The area of $\frac{\pi}{4}\leq\theta \leq\frac{\pi}{2},\quad 0\leq r\leq2$ 
Find the area 
  $\frac{\pi}{4}\leq\theta \leq\frac{\pi}{2}\;\;\;\;,0\leq r\leq2$
using double integral

Attempt:
I tried in two differnt ways and got two diffrent answers, can't find my mistakes
first methode: poolar coordinates
$$\int_{\pi/4}^{\pi/2}\int _0^2rdrd\theta=...=\pi/2$$
which is obvios wrong
seconde methode
$$\int_{0}^{\sqrt 2}\int_{x/2}^{x}dydx+\int_{\sqrt 2}^{4/\sqrt 5}\int_{x/2}^{\sqrt{4-x^2}}dydx=...\approx 1/2+0.14$$
 A: Your bounds in the second method are wrong.  They should be $$\int_{0}^{\sqrt{2}}\int_{x}^{\sqrt{4-x^2}}dydx$$
Here's what a plot of the region should look like (easy to plot from polar coordinates).

So my approach here was to first get an inequality of $y$ in terms of $x$.  So coming up from the $x$-axis, we first enter the region at the line $y=x$ (the $\pi/4$ radian line).  Then we leave one we get to the circle $r=2$, or equivalently $x^2+y^2=4$.  Thus I get the bounds on $y$.
Then $x$ is easy.  We just take the lowest and highest values of $x$ in the region.  Clearly $x$ starts at $0$ and then moves along until the very point where its on the $r=2$ circle at $\pi/4$ rad.  That's clearly at $2\cos(\pi/4) = \sqrt{2}$.
Thus the Cartesian bounds are obtained.
Note that this exercise should have been really easy to figure out geometrically.  Your region is just one eighth of a circle and thus its area is $$A = \frac{1}{8}(\pi(2)^2) = \frac{\pi}{2}$$
A: In the first quadrant,  the area is delimited by the arc $y=\sqrt{4-x^2}$ from the top, the line $y=x$ from the bottom,  $x=0$ to the left and $x=2\cos(\dfrac \pi 4)$ to the right.
Alternatively, you can integrate $y=\sqrt{4-x^2}$ between $0$ and $\sqrt 2$ then subtract the area of the triangle $A=\frac 12\sqrt 2 \sqrt 2=1$
A: $$I=\int_{0}^{\sqrt 2}\int_{x}^{\sqrt{2}}dydx+\int_{0}^{\sqrt 2}\int_{\sqrt 2}^{\sqrt{4-x^2}}dydx,$$
$$I=\int_{0}^{\sqrt 2}(\sqrt{2}-x)dx+\int_{0}^{\sqrt 2}(\sqrt{4-x^2}-\sqrt 2)dx.$$
$$I=-1+\int_{0}^{\sqrt 2}\sqrt{4-x^2}dx=-1+J.$$ Now use substitution $x=2\sin{t}$ to solve integral $J$. 
It is easy to calculate $$J=1+\frac{\pi}{2}.$$ Therefore, $$I=\frac{\pi}{2}.$$ The same result you will obtain if you use polar coordinates.
