Pauls Online Notes
Pauls Online Notes
Home / Calculus III / Line Integrals / Line Integrals of Vector Fields
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 5-4 : Line Integrals of Vector Fields

In the previous two sections we looked at line integrals of functions. In this section we are going to evaluate line integrals of vector fields. We’ll start with the vector field,

\[\vec F\left( {x,y,z} \right) = P\left( {x,y,z} \right)\vec i + Q\left( {x,y,z} \right)\vec j + R\left( {x,y,z} \right)\vec k\]

and the three-dimensional, smooth curve given by

\[\vec r\left( t \right) = x\left( t \right)\vec i + y\left( t \right)\vec j + z\left( t \right)\vec k\hspace{0.25in}\hspace{0.25in}a \le t \le b\]

The line integral of \(\vec F\) along \(C\) is

\[\int\limits_{C}{{\vec F\centerdot d\,\vec r}} = \int_{{\,a}}^{{\,b}}{{\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right)\,dt}}\]

Note the notation in the integral on the left side. That really is a dot product of the vector field and the differential really is a vector. Also, \(\vec F\left( {\vec r\left( t \right)} \right)\) is a shorthand for,

\[\vec F\left( {\vec r\left( t \right)} \right) = \vec F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\]

We can also write line integrals of vector fields as a line integral with respect to arc length as follows,

\[\int\limits_{C}{{\vec F\centerdot d\,\vec r}} = \int\limits_{C}{{\vec F\centerdot \vec T\,ds}}\]

where \(\vec T\left( t \right)\) is the unit tangent vector and is given by,

\[\vec T\left( t \right) = \frac{{\vec r'\left( t \right)}}{{\left\| {\vec r'\left( t \right)} \right\|}}\]

If we use our knowledge on how to compute line integrals with respect to arc length we can see that this second form is equivalent to the first form given above.

\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\,\vec r}} & = \int\limits_{C}{{\vec F\centerdot \vec T\,ds}}\\ & = \int_{{\,a}}^{{\,b}}{{\vec F\left( {\vec r\left( t \right)} \right)\centerdot \frac{{\vec r'\left( t \right)}}{{\left\| {\vec r'\left( t \right)} \right\|}}\,\,\left\| {\vec r'\left( t \right)} \right\|\,dt}}\\ & = \int_{{\,a}}^{{\,b}}{{\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right)\,dt}}\end{align*}\]

In general, we use the first form to compute these line integral as it is usually much easier to use. Let’s take a look at a couple of examples.

Example 1 Evaluate \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\,\vec r}}\) where \(\vec F\left( {x,y,z} \right) = 8{x^2}y\,z\,\vec i + 5z\,\vec j - 4x\,y\,\vec k\) and \(C\) is the curve given by \(\vec r\left( t \right) = t\,\vec i + {t^2}\,\vec j + {t^3}\,\vec k\), \(0 \le t \le 1\).
Show Solution

Okay, we first need the vector field evaluated along the curve.

\[\vec F\left( {\vec r\left( t \right)} \right) = 8{t^2}\left( {{t^2}} \right)\left( {{t^3}} \right)\vec i + 5{t^3}\,\vec j - 4t\left( {{t^2}} \right)\vec k = 8{t^7}\,\vec i + 5{t^3}\,\vec j - 4{t^3}\,\vec k\]

Next, we need the derivative of the parameterization.

\[\vec r'\left( t \right) = \,\vec i + 2t\,\vec j + 3{t^2}\,\vec k\]

Finally, let’s get the dot product taken care of.

\[\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) = 8{t^7} + 10{t^4} - 12{t^5}\]

The line integral is then,

\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\,\vec r}} & = \int_{{\,0}}^{{\,1}}{{8{t^7} + 10{t^4} - 12{t^5}\,dt}}\\ & = \left. {\left( {{t^8} + 2{t^5} - 2{t^6}} \right)} \right|_0^1\\ & = 1\end{align*}\]
Example 2 Evaluate \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\,\vec r}}\) where \(\vec F\left( {x,y,z} \right) = x\,z\,\vec i - y\,z\,\vec k\) and \(C\) is the line segment from \(\left( { - 1,2,0} \right)\) to \(\left( {3,0,1} \right)\).
Show Solution

We’ll first need the parameterization of the line segment. We saw how to get the parameterization of line segments in the first section on line integrals. We’ve been using the two dimensional version of this over the last couple of sections. Here is the parameterization for the line.

\[\begin{align*}\vec r\left( t \right) & = \left( {1 - t} \right)\left\langle { - 1,2,0} \right\rangle + t\left\langle {3,0,1} \right\rangle \\ & = \left\langle {4t - 1,2 - 2t,t} \right\rangle ,\hspace{1.0in} 0 \le t \le 1\end{align*}\]

So, let’s get the vector field evaluated along the curve.

\[\begin{align*}\vec F\left( {\vec r\left( t \right)} \right) & = \left( {4t - 1} \right)\left( t \right)\,\vec i - \left( {2 - 2t} \right)\left( t \right)\,\vec k\\ & = \left( {4{t^2} - t} \right)\vec i - \left( {2t - 2{t^2}} \right)\vec k\end{align*}\]

Now we need the derivative of the parameterization.

\[\vec r'\left( t \right) = \left\langle {4, - 2,1} \right\rangle \]

The dot product is then,

\[\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) = 4\left( {4{t^2} - t} \right) - \left( {2t - 2{t^2}} \right) = 18{t^2} - 6t\]

The line integral becomes,

\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\,\vec r}}& = \int_{{\,0}}^{{\,1}}{{18{t^2} - 6t\,dt}}\\ & = \left. {\left( {6{t^3} - 3{t^2}} \right)} \right|_0^1\\ & = 3\end{align*}\]

Let’s close this section out by doing one of these in general to get a nice relationship between line integrals of vector fields and line integrals with respect to \(x\), \(y\), and \(z\).

Given the vector field \(\vec F\left( {x,y,z} \right) = P\,\vec i + Q\,\vec j + R\,\vec k\) and the curve \(C\) parameterized by \(\vec r\left( t \right) = x\left( t \right)\vec i + y\left( t \right)\vec j + z\left( t \right)\vec k\), \(a \le t \le b\) the line integral is,

\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\,\vec r}} & = \int_{{\,a}}^{{\,b}}{{\left( {P\,\vec i + Q\,\vec j + R\,\vec k} \right)\centerdot \left( {x'\,\vec i + y'\,\vec j + z'\,\vec k} \right)\,dt}}\\ & = \int_{{\,a}}^{{\,b}}{{Px' + Qy' + Rz'\,dt}}\\ & = \int_{{\,a}}^{{\,b}}{{Px'\,dt}} + \int_{{\,a}}^{{\,b}}{{Qy'\,dt}} + \int_{{\,a}}^{{\,b}}{{Rz'\,dt}}\\ & = \int\limits_{C}{{P\,dx}} + \int\limits_{C}{{Q\,dy}} + \int\limits_{C}{{R\,dz}}\\ & = \int\limits_{C}{{P\,dx}} + Q\,dy + R\,dz\end{align*}\]

So, we see that,

\[\int\limits_{C}{{\vec F\centerdot d\,\vec r}} = \int\limits_{C}{{P\,dx}} + Q\,dy + R\,dz\]

Note that this gives us another method for evaluating line integrals of vector fields.

This also allows us to say the following about reversing the direction of the path with line integrals of vector fields.

Fact

\[\int\limits_{{ - C}}{{\vec F\centerdot d\,\vec r}} = - \int\limits_{C}{{\vec F\centerdot d\,\vec r}}\]

This should make some sense given that we know that this is true for line integrals with respect to \(x\), \(y\), and/or \(z\) and that line integrals of vector fields can be defined in terms of line integrals with respect to \(x\), \(y\), and \(z\).