Area of circle in polar coordinates

Click For Summary

Homework Help Overview

The discussion revolves around finding the area of a circle defined by the polar equation r=2cos(θ) using polar integration techniques. The context is situated within multivariable calculus.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the area formula for polar coordinates but expresses confusion regarding an unexpected result of 2π. Some participants question the choice of integration limits and suggest that the interval should be adjusted based on the graph of the circle.

Discussion Status

There is an acknowledgment of a potential misunderstanding regarding the integration limits, with some participants suggesting that the original poster's answer may still be valid despite the concerns raised. The conversation reflects an exploration of different interpretations of the problem.

Contextual Notes

Participants note that the original poster's calculator was in degree mode, which may have contributed to the confusion. The discussion also highlights the importance of correctly identifying the interval of integration when dealing with polar coordinates.

Malabeh

Homework Statement


r=2cos(theta) I want to find the area using polar integration.

Homework Equations


area=(1/2)r^2 from 0-pi

The Attempt at a Solution


When I plug everything in I get 2pi as the answer. I'm in multivariable calculus so this is very frustrating. What am I doing wrong, I don't know, but it's really really really really bugging me. I have an extra 2 somewhere and I don't know where.
 
Physics news on Phys.org
EDIT: I don't know how to delete a post but I got it. My calculator was in degree mode.
 
Malabeh said:
EDIT: I don't know how to delete a post but I got it. My calculator was in degree mode.
While you may have gotten the correct answer, if you plot that circle you will see that the appropriate interval of integration would be ##[-\frac \pi 2,\frac \pi 2]##, not ##[0,\pi]##.
 
LCKurtz said:
While you may have gotten the correct answer, if you plot that circle you will see that the appropriate interval of integration would be ##[-\frac \pi 2,\frac \pi 2]##, not ##[0,\pi]##.
It works both ways, does it not?
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K