Double integral for loop of rose r=cos2θ

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SUMMARY

The discussion focuses on understanding the double integral for the polar rose defined by the equation r = cos(2θ). Participants clarify the range for θ, which is determined to be ±π/4, and the limits for r, which range from 0 to cos(2θ). The importance of graphing the function to visualize the area enclosed by the curve is emphasized, particularly for those unfamiliar with polar coordinates. The conversation highlights the necessity of plotting points to comprehend the behavior of the curve.

PREREQUISITES
  • Understanding of polar coordinates and their representation
  • Familiarity with double integrals in calculus
  • Basic graphing skills for polar equations
  • Knowledge of trigonometric functions, specifically cosine
NEXT STEPS
  • Learn how to graph polar equations effectively
  • Study the properties of polar roses and their integrals
  • Explore the concept of area enclosed by polar curves
  • Practice solving double integrals involving polar coordinates
USEFUL FOR

Students studying calculus, particularly those focusing on polar coordinates and double integrals, as well as educators looking to enhance their teaching methods in these areas.

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
 

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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 :)
 

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