# Difference between revisions of "009C Sample Final 1, Problem 9"

A curve is given in polar coordinates by

${\displaystyle r=\theta }$
${\displaystyle 0\leq \theta \leq 2\pi }$

Find the length of the curve.

Foundations:
1. The formula for the arc length ${\displaystyle L}$ of a polar curve ${\displaystyle r=f(\theta )}$ with ${\displaystyle \alpha _{1}\leq \theta \leq \alpha _{2}}$ is
${\displaystyle L=\int _{\alpha _{1}}^{\alpha _{2}}{\sqrt {r^{2}+{\bigg (}{\frac {dr}{d\theta }}{\bigg )}^{2}}}d\theta .}$
2. How would you integrate ${\displaystyle \int {\sqrt {1+x^{2}}}~dx?}$
You could use trig substitution and let ${\displaystyle x=\tan \theta .}$
3. Recall that ${\displaystyle \int \sec ^{3}x~dx={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|+C.}$

Solution:

Step 1:
First, we need to calculate ${\displaystyle {\frac {dr}{d\theta }}}$.
Since ${\displaystyle r=\theta ,~{\frac {dr}{d\theta }}=1.}$
Using the formula in Foundations, we have
${\displaystyle L=\int _{0}^{2\pi }{\sqrt {\theta ^{2}+1}}d\theta .}$
Step 2:
Now, we proceed using trig substitution. Let ${\displaystyle \theta =\tan x.}$ Then, ${\displaystyle d\theta =\sec ^{2}xdx.}$
So, the integral becomes
${\displaystyle {\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {\int _{\theta =0}^{\theta =2\pi }{\sqrt {\tan ^{2}x+1}}\sec ^{2}xdx}\\&&\\&=&\displaystyle {\int _{\theta =0}^{\theta =2\pi }\sec ^{3}xdx}\\&&\\&=&\displaystyle {{\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|{\bigg |}_{\theta =0}^{\theta =2\pi }.}\\\end{array}}}$
Step 3:
Since ${\displaystyle \theta =\tan x,}$ we have ${\displaystyle x=\tan ^{-1}\theta .}$
So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {{\frac {1}{2}}\sec(\tan ^{-1}(\theta ))\theta +{\frac {1}{2}}\ln |\sec(\tan ^{-1}(\theta ))+\theta |{\bigg |}_{0}^{2\pi }}\\&&\\&=&\displaystyle {{\frac {1}{2}}\sec(\tan ^{-1}(2\pi ))2\pi +{\frac {1}{2}}\ln |\sec(\tan ^{-1}(2\pi ))+2\pi |.}\\\end{array}}}$
${\displaystyle {\frac {1}{2}}\sec(\tan ^{-1}(2\pi ))2\pi +{\frac {1}{2}}\ln |\sec(\tan ^{-1}(2\pi ))+2\pi |}$