frac{{\partial \left( {xy + 1} \right)}}{{\partial y}} - \frac{{\partial \left( {xz + 2y} \right)}}{{\partial z}}} \right)\mathbf{i} + \left( {\frac{{\partial \left( {yz} \right)}}{{\partial z}} - \frac{{\partial \left( {xy + 1} \right)}}{{\partial x}}} \right)\mathbf{j} + \left( {\frac{{\partial \left( {xz + 2y} \right)}}{{\partial x}} - \frac{{\partial \left( {yz} \right)}}{{\partial y}}} \right)\mathbf{k} = \left( {x - x} \right)\mathbf{i} + \left( {y - y} \right)\mathbf{j} + \left( {z - z} \right)\mathbf{k} = \mathbf{0}.$

Hence, the vector field $$\mathbf{F}$$ is conservative. To find its potential, we integrate $$P\left( {x,y,z} \right)$$ with respect to the variable $$x.$$

$u\left( {x,y,z} \right) = \int {P\left( {x,y,z} \right)dx} + G\left( {y,z} \right) = \int {yzdx} + G\left( {y,z} \right) = xyz + G\left( {y,z} \right).$

In the above integral the variables $$y$$ and $$z$$ are treated as constants.

Now we differentiate the potential $$u$$ with respect to the variable $$y$$ and set $$\frac{{\partial u}}{{\partial y}}$$ equal to $$Q$$ to get

$\frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {xyz + G\left( {y,z} \right)} \right] = xz + {G'_y}\left( {y,z} \right) = xz + zy.$

We see from the last equation that $${G'_y}\left( {y,z} \right) = 2y.$$

To determine $$G\left( {y,z} \right)$$ we integrate with respect to $$y$$ and add a constant function $$H\left( {z} \right).$$

$G\left( {y,z} \right) = \int {{G'_y}\left( {y,z} \right)dy} + H\left( z \right) = \int {2ydy} + H\left( z \right) = {y^2} + H\left( z \right).$

Thus, the scalar potential is

$u\left( {x,y,z} \right) = xyz + {y^2} + H\left( z \right).$

Finally,

$\frac{{\partial u}}{{\partial z}} = \frac{\partial }{{\partial z}}\left[ {xyz + {y^2} + H\left( z \right)} \right] = xy + H'\left( z \right).$

so that setting $$\frac{{\partial u}}{{\partial z}}$$ equal to $$R = xy + 1$$ yields

$xy + H'\left( z \right) = xy + 1,\;\; \Rightarrow H'\left( z \right) = 1,\;\; \Rightarrow H\left( z \right) = z + {C_0}.$

$u\left( {x,y,z} \right) = xyz + {y^2} + z + {C_0},$
where $${C_0}$$ is an arbitrary constant.