# Double Integral with Polar Method

I have the following region in the first quadrant bounded by the $$r=1$$, and $$r=\sin(2\theta)$$, with $$\frac{\pi}{4}\le\theta\le\frac{\pi}{2}$$. My question is what goes how to setup the integral, I have the following limits of integration for the double integral:$$\begin{equation}\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_\limits{1}^{\sin(2\theta)}f(r\cos(\theta),r\sin(\theta))drd\theta\end{equation}$$

I am curious to know what is the integrand in this case. Or in layman terms how do I determine the f in this case.

• Who is $f$? In general, $f(x,y)$ becomes $f(r \cos \theta, r \sin \theta)$; you've forgotten the modulus of the jacobian $r$. Apr 30, 2020 at 0:48
• @Dunkelheit I trying to figure out what is the integrad, or the function inside the integral. Apr 30, 2020 at 0:53

The answer in the back of the textbook is the following $$\frac{\pi}{16}$$. I realized that my mistake is in the limits of integration, in choosing what function is in the integrand boils down to a simple manner. The integrand is implicitly defined in this problem as $$r$$. I first had to realize that it is the other radius function to the inner radius function as the limits of the function. So to effect the integral can be defined as follows:\begin{align}\iint_RrdA&=\frac{\pi}{16} \\ \displaystyle{\int_{\frac{\pi}4}^{\frac{\pi}{2}}\int_{\sin(2\theta)}^{1}}r\ dr d\theta&=\frac{\pi}{16} \end{align}