Find the exact length of the polar curve. r=\theta^{2},\ 0\le\theta\le2

Find the exact length of the polar curve.
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Opeance1951
Step 1
It is given that
$r={\theta }^{2}$
Differentiate using the power rule
$\frac{dr}{d\theta }=2\cdot {\theta }^{2-1}$
$\frac{dr}{d\theta }=2\theta$
Formula 6 from this section:
Length of a curve with polar equation is given by
$L={\int }_{a}^{b}\sqrt{{r}^{2}+{\left[\frac{dr}{d\theta }\right]}^{2}}d\theta$
$={\int }_{0}^{2\pi }\sqrt{{\left({\theta }^{2}\right)}^{2}+{\left(2\theta \right)}^{2}}d\theta$
$={\int }_{0}^{2\pi }\sqrt{{\theta }^{4}+4{\theta }^{2}}d\theta$
Note that: $\sqrt{{\theta }^{4}+4{\theta }^{2}}=\sqrt{{\theta }^{2}\left({\theta }^{2}+4\right)}=\sqrt{{\theta }^{2}}\cdot \sqrt{{\theta }^{2}+4}=\theta \cdot \sqrt{{\theta }^{2}+4}$
Since we are integrating over is positive and $\sqrt{{\theta }^{2}=|\theta |=\theta }$
$={\int }_{0}^{2\pi }\theta \cdot \sqrt{{\theta }^{2}+4}d\theta$

We will now substitute
${\theta }^{2}+\mathrm{}=u$
Step 2
On differentiation, we get
$2\theta ,d\theta =du⇒\theta d\theta =\frac{1}{2}du$
The limits of integration will change from ${\int }_{0}^{2\pi }$ to ${\int }_{{0}^{2}+4}^{4{\pi }^{2}+4}$
$={\int }_{4}^{4{\pi }^{2}+4}\sqrt{u}\left[\frac{1}{2}du\right]$
$=\frac{1}{2}{\int }_{4}^{4{\pi }^{2}+4}{u}^{\frac{1}{2}}du$
$=\frac{1}{2}{\left[\frac{{u}^{\left(\frac{1}{2}\right)+1}}{\left(\frac{1}{2}\right)+1}\right]}_{4}^{4{\pi }^{2}+4}$
Not exactly what you’re looking for?
huckelig75
Step 1
Given: $r={\theta }^{2}$
$l={\int }_{a}^{b}\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}d\theta }$ - formula for length of a curve.
$\frac{dr}{d\theta }=2\theta$
$L={\int }_{0}^{2\pi }\sqrt{{\left({\theta }^{2}\right)}^{2}+{\left(2\theta \right)}^{2}}d\theta$
$={\int }_{0}^{2\pi }\sqrt{{\theta }^{4}+4{\theta }^{2}}d\theta$
$={\int }_{0}^{2\pi }\sqrt{{\theta }^{2}\left({\theta }^{2}+4\right)}d\theta$
$={\int }_{0}^{2\pi }\theta \sqrt{{\theta }^{2}+4}d\theta$
Substitution:
$t={\theta }^{2}+4$
$dt=2\theta d\theta \to \theta d\theta =\frac{dt}{2}$
$0\to 4$
$2\pi \to 4{\pi }^{2}+4$
$L=\frac{1}{2}{\int }_{4}^{4{\pi }^{2}+4}\sqrt{tdt}=\frac{1}{2}\left(\frac{2}{3}{t}^{\frac{3}{2}}\right){\mid }_{4}^{4{\pi }^{2}+4}$
$=\frac{1}{3}\left({\left(4{\pi }^{2}+4\right)}^{\frac{3}{2}}-{4}^{\frac{3}{2}}\right)\approx 92.89$
$L\approx 92.89$
Not exactly what you’re looking for?
user_27qwe

Step 1

The length of a polar function $r=f\left(\theta \right)$ is:

$s={\int }_{{\theta }_{1}}^{{\theta }_{2}}\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}}d\theta$ sub in $r={\theta }^{2}$ so $\frac{dr}{d\theta }=2\theta$

$={\int }_{0}^{2\pi }\sqrt{{\theta }^{4}+4{\theta }^{2}}d\theta$ factor $\theta$

$={\int }_{0}^{2\pi }\theta \sqrt{{\theta }^{2}+4}d\theta$ let $u={\theta }^{2}+4$ so $du=2\theta d\theta$

$=\frac{1}{2}{\int }_{4}^{4\left({\pi }^{2}+1\right)}{u}^{\frac{1}{2}}du$

$=\frac{1}{2}\left(\frac{2}{3}\right){\left({u}^{\frac{3}{2}}\right)}_{4}^{4\left({\pi }^{2}+1\right)}$

$=\frac{1}{3}\left({4}^{\frac{3}{2}}\left({\pi }^{2}+1{\right)}^{\frac{3}{2}}-{4}^{\frac{3}{2}}\right)$

$={\frac{4}{3}}^{\frac{3}{2}}\left(\left({\pi }^{2}+1{\right)}^{\frac{3}{2}}-1\right)$

The length of the curve is $={\frac{4}{3}}^{\frac{3}{2}}\left(\left({\pi }^{2}+1{\right)}^{\frac{3}{2}}-1\right)$