Question about integrals in polar coordinates

Question

I've just made on question where is asked the area of a region enclosed by one loop of rose $r=\cos3\theta$ and had one uncertained. In this case, the figure is the following:

Suppose if it is asked to me to find the area of a certain region, in which $r=f(\theta)$, and the figure is something like this:

My $\theta$ variation is between the brown lines (exterior) or green lines (interior)?

Answer 1

In your 2nd example, the bounds for $\theta$ would be the brown lines.

If you set up an integral with the green lines as the bounds for $\theta$, you would only get the area of the red region that is in between the green lines.

Note though that you the bounds for $r$ (which will depend on $\theta$) might not be a simple function of $\theta$, so you may have to split the integral into pieces, one for the part of the region in between the green lines, and another for the parts of the region outside the green lines.

Answer 2

simply ,you can think it as that $dr$ will span the whole curve and the angle that it sweeps is $d\theta $ so you can integrate over both of them .as in above case $\int_{-\pi/6}^{\pi/6}\int_{0}^{r} dr d\theta $= $\int_{-\pi/6}^{\pi/6}r d\theta $= $\int_{-\pi/6}^{\pi/6}cos(3\theta) d\theta $.