Finding Area of Circle and Cardoid Intersection

Click For Summary
The discussion focuses on finding the area of the region inside the circle defined by r = 6cos(θ) but outside the cardioid r = 2 + 2cos(θ). The intersection points of the two curves are determined to be at θ = ±π/3. An integral is set up to calculate the area, expressed as A = 2 x (1/2) ∫ [(6·cosθ)² - (2 + 2·cosθ)²] dθ. Participants suggest using math software to visualize the graphs for better understanding and provide hints for completing the integral, particularly recommending a trigonometric identity for integrating cos²(θ). The discussion emphasizes the importance of setting up the correct integral based on the graphical representation of the curves.
rcmango
Messages
232
Reaction score
0

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
 
Physics news on Phys.org
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
 
Last edited by a moderator:
No, I will not do the integral for you. I will give you a hint: to integrate cos^2(\theta), use the trig identity cos^2(\theta)= \frac{1}{2}(cos(2\theta)+ 1).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
3K