# Cardoid area

## Homework Statement

Find the area of the region that is inside the circle r = 6cos(theta) but outside the cardoid r = 2 + 2cos(theta)

r = 6·cosθ
r = 2 + 2·cosθ

## The Attempt at a Solution

intersections of the two curves.

6·cosθ = 2 + 2·cosθ → 4·cosθ = 2

cosθ = 1/2 → θ = ±π/3

Can someone finish the integral for me, i'm not good at integrals, this is as far as i can get. thanks for any help.

A = 2 x (1/2) ∫ [(6·cosθ)² - (2 + 2·cosθ)²] dθ

## Answers and Replies

dextercioby
Science Advisor
Homework Helper
You still need a picture. Use math software and ask it to draw both graphs in polar coordinates (they should be $\left(\rho,\varphi\right)$, not "r" and "theta") and just then you can set up a correct integral.

heres what i got so far.. can u please help finish this problem.

heres where i'm at: http://img109.imageshack.us/img109/4070/untitledxz9.jpg [Broken]

Last edited by a moderator:
HallsofIvy
Science Advisor
Homework Helper
No, I will not do the integral for you. I will give you a hint: to integrate $cos^2(\theta)$, use the trig identity [itex]cos^2(\theta)= \frac{1}{2}(cos(2\theta)+ 1).