Multivariable calculus and vector analysis

Path integral examples

Example 1

Let

$$\begin{array}{lll}\hfill \mathbf{c}\left(t\right)=\left(3t-2,t+1\right),1\le t\le 2& & \hfill \end{array}$$

be a parameterization of a wire. Let the density of the wire at point $\left(x,y\right)$ be given by

$$\begin{array}{lll}\hfill f\left(x,y\right)=x+y.& & \hfill \end{array}$$

Compute the mass of the wire.

Solution: Mass is the integral of the density along the wire. So we must compute ${\int \; <!--nolimits\; -->}_{\mathbf{c}}fds$. Since,

$$\begin{array}{llll}\hfill {\mathbf{c}}^{\text{'}}\left(t\right)& =\left(3,1\right),& \hfill & \\ \hfill \left|\right|{\mathbf{c}}^{\text{'}}\left(t\right)\left|\right|& =\sqrt{3^2+1^2}=\sqrt{10},& \hfill & \\ \hfill f\left(\mathbf{c}\left(t\right)\right)& =\left(3t-2\right)+\left(t+1\right)=4t-1,& \hfill & \end{array}$$

the integral is

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_{\mathbf{c}}fds& ={\int \; <!--nolimits\; -->}_a^{b}f\left(\mathbf{c}\left(t\right)\right)\left|\right|{\mathbf{c}}^{\text{'}}\left(t\right)\left|\right|dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_1^2\left(4t-1\right)\sqrt{10}dt& \hfill & \\ \hfill & =\left(2t^2-t\right){\left.\sqrt{10}\right|}_1^2& \hfill & \\ \hfill & =\left(8-2-\left(2-1\right)\right)\sqrt{10}=5\sqrt{10}& \hfill & \end{array}$$

So, if $f$ were given in grams/cm and $\mathbf{c}\left(t\right)$ were given in cm, then the mass of the wire would $5\sqrt{10}$ grams.

Example 2

Neither the length of the wire nor its mass can depend on parametrization. Check that you get the same answer with the parametrization

$$\begin{array}{lll}\hfill \mathbf{p}\left(t\right)=\left(9t-2,3t+1\right),1∕3\le t\le 2∕3.& & \hfill \end{array}$$

Not that this is the same wire as in Example 1, which is a straight line from the point $\left(1,2\right)$ to $\left(4,3\right)$.

Solution: We repeat the same calculations as in Example 1. (Recall that $f\left(x,y\right)=x+y$.) Since

$$\begin{array}{llll}\hfill {\mathbf{p}}^{\text{'}}\left(t\right)& =\left(9,3\right),& \hfill & \\ \hfill \left|\right|{\mathbf{p}}^{\text{'}}\left(t\right)\left|\right|& =\sqrt{9^2+3^2}=\sqrt{90}=3\sqrt{10},& \hfill & \\ \hfill f\left(\mathbf{p}\left(t\right)\right)& =\left(9t-2\right)+\left(3t+1\right)=12t-1,& \hfill & \end{array}$$

the integral is

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_{\mathbf{p}}fds& ={\int \; <!--nolimits\; -->}_{1∕3}^{2∕3}f\left(\mathbf{p}\left(t\right)\right)\left|\right|{\mathbf{p}}^{\text{'}}\left(t\right)\left|\right|dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_{1∕3}^{2∕3}\left(12t-1\right)3\sqrt{10}dt& \hfill & \\ \hfill & =\left(6t^2-t\right){\left.3\sqrt{10}\right|}_{1∕3}^{2∕3}& \hfill & \\ \hfill & =\left[6{\left(\frac{2}{3}\right)}^2-\frac{2}{3}-\left(6{\left(\frac{1}{3}\right)}^2-\frac{1}{3}\right)\right]3\sqrt{10}& \hfill & \\ \hfill & =\left[\frac{24}{9}-\frac{2}{3}-\frac{6}{9}+\frac{1}{3}\right]3\sqrt{10}=\frac{5}{3}3\sqrt{10}=5\sqrt{10},& \hfill & \end{array}$$

which matches Example 1.

Note that the “speed” of the parametrization $\left|\right|{\mathbf{p}}^{\text{'}}\left(t\right)\left|\right|=3\sqrt{10}$ was three times greater than the speed $\left|\right|{\mathbf{c}}^{\text{'}}\left(t\right)\left|\right|=\sqrt{10}$ of Example 1. But the range of integration $1∕3\le t\le 2∕3$ was one third the range of integration from Example 1.

Example 3

Here’s another parametrization of the same straight-line wire from $\left(1,2\right)$ to $\left(4,3\right)$. This time, the “speed” of the parametrization is not constant, but depends on t:

$$\begin{array}{lll}\hfill \mathbf{q}\left(t\right)=\left(3t^2-2,t^2+1\right),1\le t\le \sqrt{2}.& & \hfill \end{array}$$

Still using the density $f\left(x,y\right)=x+y$, calculate the mass of the wire.

Solution: Since

$$\begin{array}{llll}\hfill {\mathbf{q}}^{\text{'}}\left(t\right)& =\left(6t,2t\right),& \hfill & \\ \hfill \left|\right|{\mathbf{q}}^{\text{'}}\left(t\right)\left|\right|& =\sqrt{{\left(6t\right)}^2+{\left(2t\right)}^2}=\sqrt{40t^2}=2t\sqrt{10},& \hfill & \\ \hfill f\left(\mathbf{q}\left(t\right)\right)& =\left(3t^2-2\right)+\left(t^2+1\right)=4t^2-1.& \hfill & \end{array}$$

the integral is

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_{\mathbf{q}}fds& ={\int \; <!--nolimits\; -->}_1^{\sqrt{2}}f\left(\mathbf{q}\left(t\right)\right)\left|\right|{\mathbf{q}}^{\text{'}}\left(t\right)\left|\right|dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_1^{\sqrt{2}}\left(4t^2-1\right)2t\sqrt{10}dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_1^{\sqrt{2}}\left(8t^3-2t\right)\sqrt{10}dt& \hfill & \\ \hfill & =\left(2t^4-t^2\right){\left.\sqrt{10}\right|}_1^{\sqrt{2}}& \hfill & \\ \hfill & =\left[2{\left(\sqrt{2}\right)}^4-{\left(\sqrt{2}\right)}^2-2+1\right]\sqrt{10}=5\sqrt{10},& \hfill & \end{array}$$

which matches Examples 1 and 2.

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