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
\[xy + H'\left( z \right) = xy + 1,\;\; \Rightarrow H'\left( z \right) = 1,\;\; \Rightarrow H\left( z \right) = z + {C_0}.\]
The final answer is
\[u\left( {x,y,z} \right) = xyz + {y^2} + z + {C_0},\]
where \({C_0}\) is an arbitrary constant.
