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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 $\text{r}\left(t\right)=\u27e8x\left(t\right),y\left(t\right),z\left(t\right)\u27e9$ for $a\le t\le b$ be a parameterization of $C.$ Since we are assuming that $C$ is smooth, ${r}^{\prime}\left(t\right)=\u27e8{x}^{\prime}\left(t\right),{y}^{\prime}\left(t\right),{z}^{\prime}\left(t\right)\u27e9$ is continuous for all $t$ in $\left[a,b\right].$ In particular, $x\text{\u2032}\left(t\right),y\text{\u2032}\left(t\right),$ and $z\text{\u2032}\left(t\right)$ exist for all $t$ in $\left[a,b\right].$ According to the arc length formula, we have
If width $\text{\Delta}{t}_{i}={t}_{i}-{t}_{i-1}$ is small, then function ${\int}_{{t}_{i-1}}^{{t}_{i}}\Vert {r}^{\prime}\left(t\right)\Vert dt\approx \Vert {r}^{\prime}\left({t}_{i}^{*}\right)\Vert \text{\Delta}{t}_{i},$ $\Vert {r}^{\prime}\left(t\right)\Vert $ is almost constant over the interval $\left[{t}_{i-1},{t}_{i}\right].$ Therefore,
and we have
See [link] .
Note that
In other words, as the widths of intervals $[{t}_{i-1},{t}_{i}]$ shrink to zero, the sum $\sum _{i=1}^{n}f(\text{r}({t}_{i}^{*}))}\Vert {r}^{\prime}({t}_{i}^{*})\Vert \text{\Delta}{t}_{i$ converges to the integral ${\int}_{a}^{b}f(\text{r}(t))}\Vert {r}^{\prime}(t)\Vert dt.$ Therefore, we have the following theorem.
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
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
we obtain the following theorem, which we use to compute scalar line integrals.
Let $f$ be a continuous function with a domain that includes the smooth curve C with parameterization $\text{r}\left(t\right)=\u27e8x\left(t\right),y\left(t\right),z\left(t\right)\u27e9,a\le t\le b.$ Then
Similarly,
if C is a planar curve and $f$ is a function of two variables.
Note that a consequence of this theorem is the equation $ds=\Vert {r}^{\prime}\left(t\right)\Vert dt.$ In other words, the change in arc length can be viewed as a change in the t domain, scaled by the magnitude of vector ${r}^{\prime}\left(t\right).$
Find the value of integral ${\int}_{C}\left({x}^{2}+{y}^{2}+z\right)ds,$ where $C$ is part of the helix parameterized by $\text{r}\left(t\right)=\u27e8\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t,t\u27e9,$ $0\le t\le 2\pi .$
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
Therefore,
Notice that [link] translated the original difficult line integral into a manageable single-variable integral. Since
we have
Evaluate ${\int}_{C}\left({x}^{2}+{y}^{2}+z\right)ds,$ where C is the curve with parameterization $\text{r}\left(t\right)=\u27e8\text{sin}\left(3t\right),\text{cos}\left(3t\right)\u27e9,0\le t\le \frac{\pi}{4}.$
$\frac{1}{3}+\frac{\sqrt{2}}{6}+\frac{3\pi}{4}$
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