6.2 Line integrals  (Page 3/20)

 Page 3 / 20

Find the value of ${\int }_{C}\left(x=y\right)ds,$ where $C$ is the curve parameterized by $x=t,$ $y=t,$ $0\le t\le 1.$

$\sqrt{2}$

Note that in a scalar line integral, the integration is done with respect to arc length s , which can make a scalar line integral difficult to calculate. To make the calculations easier, we can translate ${\int }_{C}fds$ to an integral with a variable of integration that is t .

Let $\text{r}\left(t\right)=⟨x\left(t\right),y\left(t\right),z\left(t\right)⟩$ for $a\le t\le b$ be a parameterization of $C.$ Since we are assuming that $C$ is smooth, ${r}^{\prime }\left(t\right)=⟨{x}^{\prime }\left(t\right),{y}^{\prime }\left(t\right),{z}^{\prime }\left(t\right)⟩$ is continuous for all $t$ in $\left[a,b\right].$ In particular, $x\text{′}\left(t\right),y\text{′}\left(t\right),$ and $z\text{′}\left(t\right)$ exist for all $t$ in $\left[a,b\right].$ According to the arc length formula, we have

$\text{length}\left({C}_{i}\right)=\text{Δ}{s}_{i}={\int }_{{t}_{i-1}}^{{t}_{i}}‖{r}^{\prime }\left(t\right)‖dt.$

If width $\text{Δ}{t}_{i}={t}_{i}-{t}_{i-1}$ is small, then function ${\int }_{{t}_{i-1}}^{{t}_{i}}‖{r}^{\prime }\left(t\right)‖dt\approx ‖{r}^{\prime }\left({t}_{i}^{*}\right)‖\text{Δ}{t}_{i},$ $‖{r}^{\prime }\left(t\right)‖$ is almost constant over the interval $\left[{t}_{i-1},{t}_{i}\right].$ Therefore,

${\int }_{{t}_{i-1}}^{{t}_{i}}‖{r}^{\prime }\left(t\right)‖dt\approx ‖{r}^{\prime }\left({t}_{i}^{*}\right)‖\text{Δ}{t}_{i},$

and we have

$\sum _{i=1}^{n}f\left(\text{r}\left({t}_{i}^{*}\right)\right)\text{Δ}{s}_{i}=\sum _{i=1}^{n}f\left(\text{r}\left({t}_{i}^{*}\right)\right)‖{r}^{\prime }\left({t}_{i}^{*}\right)‖\text{Δ}{t}_{i}.$

Note that

$\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}f\left(\text{r}\left({t}_{i}^{*}\right)\right)‖{r}^{\prime }\left({t}_{i}^{*}\right)‖\text{Δ}{t}_{i}={\int }_{a}^{b}f\left(\text{r}\left(t\right)\right)‖{r}^{\prime }\left(t\right)‖dt.$

In other words, as the widths of intervals $\left[{t}_{i-1},{t}_{i}\right]$ shrink to zero, the sum $\sum _{i=1}^{n}f\left(\text{r}\left({t}_{i}^{*}\right)\right)‖{r}^{\prime }\left({t}_{i}^{*}\right)‖\text{Δ}{t}_{i}$ converges to the integral ${\int }_{a}^{b}f\left(\text{r}\left(t\right)\right)‖{r}^{\prime }\left(t\right)‖dt.$ Therefore, we have the following theorem.

Evaluating a scalar line integral

Let $f$ be a continuous function with a domain that includes the smooth curve $C$ with parameterization $\text{r}\left(t\right),a\le t\le b.$ Then

${\int }_{C}fds={\int }_{a}^{b}f\left(\text{r}\left(t\right)\right)‖{r}^{\prime }\left(t\right)‖dt.$

Although we have labeled [link] as an equation, it is more accurately considered an approximation because we can show that the left-hand side of [link] approaches the right-hand side as $n\to \infty .$ In other words, letting the widths of the pieces shrink to zero makes the right-hand sum arbitrarily close to the left-hand sum. Since

$‖{r}^{\prime }\left(t\right)‖=\sqrt{{\left(x\prime \left(t\right)\right)}^{2}+{\left(y\prime \left(t\right)\right)}^{2}+{\left(z\prime \left(t\right)\right)}^{2},}$

we obtain the following theorem, which we use to compute scalar line integrals.

Scalar line integral calculation

Let $f$ be a continuous function with a domain that includes the smooth curve C with parameterization $\text{r}\left(t\right)=⟨x\left(t\right),y\left(t\right),z\left(t\right)⟩,a\le t\le b.$ Then

${\int }_{C}f\left(x,y,z\right)ds={\int }_{a}^{b}f\left(\text{r}\left(t\right)\right)=\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}+{\left({z}^{\prime }\left(t\right)\right)}^{2}}dt.$

Similarly,

${\int }_{C}f\left(x,y\right)ds={\int }_{a}^{b}f\left(\text{r}\left(t\right)\right)=\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt$

if C is a planar curve and $f$ is a function of two variables.

Note that a consequence of this theorem is the equation $ds=‖{r}^{\prime }\left(t\right)‖dt.$ In other words, the change in arc length can be viewed as a change in the t domain, scaled by the magnitude of vector ${r}^{\prime }\left(t\right).$

Evaluating a line integral

Find the value of integral ${\int }_{C}\left({x}^{2}+{y}^{2}+z\right)ds,$ where $C$ is part of the helix parameterized by $\text{r}\left(t\right)=⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t,t⟩,$ $0\le t\le 2\pi .$

To compute a scalar line integral, we start by converting the variable of integration from arc length s to t . Then, we can use [link] to compute the integral with respect to t . Note that $f\left(\text{r}\left(t\right)\right)={\text{cos}}^{2}t+{\text{sin}}^{2}t+t=1+t$ and

$\begin{array}{cc}\hfill \sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}+{\left({z}^{\prime }\left(t\right)\right)}^{2}}& =\sqrt{{\left(\text{−}\text{sin}\left(t\right)\right)}^{2}+{\text{cos}}^{2}\left(t\right)+1}\hfill \\ & =\sqrt{2}.\hfill \end{array}$

Therefore,

${\int }_{C}\left({x}^{2}+{y}^{2}+z\right)ds={\int }_{0}^{2\pi }\left(1+t\right)\sqrt{2}dt.$

Notice that [link] translated the original difficult line integral into a manageable single-variable integral. Since

$\begin{array}{cc}\hfill {\int }_{0}^{2\pi }\left(1+t\right)\sqrt{2}dt& ={\left[\sqrt{2}t+\frac{\sqrt{2}{t}^{2}}{2}\right]}_{0}^{2\pi }\hfill \\ & =2\sqrt{2}\pi +2\sqrt{2}{\pi }^{2},\hfill \end{array}$

we have

${\int }_{C}\left({x}^{2}+{y}^{2}+z\right)ds=2\sqrt{2}\pi +2\sqrt{2}{\pi }^{2}.$

Evaluate ${\int }_{C}\left({x}^{2}+{y}^{2}+z\right)ds,$ where C is the curve with parameterization $\text{r}\left(t\right)=⟨\text{sin}\left(3t\right),\text{cos}\left(3t\right)⟩,0\le t\le \frac{\pi }{4}.$

$\frac{1}{3}+\frac{\sqrt{2}}{6}+\frac{3\pi }{4}$

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