Shared area of $2$ circles (polar integral) Given $x^2+y^2=4y$ and $x^2+y^2=4x$ find the shared area  of the 2 circles
What I tried was :
so first I transformed them to polar coordinates
I got $r=4cos\theta$ for $x^2+y^2=4x$ and $r=4sin\theta$ for $x^2+y^2=4y$
After that I did $4cos\theta=4sin\theta$ then $\theta=\frac{\pi}{4}$ I am not sure but according to this I believe that $\frac{\pi}{4}≤\theta≤\frac{\pi}{2}$ lastly I did the integral
$\int_\frac{\pi}{4}^\frac{\pi}{2}\frac{1}{2}((4cos\theta)^2-(4sin\theta)^2) d\theta $
after integrating I got $-4$ which is a wrong answer.
what am I doing wrong? thanks for any help and tips!
Edit: I forgot to mention that I havent studied double integrals yet , I learned polar system and this formula $\int_\alpha^\beta\frac{1}{2}(r(\theta))^2 d\theta $


Please note that for $0 \leq \theta \leq \frac{\pi}{4}$, you are bound by circle $4 \sin\theta $ and for $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$, you are bound by circle $4 \cos\theta$.
So the area is given by,  $A =  \displaystyle 2 \int_0^{\pi/4} \int_0^{4\sin\theta} r \ dr \ d\theta$
or $ \ \displaystyle 2 \int_0^{\pi/4} \frac{1}{2}(4\sin\theta)^2 \ d\theta$
You can see that the area between $0 \leq \theta \leq \frac{\pi}{4}$ and between $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$ are same. So we multiply by $2$ or use the fact that $\sin\theta = \cos (\frac{\pi}{2} - \theta)$.
