Calculus

Line Integrals

Path Independence of Line Integrals

Page 1
Problem 1
Page 2
Problems 2-5

Definitions

The line integral of a vector function \(\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\) is said to be path independent, if and only if \(P,\) \(Q\) and \(R\) are continuous in a domain \(D,\) and if there exists some scalar function \(u = u\left( {x,y,z} \right)\) in \(D\) such that
\[
{\mathbf{F} = \text{grad}\,u\;\;\;}\kern-0.3pt{\text{ or }\;\;\frac{{\partial u}}{{\partial x}} = P,\;\;\;}\kern-0.3pt
{\frac{{\partial u}}{{\partial y}} = Q,\;\;\;}\kern-0.3pt
{\frac{{\partial u}}{{\partial z}} = R.}
\] If this is the case, then the line integral of \(\mathbf{F}\) along the curve \(C\) from \(A\) to \(B\) is given by the formula
\[
{\int\limits_C {\mathbf{F}\left( \mathbf{r} \right) \cdot d\mathbf{r}} }
= {\int\limits_C {Pdx + Qdy + Rdz} }
= {u\left( B \right) – u\left( A \right).}
\] (This result for line integrals is analogous to the Fundamental Theorem of Calculus for functions of one variable).

Hence, if the line integral is path independent, then for any closed contour \(C\)
\[\oint\limits_C {\mathbf{F}\left( \mathbf{r} \right) \cdot d\mathbf{r} = 0}.\] A vector field of the form \(\mathbf{F} = \text{grad}\,u\) is called a conservative field, and the function \(u = u\left( {x,y,z} \right)\) is called a scalar potential.

A Test for a Conservative Field

The line integral of a vector function \(\mathbf{F} = P\mathbf{i} + Q\mathbf{j} \) \(+ R\mathbf{k}\) is path independent if and only if
\[
{\text{rot}\,\mathbf{F} }={ \left| {\begin{array}{*{20}{c}}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\
P&Q&R
\end{array}} \right| }={ \mathbf{0}.}
\] It’s implied that each component of \(\mathbf{F}\) has continuous partial derivatives of variables \(x, y\) and \(z.\)

If the line integral is taken in the \(xy\)-plane, then the following formula is valid:
\[{\int\limits_C {Pdx + Qdy} }={ u\left( B \right) – u\left( A \right).}\] In this case, the test for determining if a vector field is conservative can be written in the form
\[\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}.\] The considered test is the necessary requirement, but generally speaking, it is not sufficient condition for a vector field to be conservative. However, this test is sufficient, if the region of integration \(D\) is simply connected.

Solved Problems

Click on problem description to see solution.

 Example 1

Evaluate the line integral \(\int\limits_{AB} {\left( {x + y} \right)dx + xdy} \) for two paths of integration:

  1. \(AB\) is the line segment from \(A\left( {0,0} \right)\) to \(B\left( {1,1} \right)\);
  2. \(AB\) is the parabola \(y = {x^2}\) from \(A\left( {0,0} \right)\) to \(B\left( {1,1} \right)\).

 Example 2

Show that the line integral \(\int\limits_{AB} {\left( {3{x^2}y + y} \right)dx }\) \(+{ \left( {{x^3} + x} \right)dy}\) is path independent and calculate this integral. The coordinates of the points \(A, B\) are \(A\left( {1,2} \right),\) \(B\left( {4,5} \right).\)

 Example 3

Determine if the vector field \(\mathbf{F} =\) \(\left( {yz,xz,xy} \right)\) 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.

 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.

Example 1.

Evaluate the line integral \(\int\limits_{AB} {\left( {x + y} \right)dx + xdy} \) for two paths of integration:

  1. \(AB\) is the line segment from \(A\left( {0,0} \right)\) to \(B\left( {1,1} \right)\);
  2. \(AB\) is the parabola \(y = {x^2}\) from \(A\left( {0,0} \right)\) to \(B\left( {1,1} \right)\).

Solution.

Consider the first case. Obviously, the equation of the line is \(y = x.\) Then using the formula
\[
{\int\limits_{AB} {P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy} \text{ = }}\kern0pt
{\int\limits_a^b {\left[ {P\left( {x,y} \right) + Q\left( {x,y} \right)\frac{{dy}}{{dx}}} \right]dx},}
\] we obtain
\[
{{I_1} }={ \int\limits_{AB} {\left( {x + y} \right)dx + xdy} }
= {\int\limits_0^1 {\left( {x + x + x \cdot 1} \right)dx} }
= {\int\limits_0^1 {3xdx} }
= {3\left[ {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1} \right] }
= {\frac{3}{2}.}
\] If the curve \(AB\) is parabola \(y = {x^2},\) we have
\[
{{I_2} }={ \int\limits_{AB} {\left( {x + y} \right)dx + xdy} }
= {\int\limits_0^1 {\left( {x + {x^2} + x \cdot 2x} \right)dx} }
= {\int\limits_0^1 {\left( {x + 3{x^2}} \right)dx} }
= {\left. {\left( {\frac{{{x^2}}}{2} + \frac{{3{x^3}}}{3}} \right)} \right|_0^1 }
= {\frac{1}{2} + 1 }={ \frac{3}{2},}
\] i.e. we have obtained the same answer.

Apply the test \({\large\frac{{\partial P}}{{\partial y}}\normalsize} = {\large\frac{{\partial Q}}{{\partial x}}\normalsize}\) to determine if the vector field is conservative.
\[
{\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}},\;\;}\Rightarrow
{\frac{{\partial \left( {x + y} \right)}}{{\partial y}} = \frac{{\partial x}}{{\partial x}},\;\;}\Rightarrow
{1 \equiv 1.}
\] As it can be seen, the vector field \(\mathbf{F} = \left( {x + y,x} \right)\) is conservative. This explains the result that the line integral is path independent.

Page 1
Problem 1
Page 2
Problems 2-5