Double integral for loop of rose r=cos2θ

[mrcleanhands]

Homework Statement

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

 

[LCKurtz]

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?

 

[LCKurtz]

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.

 

[mrcleanhands]

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 :)