# Double integral for loop of rose r=cos2θ

1. Apr 25, 2013

### mrcleanhands

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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

#### Attached Files:

• ###### annoying.jpg
File size:
17.1 KB
Views:
360
2. Apr 25, 2013

### LCKurtz

Where? I don't see it.

Did you draw the graph or have one to look at?

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: May 6, 2017
3. Apr 25, 2013

### 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.

4. Apr 27, 2013

### 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 :)