Double integral for loop of rose r=cos2θ

  • #1

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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Answers and Replies

  • #2
LCKurtz
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Homework Statement



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

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?
 
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  • #3
LCKurtz
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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.
 
  • #4
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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