Finding Area Between Polar Curves - Help Understanding Bounding

  • Context: Undergrad 
  • Thread starter Thread starter itzela
  • Start date Start date
  • Tags Tags
    Area Curves Polar
Click For Summary
SUMMARY

This discussion focuses on the process of finding the area between two polar curves, specifically addressing the challenges of determining the correct bounds for integration. The user encountered issues leading to a negative area result, indicating a misunderstanding of how to set the bounds for each function. The integral expressions provided include \(\frac{A}{2}=\frac{1}{2}\int_{\frac{\pi}{3}}^{\pi}(1+\cos(\theta))^2 d\theta - \frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(3\cos(\theta))^2 d\theta, highlighting the need for distinct bounds for each curve.

PREREQUISITES
  • Understanding of polar coordinates and polar curves
  • Knowledge of integral calculus, specifically integration techniques
  • Familiarity with the concept of area between curves
  • Ability to manipulate trigonometric functions in integrals
NEXT STEPS
  • Study the method for finding areas between polar curves
  • Learn about setting appropriate bounds for integration in polar coordinates
  • Explore examples of polar integrals to reinforce understanding
  • Review trigonometric identities and their applications in integration
USEFUL FOR

Students and educators in calculus, particularly those focusing on polar coordinates and integration techniques, as well as anyone seeking to improve their understanding of area calculations between curves.

itzela
Messages
34
Reaction score
0
I am having trouble finding the area between 2 polar curves... I have the procedure down, but the bounds are throwing me off. Any help with understanding how to bound would be great appreciated!

I have attatched one problem that I am having hard time with and the work I have done. I know that I am doing something wrong because I am getting a negative number for an area (which shouldn't be).
 

Attachments

  • problem.JPG
    problem.JPG
    20.2 KB · Views: 552
Physics news on Phys.org
You can't write that as one integral in that way. The Functions do not respond to the parameters the same way, different bounds are needed for each.
[tex]\frac{A}{2}=\frac{1}{2}\int_{\frac{\pi}{3}}^{\pi}(1+\cos(\theta))^2 d\theta - \frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(3\cos(\theta))^2 d\theta[/tex]
 
Last edited:
Thanks :smile:
After working on it for a little while i arrived at the same expression... and i was a little unsure about it, but you confirmed it!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K