 # Find the length of the curve over the given interval. zeotropojd 2021-12-19 Answered
Find the length of the curve over the given interval. $r=8+8\mathrm{cos}\theta$ on the interval $0\le \theta \le \pi$
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Ronnie Schechter
Step 1
The given function is
$r=8+8\mathrm{cos}\theta$ on the interval $0\le \theta \le \pi$
The length of the polar curve is given by
$L={\int }_{a}^{b}\sqrt{{\left(r\left(\theta \right)\right)}^{2}+{\left({r}^{\prime }\left(\theta \right)\right)}^{2}}d\theta$
First, find the derivative:
${r}^{\prime }\left(\theta \right){\left(8\mathrm{cos}\left(\theta \right)+8\right)}^{\prime }=-8\mathrm{sin}\left(\theta \right)$
Step 2
Finally, calculate the integral
$L={\int }_{0}^{\pi }\sqrt{{\left(8\mathrm{cos}\left(\theta \right)+8\right)}^{2}+{\left(-8\mathrm{sin}\left(\theta \right)\right)}^{2}}d\theta$
$={\int }_{0}^{\pi }8\sqrt{2}\sqrt{\mathrm{cos}\left(\theta \right)+1}d\theta$
$={\int }_{0}^{\pi }8\sqrt{2}\sqrt{2{\mathrm{cos}}^{2}\left(\frac{\theta }{2}\right)}d\theta$
$={\int }_{0}^{\pi }16\mathrm{cos}\left(\frac{\theta }{2}\right)d\theta$
$=32\left(\mathrm{sin}\left(\frac{\theta }{2}\right)\right){|{}_{\left\{\left(\theta =\pi \right)\right\}}-32\left(\mathrm{sin}\left(\frac{\theta }{2}\right)\right)|}_{\left(\theta =0\right)}$
$=32$

We have step-by-step solutions for your answer! Nadine Salcido
Step 1
The definite integral that represents the arc length given by
Arc length $={\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}}d\theta$
$={\int }_{0}^{\pi }\sqrt{{\left(8+8\mathrm{cos}\theta \right)}^{2}+64{\mathrm{sin}}^{2}\theta }d\theta$
$=8{\int }_{0}^{\pi }\sqrt{1+2\mathrm{cos}\theta +{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta }d\theta$
$=8{\int }_{0}^{\pi }\sqrt{2+2\mathrm{cos}\theta }d\theta$
$=8{\int }_{0}^{\pi }\sqrt{4{\mathrm{cos}}^{2}\left(\frac{\theta }{2}\right)}d\theta$
$=16{\int }_{0}^{\pi }\mathrm{cos}\left(\frac{\theta }{2}\right)d\theta$
$=32\mathrm{sin}\left(\frac{\theta }{2}\right)Big{\mid }_{0}^{\pi }$
$=32$

We have step-by-step solutions for your answer!