2021-09-07

Calculus: Line Integrals

Jeffrey Jordon

Explanation
Line integrals can be used to evaluate integrals of two- or three-dimensional curves
The curve C is expressed by the following parametric equations given by

$x={t}^{2}$ and $y=2t$

Thus, to evaluate the following integral given by

$I={\int }_{c}xyds$

we use the following line integral formula given by

${\int }_{c}f\left(x,y\right)ds={\int }_{a}^{b}f\left(x\left(t\right),y\left(t\right)\right)|{r}^{\prime }|dt$

$={\int }_{a}^{b}f\left(x\left(t\right),y\left(t\right)\right)|\frac{dr}{dt}|dt$

=$={\int }_{a}^{b}f\left(x\left(t\right),y\left(t\right)\right)\sqrt{\left(\frac{dx}{xt}{\right)}^{2}+\left(\frac{dy}{xt}{\right)}^{2}}dt$

Since $x\left(t\right)={t}^{2}$ and $y\left(t\right)=2t$

We want to evaluate ${\int }_{c}xyds$ as follows

Firstly, we must find the integrand $f\left(x\left(t\right),y\left(t\right)\right)$ as follows

$f\left(x\left(t\right),y\left(t\right)\right)=xy$

$={t}^{2}\left(2t\right)$

=$2{t}^{3}$

Also, the derivatives of $x\left(t\right)$ and $y\left(t\right)$ must be

$\frac{d}{dt}x\left(t\right)=\frac{d}{dt}\left({t}^{2}\right)=2t$ and $\frac{d}{dt}y\left(t\right)=\frac{d}{dt}\left(2t\right)=2$

Since $0\le t\le 5$, so that the upper and lower limits of the integral must be

$a=0$ and $b=5$

Thus, using the line integral formula, we get

${\int }_{c}xyds={\int }_{a}^{b}f\left(x\left(t\right),y\left(t\right)\right)\sqrt{\left(\frac{dx}{xt}{\right)}^{2}+\left(\frac{dy}{xt}{\right)}^{2}}dt$

$={\int }_{0}^{5}\left(2{t}^{3}\right)\sqrt{\left(2t{\right)}^{2}+\left(2{\right)}^{2}}dt$

$=2{\int }_{0}^{5}{t}^{3}\sqrt{4{t}^{2}+4}dt=2{\int }_{0}^{5}{t}^{3}\sqrt{4\left({t}^{2}+1\right)}dt$

$=2{\int }_{0}^{5}2{t}^{3}\sqrt{\left({t}^{2}+1\right)}dt$

$=4{\int }_{0}^{5}{t}^{3}\sqrt{\left({t}^{2}+1\right)}dt$

To evaluate the following integral ${\int }_{0}^{5}{t}^{3}\sqrt{\left({t}^{2}+1\right)}dt$, we use the method of substitution as follows.

Let $u={t}^{2}+1$, so that

Differentiable both sides of the following equation $u={t}^{2}+1$, implies that

$du=2tdt$ and $tdt=\frac{du}{2}$

Also, we have ${t}^{2}=u-1$

Also, we find that the upper and lower limits of the integral must be

$t=0⇒u={0}^{2}+1=1$ and $t=5⇒u={5}^{2}+1=26$

Thus, the line integral becomes

${\int }_{c}xyds=4{\int }_{t=0}^{5}{t}^{2}\sqrt{{t}^{2}+1}tdt$

$4{\int }_{t=0}^{26}\left(u-1\right)\sqrt{u}\left(\frac{du}{2}\right)$

$=\frac{4}{2}{\int }_{t=0}^{26}\left(u-1\right)\sqrt{u}du$

$=2{\int }_{t=0}^{26}\left({u}^{\frac{3}{2}-\frac{1}{2}}\right)du$

Thus, the following line integral implies that

${\int }_{c}xyds=2\left[\frac{{u}^{\frac{5}{2}}}{5/2}-\frac{{u}^{\frac{3}{2}}}{3/2}{\right]}_{u=1}^{26}$

$=2\left[\frac{2{u}^{\frac{5}{2}}}{5}-\frac{2{u}^{\frac{3}{2}}}{3}{\right]}_{u=1}^{26}$

$=2\left[\left(\frac{2\left(26{\right)}^{\frac{5}{2}}}{5}-\frac{2\left(26{\right)}^{\frac{3}{2}}}{3}\right)-\left(\frac{2\left(1{\right)}^{\frac{5}{2}}}{5}-\frac{2\left(1{\right)}^{\frac{3}{2}}}{3}\right)\right]$

$=\frac{8}{15}\left(949\sqrt{26}+1\right)$

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