Calculating the Area of a Region Bounded by a Cardioid and Circle

  • Thread starter Thread starter MozAngeles
  • Start date Start date
  • Tags Tags
    Area
Click For Summary

Homework Help Overview

The problem involves calculating the area of a region bounded by a cardioid defined by the equation r=2(1+sin(θ)) and a circle defined by r=2sin(θ). The original poster presents their attempt at a solution using integration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the area using integrals and provides specific bounds for integration. Some participants question the choice of bounds for the integrals, while others defend them based on the geometric interpretation of the regions involved.

Discussion Status

The discussion includes verification of the original poster's calculations and interpretations of the integration bounds. Some participants express agreement with the approach, while others raise questions about the reasoning behind the selected bounds.

Contextual Notes

There is a mention of the problem being from a textbook and the original poster's intent to verify their answer in preparation for a test. The discussion reflects varying interpretations of the integration limits based on the geometric properties of the cardioid and circle.

MozAngeles
Messages
100
Reaction score
0

Homework Statement



Inside the cardioid r=2(1+sin(theta)) and outside the circle r=2sin(theta)

Homework Equations


A= ∫ (from a..b) 1/2 f(θ) 2


The Attempt at a Solution


A= 2∫(from π/2..3π/2) 1/2 [2(1+sin(θ)]2 dθ- 2∫(from0..π/2) 1/2 (2sinθ)2

after working that out i got the answer to be 5pi.

i need a verification for my answer because it is an even problem in the book and I'm studying for my test.
 
Physics news on Phys.org
It looks OK except for the bounds of integration.

Why is the first integral taken over [\pi/2,3\pi/2], and why is the second integral taken over [0,\pi/2]?
 
The bounds of integration are OK. The range -pi/2 <= theta <= pi/2 gives the right half of the cardioid; equivalently, the range pi/2 <= theta <= 3pi/2 gives the left half. The range 0 <= theta <= pi/2 gives the right half of the circle; equivalently, 0 <= theta <= Pi gives the whole circle. (The whole range 0 <= theta <= 2*Pi goes around the circle twice, so would give twice the desired area.) The answer 5*pi is correct.

RGV
 
You're right, I didn't notice the extra factor of 2.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K