Double integral for loop of rose r=cos2θ

Click For Summary

Homework Help Overview

The discussion revolves around the evaluation of a double integral involving polar coordinates, specifically for the rose curve defined by r = cos(2θ). Participants are exploring the range of θ and the relationship between r and the curve.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning how the range of θ, specifically ±π/4, is determined. There is confusion regarding the limits for r and its definition in relation to the curve. Some suggest plotting points to better understand the graph of the function.

Discussion Status

The discussion is ongoing, with participants actively seeking clarification on the graphical representation and the implications of polar coordinates. Some have begun to understand the relationship between r and the curve, while others are still grappling with the concepts.

Contextual Notes

There is a mention of a solution being provided, but it is not visible to all participants. The discussion also highlights a lack of familiarity with polar coordinates among some participants, which may affect their understanding of the problem.

mrcleanhands

Homework Statement



attachment.php?attachmentid=58209&stc=1&d=1366915312.jpg


Homework Equations





The Attempt at a Solution


Solution is given.

I don't understand how +-∏/4 is found as a range for θ
Also why is 0 <= r <= cos2θ

r is always r which is defined as cos2θ
 

Attachments

  • annoying.jpg
    annoying.jpg
    17.1 KB · Views: 2,487
Physics news on Phys.org
mrcleanhands said:

Homework Statement



https://docs.google.com/file/d/0BztURiLWDqMyaVplX0tMYUdNNkE/edit?usp=sharing

Homework Equations





The Attempt at a Solution


Solution is given.
Where? I don't see it.

I don't understand how +-∏/4 is found as a range for θ
Did you draw the graph or have one to look at?

Also why is 0 <= r <= cos2θ

r is always r which is defined as cos2θ

That is r on the curve. If you want the area enclosed, r goes from r=0 to r on the curve, no?
 
Last edited by a moderator:
I see the graph has now appeared. Plot a few points by hand, say for ##\theta = -\frac {\pi}4,\,-\frac {\pi}6,\,0,\,\frac {\pi}6,\,\frac {\pi}4## and see if you don't get that right hand loop.
 
Okay I got it. I haven't done much work with polar co-ordinates so I didn't really process that x^2+y^2=r and the circles radius is now going to vary according to cos...

Thanks :)
 

Similar threads

Integral problem: 1/((x^2)(sqrt(4-x^2)))
  • · Replies 3 ·
Replies
3
Views
14K
Why do you get the area of a whole circle when you integrate
  • · Replies 21 ·
Replies
21
Views
3K
Integral with trig substitution
  • · Replies 3 ·
Replies
3
Views
2K
Indefinite Integral with integration by parts
  • · Replies 2 ·
Replies
2
Views
2K
Symmetry of an Integral of a Dot product
  • · Replies 9 ·
Replies
9
Views
2K
Integrating with a given substitution
  • · Replies 4 ·
Replies
4
Views
2K
Problem with a double integral
  • · Replies 4 ·
Replies
4
Views
1K
Change of Variables: Integrals w/Polar Coordinates
  • · Replies 2 ·
Replies
2
Views
2K
How to Solve Problems Involving AC Voltage Using Integral Calculus
  • · Replies 3 ·
Replies
3
Views
2K
Finding population density with double integrals
  • · Replies 11 ·
Replies
11
Views
4K