Line Integrals of Vector Fields

Definition

Suppose that a curve C is defined by the vector function r = r (s), 0 ≤ s ≤ S, where s is the arc length of the curve. Then the derivative of the vector function

\[\frac{{d\mathbf{r}}}{{dt}} = \boldsymbol{\tau} = \left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right)\]

is the unit vector of the tangent line to this curve (Figure 1).

The derivative of the vector function — Figure 1.

Here \(\alpha, \beta\) and \(\gamma\) are the angles between the tangent line and the positive axis \(Ox, Oy\) and \(Oz,\) respectively.

We introduce the vector function \(\mathbf{F}\left( {P,Q,R} \right)\) defined over the curve \(C\) so that for the scalar function

\[\mathbf{F} \cdot \boldsymbol{\tau} = P\cos \alpha + Q\cos \beta + R\cos \gamma ,\]

the line integral \(\int\limits_C {\left( {\mathbf{F} \cdot \boldsymbol{\tau}} \right)ds}\) exists. Such an integral \(\int\limits_C {\left( {\mathbf{F} \cdot \boldsymbol{\tau}} \right)ds}\) is called the line integral of the vector field \(\mathbf{F}\) along the curve \(C\) and is denoted as

\[\int\limits_C {Pdx + Qdy + Rdz} .\]

Thus, by definition,

\[\int\limits_C {Pdx + Qdy + Rdz} = \int\limits_0^S {\left( {P\cos \alpha + Q\cos \beta + R\cos \gamma } \right)ds},\]

where \(\boldsymbol{\tau} \left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right)\) is the unit vector of the tangent line to the curve \(C.\)

The latter formula can be written in the vector form:

\[\int\limits_C {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} = \int\limits_0^S {\left( {\mathbf{F}\left( {\mathbf{r}\left( s \right)} \right) \cdot \boldsymbol{\tau} } \right)ds} ,\]

where \(d\mathbf{r} = \left( {dx,dy,dz} \right).\)

If a curve \(C\) lies in the \(xy\)-plane and \(R = 0\), we can write:

\[\int\limits_C {Pdx + Qdy} = \int\limits_0^S {\left( {P\cos \alpha + Q\cos \beta } \right)ds} .\]

Properties of Line Integrals of Vector Fields

The line integral of vector function has the following properties:

Let \(C\) denote the curve \(AB\) which is traversed from \(A\) to \(B,\) and let \(-C\) denote the curve \(BA\) with the opposite orientation − from \(B\) to \(A.\) Then
\[\int\limits_{ - C} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} = - \int\limits_C {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} ;\]
If \(C\) is the union of the curves \({C_1}\) and \({C_2}\) (Figure \(2\)), then
\[\int\limits_C {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} = \int\limits_{{C_1} \cup {C_2}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} = \int\limits_{{C_1}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} + \int\limits_{{C_2}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} ;\]

Union of the two curves C1 and C2 — Figure 2.

If the curve \(C\) is parameterized by \(\mathbf{r}\left( t \right) = \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right),\) \(\alpha \le t \le \beta ,\) then
\[\int\limits_C {Pdx + Qdy + Rdz} = \int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dx}}{{dt}} + Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dy}}{{dt}} + R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dz}}{{dt}}} \right]dt} .\]
If \(C\) lies in the \(xy\)-plane and is given by the equation \(y = f\left( x \right)\) (in this case \(R = 0\) and \(t = x\)), then the latter formula can be written as
\[\int\limits_C {Pdx + Qdy} = \int\limits_a^b {\Big[ {P\left( {x,f\left( x \right)} \right) + Q\left( {x,f\left( x \right)} \right)\frac{{df}}{{dx}}} \Big]dx} .\]

Solved Problems

Example 1.

Evaluate the integral \[\int\limits_C {ydx - xdy}\] over the curve \(C\) parameterized by

\[\mathbf{r}\left( t \right) = \left( {\cos t,\sin t} \right), 0 \le t \le {\frac{\pi }{2}}.\]

Solution.

Using the formula

\[\int\limits_C {Pdx + Qdy} = \int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right)} \right)\frac{{dx}}{{dt}} + Q\left( {x\left( t \right),y\left( t \right)} \right)\frac{{dy}}{{dt}}} \right]dt} ,\]

we find the answer:

\[\int\limits_C {ydx - xdy} = \int\limits_0^{\frac{\pi }{2}} {\left( {y\frac{{dx}}{{dt}} - x\frac{{dy}}{{dt}}} \right)dt} = \int\limits_0^{\frac{\pi }{2}} {\left( {\sin t\frac{{d\left( {\cos t} \right)}}{{dt}} - \cos t\frac{{d\left( {\sin t} \right)}}{{dt}}} \right)dt} = \int\limits_0^{\frac{\pi }{2}} {\left( {\sin t\left( { - \sin t} \right) - \cos t\cos t} \right)dt} = - \int\limits_0^{\frac{\pi }{2}} {\left( {{{\sin }^2}t + {{\cos }^2}t} \right)dt} = - \int\limits_0^{\frac{\pi }{2}} {dt} = - \frac{\pi }{2}.\]

Example 2.

Evaluate the line integral \[\int\limits_C {xdy - ydx} \] along the curve \(C\) defined by the equation \(y = {x^3}\) from the origin \(\left( {0,0} \right)\) to \(\left( {2,8} \right).\)

Solution.

To find the given integral, we use the formula

\[\int\limits_C {Pdx + Qdy} = \int\limits_a^b {\left[ {P\left( {x,f\left( x \right)} \right) + Q\left( {x,f\left( x \right)} \right)\frac{{df}}{{dx}}} \right]dx} .\]

Substituting \(y = {x^3}\) and \(dy = 3{x^2}dx\) in the integrand, we obtain

\[\int\limits_C {xdy - ydx} = \int\limits_0^2 {x \cdot 3{x^2}dx - {x^3}dx} = \int\limits_0^2 {2{x^3}dx} = 2\left[ {\left. {\left( {\frac{{{x^4}}}{4}} \right)} \right|_0^8} \right] = 8.\]

See more problems on Page 2.