Polar area integrals for r=a(2+cos θ) and r=2(1-sin θ)√cos θ

[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

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

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.