Finding Area of Circle and Cardoid Intersection

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Homework Help Overview

The problem involves finding the area of the region that lies inside a circle defined by the equation r = 6cos(θ) and outside a cardioid defined by r = 2 + 2cos(θ). The discussion centers around the setup of the integral needed to calculate this area.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the intersections of the two curves and sets up an integral based on their equations. Some participants suggest visualizing the curves to aid in setting up the integral correctly, while others provide hints regarding integration techniques.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some guidance has been offered regarding the use of trigonometric identities for integration, but there is no explicit consensus on how to proceed with the integral itself.

Contextual Notes

Participants note the importance of visualizing the curves in polar coordinates to correctly set up the integral, indicating that a graphical representation may be necessary for further progress.

rcmango
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Homework Statement



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

Homework Equations



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
 
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You still need a picture. Use math software and ask it to draw both graphs in polar coordinates (they should be [itex]\left(\rho,\varphi\right)[/itex], 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
 
Last edited by a moderator:
No, I will not do the integral for you. I will give you a hint: to integrate [itex]cos^2(\theta)[/itex], use the trig identity [itex]cos^2(\theta)= \frac{1}{2}(cos(2\theta)+ 1).[/itex]
 

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