(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

a)

[tex]r=a(2+ cos(\theta))[/tex]

Find the area of the region enclosed by the curve giving answers in terms of [tex]\pi[/tex]

and [tex]a[/tex]

b) Show that the area enclosed by the loop [tex]r=2(1-sin(\theta))\sqrt{cos(\theta)}[/tex] is [tex]\frac{16}{3}[/tex] and show that the initial line divides the area enclosed by the loop in the ratio 1:7 [tex](\theta[/tex] is between [tex]-\frac{\pi}{2} and \frac{\pi}{2})[/tex]

2. Relevant equations

[tex]

A=\frac{1}{2} \int_{\beta}^{\alpha} r^2 d\theta

[/tex]

3. The attempt at a solution

a) I tried using [tex]\theta=0, \pi, \frac{\pi}{2}, \frac{3\pi}{2}[/tex] as limits and adding the quadrants up but comes out to be the incorrect answer. (textbook answer: [tex]\frac{9a^2\pi}{2}[/tex])

b) I'm having a hard time integrating the [tex] cos(\theta)cos(2\theta)[/tex] part after squaring r. I know that one of the limits is [tex]\frac{\pi}{2}[/tex] but I'm not sure what the other should be.

Thanks for your help,

Charismaztex

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# Homework Help: Finding the area enclosed by curves in polar form

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