# 6.2 Line integrals  (Page 4/20)

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## Independence of parameterization

Find the value of integral ${\int }_{C}\left({x}^{2}+{y}^{2}+z\right)ds,$ where $C$ is part of the helix parameterized by $\text{r}\left(t\right)=⟨\text{cos}\left(2t\right),\text{sin}\left(2t\right),2t⟩,0\le t\le \pi .$ Notice that this function and curve are the same as in the previous example; the only difference is that the curve has been reparameterized so that time runs twice as fast.

As with the previous example, we use [link] to compute the integral with respect to t . Note that $f\left(\text{r}\left(t\right)\right)={\text{cos}}^{2}\left(2t\right)+{\text{sin}}^{2}\left(2t\right)+2t=2t+1$ and

$\begin{array}{cc}\hfill \sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}+{\left(z\prime \left(t\right)\right)}^{2}}& =\sqrt{\left(\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t+\text{cos}\phantom{\rule{0.2em}{0ex}}t+4\right)}\hfill \\ & =2\sqrt{2}\hfill \end{array}$

so we have

$\begin{array}{cc}\hfill {\int }_{C}\left({x}^{2}+{y}^{2}+z\right)ds& =2\sqrt{2}{\int }_{0}^{\pi }\left(1+2t\right)dt\hfill \\ & =2\sqrt{2}{\left[t+{t}^{2}\right]}_{0}^{\pi }\hfill \\ & =2\sqrt{2}\left(\pi +{\pi }^{2}\right).\hfill \end{array}$

Notice that this agrees with the answer in the previous example. Changing the parameterization did not change the value of the line integral. Scalar line integrals are independent of parameterization, as long as the curve is traversed exactly once by the parameterization.

Evaluate line integral ${\int }_{C}\left({x}^{2}+yz\right)ds,$ where $C$ is the line with parameterization $\text{r}\left(t\right)=⟨2t,5t,\text{−}t⟩,0\le t\le 10.$ Reparameterize C with parameterization $\text{s}\left(t\right)=⟨4t,10t,-2t⟩,0\le t\le 5,$ recalculate line integral ${\int }_{C}\left({x}^{2}+yz\right)ds,$ and notice that the change of parameterization had no effect on the value of the integral.

Both line integrals equal $-\frac{1000\sqrt{30}}{3}.$

Now that we can evaluate line integrals, we can use them to calculate arc length. If $f\left(x,y,z\right)=1,$ then

$\begin{array}{cc}\hfill {\int }_{C}f\left(x,y,z\right)ds& =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}f\left({t}_{i}^{*}\right)\text{Δ}{s}_{i}\hfill \\ & =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\text{Δ}{s}_{i}\hfill \\ & =\underset{n\to \infty }{\text{lim}}\text{length}\left(C\right)\hfill \\ & =\text{length}\left(C\right).\hfill \end{array}$

Therefore, ${\int }_{C}1ds$ is the arc length of $C.$

## Calculating arc length

A wire has a shape that can be modeled with the parameterization $\text{r}\left(t\right)=⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t,\sqrt{t}⟩,0\le t\le 4\pi .$ Find the length of the wire.

The length of the wire is given by ${\int }_{C}1ds,$ where C is the curve with parameterization r . Therefore,

$\begin{array}{cc}\hfill \text{The length of the wire}& ={\int }_{C}1ds\hfill \\ & ={\int }_{0}^{4\pi }‖{r}^{\prime }\left(t\right)‖dt\hfill \\ & ={\int }_{0}^{4\pi }\sqrt{{\left(\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t\right)}^{2}+{\text{cos}}^{2}t+t}dt\hfill \\ & ={\int }_{0}^{4\pi }\sqrt{1+t}dt\hfill \\ & ={\left[\frac{2{\left(1+t\right)}^{3\text{/}2}}{3}\right]}_{0}^{4\pi }\hfill \\ & =\frac{2}{3}\left({\left(1+4\pi \right)}^{3\text{/}2}-1\right).\hfill \end{array}$

Find the length of a wire with parameterization $\text{r}\left(t\right)=⟨3t+1,4-2t,5+2t⟩,0\le t\le 4.$

$4\sqrt{17}$

## Vector line integrals

The second type of line integrals are vector line integrals, in which we integrate along a curve through a vector field. For example, let

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

be a continuous vector field in ${ℝ}^{3}$ that represents a force on a particle, and let C be a smooth curve in ${ℝ}^{3}$ contained in the domain of $\text{F}.$ How would we compute the work done by $\text{F}$ in moving a particle along C ?

To answer this question, first note that a particle could travel in two directions along a curve: a forward direction and a backward direction. The work done by the vector field depends on the direction in which the particle is moving. Therefore, we must specify a direction along curve C ; such a specified direction is called an orientation of a curve    . The specified direction is the positive direction along C ; the opposite direction is the negative direction along C . When C has been given an orientation, C is called an oriented curve ( [link] ). The work done on the particle depends on the direction along the curve in which the particle is moving.

A closed curve    is one for which there exists a parameterization $\text{r}\left(t\right),$ $a\le t\le b,$ such that $\text{r}\left(a\right)=\text{r}\left(b\right),$ and the curve is traversed exactly once. In other words, the parameterization is one-to-one on the domain $\left(a,b\right).$

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