Finding the area between : $\frac{\pi}{3}\leq \theta \leq \frac{2 \pi}{3}, 0 \leq r \leq 1$ using polar coordinates 
I need to find the area between : $\frac{\pi}{3}\leq \theta \leq \frac{2 \pi}{3}, 0 \leq r \leq 1$ using polar coordinates $x=r\cos \theta , \, y=r\sin \theta$

$$x^2+y^2=r \Longrightarrow r=1$$
My attempt:
$$\iint (r\cdot J )\, dr \,d\theta$$
Using double integral: 
$$\color{blue}{J=r=1}:$$ 
$$\int\limits_{y=\frac{\pi}{3}}^{y=\frac{2\pi}{3}}\left[~\int\limits_{x=0}^{x=1} r^2\,dr \right]d\theta=\int\limits_{y=\frac{\pi}{3}}^{y=\frac{2\pi}{3}} \frac{r^3}{3} \,d\theta=\frac{r^4}{12}\bigg|_\frac{\pi}{3}^\frac{2\pi}{3}=\boxed{\color{blue} {\frac{\pi}{36}}} $$ 
The answer should be : $\boxed{\color{red} {\frac{\pi}{6}}} $

Where am I wrong?


Here's how it should look:
$$\int_{\pi/3}^{2 \pi / 3} \int_0^1 \color{blue}{r} \, dr \, d \theta = \int_{\pi/3}^{2 \pi/3} \left. \color{blue}{\frac{r^2}{2}} \right|_0^1 d \theta$$
$$= \int_{\pi/3}^{2 \pi/3} \frac{1}{2} \, d \theta = \left. \frac{1}{2} \theta \right|_{\pi/3}^{2\pi/3} = \color{red}{\frac{\pi}{6}}$$
You did two things wrong: One, you had $r^2$ instead of $r$. 
Second, you did not evaluate inner definite integral; instead you took the indefinite integral and then used that result in the outer integral.
Visually, the inner integral represents a straight line from the origin that is $1$ unit long. Then, your $d \theta$ integral takes that straight line and "sweeps" along the origin at angle $\frac{\pi}{3}$ to the angle $\frac{2 \pi}{3}$. This is how the area of the region is found.

I am unfamiliar with the $J$ notation. If you could explain it I could tell you why it gave you a wrong answer.
