Multivariable calculus and vector analysis

Line integral examples

Example 1

If a force is given by

$$\begin{array}{lll}\hfill \mathbf{F}\left(x,y\right)=\left(0,x\right),& & \hfill \end{array}$$

compute the work done by the force field on a particle that moves along the curve $C$ that is the counter-clockwise quarter unit circle in the first quadrant. In the below picture, the curve $C$ is plotted by the long green curved arrow. The vector field $\mathbf{F}$ is represented by the vertical black arrows.

Solution: First, can you see what the sign of the integral should be? Notice that curve and the vector field are mostly going in the same direction. The tangent vector to the curve and the vector field always make an angle less that $\pi ∕2$. So, their dot product must be positive everywhere along the curve, which means the integral must be positive.

The first step to computing the integral is to parametrize the curve $C$. We’ll parametrize it by

$$\begin{array}{lll}\hfill \mathbf{c}\left(t\right)=\left(cost,sint\right),0\le t\le \frac{\pi }{2}.& & \hfill \end{array}$$

so that the derivative is

$$\begin{array}{lll}\hfill {\mathbf{c}}^{\text{'}}\left(t\right)=\left(-sint,cost\right).& & \hfill \end{array}$$

The work is the line integral

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}& ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\mathbf{F}\left(\mathbf{c}\left(t\right)\right)\cdot {\mathbf{c}}^{\text{'}}\left(t\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\mathbf{F}\left(cost,sint\right)\cdot \left(-sint,cost\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\left(0,cost\right)\cdot \left(-sint,cost\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}{cos}^{2}tdt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\frac{1}{2}\left(1+cos2t\right)dt& \hfill & \\ \hfill & =\frac{1}{2}{\left.\left(t+\frac{sin2t}{2}\right)\right|}_0^{\pi ∕2}=\frac{1}{2}\left(\frac{\pi }{2}-0\right)=\frac{\pi }{4}.& \hfill & \end{array}$$

It’s a good thing the answer is positive, so it agrees with what we read off the picture.

Example 1’

The integral ${\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}$ does not depend on parameterization, as long as we parametrize $C$ to go in the counter-clockwise direction. To see this, calculate the integral using the parametrization

$$\begin{array}{lll}\hfill \mathbf{p}\left(t\right)=\left(\sqrt{1-t^2},t\right),0\le t\le 1.& & \hfill \end{array}$$

Solution: Since

$$\begin{array}{lll}\hfill {\mathbf{p}}^{\text{'}}\left(t\right)=\left(\frac{-t}{\sqrt{1-t^2}},1\right),& & \hfill \end{array}$$

the work is

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}& ={\int \; <!--nolimits\; -->}_0^1\mathbf{F}\left(\mathbf{p}\left(t\right)\right)\cdot {\mathbf{p}}^{\text{'}}\left(t\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\mathbf{F}\left(\sqrt{1-t^2},t\right)\cdot \left(\frac{-t}{\sqrt{1-t^2}},1\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\left(0,\sqrt{1-t^2}\right)\cdot \left(\frac{-t}{\sqrt{1-t^2}},1\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\sqrt{1-t^2}dt.& \hfill & \end{array}$$

To complete this integral, use the $u$-substitution $t=sinu$, so that $dt=cosudu$. Since $t$ goes from 0 to 1, $u$ must go from 0 to $\pi ∕2$. Thus, the integral becomes

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}& ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\sqrt{1-{sin}^{2}u}cosudu& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\sqrt{{cos}^{2}u}cosudu& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}{cos}^{2}udu.& \hfill & \end{array}$$

This is exactly the same integral we had for Example 1 (if replace $t$ with $u$ in the third line from the bottom of Example 1). So we immediately see that the work is, again, $\pi ∕4$.

Example 2

The line integral does, in general depend on the path. The path for Example 1 started at (1,0) and ended at (0,1). What is the integral

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}& & \hfill \end{array}$$

if $C$ is the following different path (shown in blue) from (1,0) to (0,1)?

Solution: We can find this answer without doing any computations.

The first part of the path is moving left along the $x$-axis. In that case, the path is orthogonal to the vector field, which points up. Therefore, the dot product between the vector field and the tangent vector is always zero. The force does no work on the particle during this first segment.

The second part of the path is moving straight up along the $y$-axis. Along the $y$-axis, $x=0$, so that $\mathbf{F}\left(x,y\right)=\left(0,x\right)=\left(0,0\right)$. The vector field is zero, and the integral of zero is zero. We can conclude that

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}=0.& & \hfill \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

4a. Vector field basics

4b. Parametrized curves

4c. Parametrized surfaces

4d. Path integrals and line integrals

• Path integral of a scalar-valued function
• Path integral examples
• Line integral of a vector field
• Alternate notation for line integrals
• Line integrals as circulation
• Line integral examples

4e. Surface integrals

5. The fundamental theorems of vector calculus

6. Review material