Finding the area enclosed by curves in polar form

[Charismaztex]

Homework Statement

a)
r=a(2+ cos(\theta))
Find the area of the region enclosed by the curve giving answers in terms of \pi
and a

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

Homework Equations

<br /> A=\frac{1}{2} \int_{\beta}^{\alpha} r^2 d\theta<br />

The Attempt at a Solution

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

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

Thanks for your help,
Charismaztex

 

[LCKurtz]

For (a) your limits should be from 0 to 2\pi.

For (b), since \cos\theta is negative on part of its domain, you want to use the interval [-\pi/2,\pi/2]. I don't see any need for a \cos{2\theta}. When you square r, you get some sines times a cosine, so a u substitution should work.

 

[Charismaztex]

So what would be the best way to integrate sin^2(\theta)cos(\theta)?

 

[LCKurtz]

So what would be the best way to integrate sin^2(\theta)cos(\theta)?

u substitution. Try u = \sin\theta

 

[Charismaztex]

Ahh yes, that worked fine! so I've got the area to be\frac{16}{3}, what intervals do you recommend (the part I normally struggle with) to calculate the separate area in order to get the ratio?

 

[LCKurtz]

Ahh yes, that worked fine! so I've got the area to be\frac{16}{3}, what intervals do you recommend (the part I normally struggle with) to calculate the separate area in order to get the ratio?

The problem itself says "show the initial line" divides the area. I take that to mean the x axis. Try it.