Path Independence of Line Integrals

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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

\[\mathbf{F} = \text{grad}\,u\;\; \text{ or }\;\;\frac{{\partial u}}{{\partial x}} = P,\;\; \frac{{\partial u}}{{\partial y}} = Q,\;\; \frac{{\partial u}}{{\partial z}} = R.\]

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 holds:

\[\int\limits_C {Pdx + Qdy} = u\left( B \right) - u\left( A \right).\]

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 or tap a problem to see the solution.

Example 1

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

\(AB\) is the line segment from \(A\left( {0,0} \right)\) to \(B\left( {1,1} \right)\);
\(AB\) 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 1.

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

\(AB\) is the line segment from \(A\left( {0,0} \right)\) to \(B\left( {1,1} \right)\);
\(AB\) 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

\[\int\limits_{AB} {P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy} = \int\limits_a^b {\left[ {P\left( {x,y} \right) + Q\left( {x,y} \right)\frac{{dy}}{{dx}}} \right]dx},\]

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

\[{I_2} = \int\limits_{AB} {\left( {x + y} \right)dx + xdy} = \int\limits_0^1 {\left( {x + {x^2} + x \cdot 2x} \right)dx} = \int\limits_0^1 {\left( {x + 3{x^2}} \right)dx} = \left. {\left( {\frac{{{x^2}}}{2} + \frac{{3{x^3}}}{3}} \right)} \right|_0^1 = \frac{1}{2} + 1 = \frac{3}{2},\]

that is we have obtained the same answer.

Apply the test \({\frac{{\partial P}}{{\partial y}}} = {\frac{{\partial Q}}{{\partial x}}}\) 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.

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 \(A\left( {1,2} \right),\) \(B\left( {4,5} \right).\)

Solution.

Since the components of the vector field

\[P\left( {x,y} \right) = 3{x^2}y + y,\;\; Q\left( {x,y} \right) = {x^3} + x\]

and their partial derivatives

\[\frac{{\partial P}}{{\partial y}} = \frac{{\partial \left( {3{x^2}y + y} \right)}}{{\partial y}} = 3{x^2} + 1,\;\;\; \frac{{\partial Q}}{{\partial x}} = \frac{{\partial \left( {{x^3} + x} \right)}}{{\partial x}} = 3{x^2} + 1\]

are continuous, and the test \(\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}\) is satisfied, we conclude that the vector field

\[\mathbf{F} = \left( {3{x^2}y + y,{x^3} + x} \right)\]

is conservative and, hence, the line integral is path independent. To evaluate the line integral, we notice that

\[\left( {3{x^2}y + y} \right)dx + \left( {{x^3} + x} \right)dy = \left( {3{x^2}ydx + {x^3}dy} \right) + \left( {ydx + xdy} \right) = d\left( {{x^3}y} \right) + d\left( {xy} \right) = d\left( {{x^3}y + xy} \right) = du,\]

so the scalar potential is \(u = {{x^3}y + xy}.\) Then by the formula

\[\int\limits_{AB} {Pdx + Qdy} = u\left( B \right) - u\left( A \right),\]

we find the integral:

\[I = \int\limits_{AB} {\left( {3{x^2}y + y} \right)dx + \left( {{x^3} + x} \right)dy} = u\left( B \right) - u\left( A \right) = \left( {{4^3} \cdot 5 + 4 \cdot 5} \right) - \left( {{1^3} \cdot 2 + 1 \cdot 2} \right) = 336.\]