Area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ I am trying to find the area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ using Double integrl in Polar coordinates.

Now as shown in the figure the angle $\theta$ takes values from $0$ to $\frac{\pi}{2}$.
Now $r$ depends on $\theta$.
From $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$,substituting $x=r\cos(\theta)$ and $y=r\sin(\theta)$ we get
$$r=\sin(\theta)$$ and like-wise substituting $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ we get
$$r=\cos(\theta)$$
So the double integral which gives the area is:
$$A=\int_{\theta=0}^{\pi / 2} \int_{r=\sin \theta}^{\cos \theta} r d r d \theta \text { . }$$
But i am getting answer as ZERO.

Try setting your integral as
$A=2\int_0^\frac{\pi}{4}\int_{\sin\theta}^{\cos\theta}rdrd\theta$
