Evaluate $\iint dy\,dx;\frac{\pi}{4}\leq\theta \leq\frac{3\pi}{4};0\leq r\leq2$ I need to evaluate 
$\displaystyle\iint \color{red}{dydx}\;\;\;,\frac{\pi}{4}\leq\theta \leq\frac{3\pi}{4}\;\;\;\;,0\leq r\leq2$ 
$\color{blue}{\text{without using polar coordinates}}$.
My attempt:
$$\int_{x=-2}^{x=0}\int_{y=-x}^{\sqrt{4-x^2}}dydx+\int_{x=0}^{x=2}\int_{y=x}^{y=\sqrt{4-x^2}}dydx$$

Is the correct?


 A: I guess you need to evaluate the integral $$\iint_A1\,dxdy,$$ where $A=\psi(\{(r, \theta)\in[0, \infty)\times[0,2\pi]:\pi/4\leq\theta\leq3/4\,\pi \quad\text{and}\quad0\leq r\leq2\})$ and $\psi$ is the polar coordinates map.
Then $A=\{(x,y)\in \Bbb R^2: -\sqrt2\leq x\leq\sqrt2\quad \text{and}\quad |x|\leq y\leq\sqrt{4-x^2}\}$.
So by noticing that $A$ is a $x$-simple region and by applying the reduction formula, your integral becomes $$\int_{-\sqrt2}^\sqrt2\left(\int_{|x|}^{\sqrt{4-x^2}}1\,dy\right)\,dx,$$
which can be split into $$\int_{-\sqrt2}^0\left(\int_{-x}^{\sqrt{4-x^2}}1\,dy\right)\,dx+\int_{0}^\sqrt2\left(\int_{x}^{\sqrt{4-x^2}}1\,dy\right)\,dx $$ to get rid of the absolute value sign.
A: There is a mistake in your reasoning, I'll illustrate this with the double integral on the right. Letting $x$ vary from $0$ to $2$ will give  

where green counts positive and red counts negative.
A: At $\theta=\pi/4$ and $y<\sqrt{2}$, $x=y$. Similarly, at $\theta=3\pi/4$ and $y<\sqrt{2}$, $x=-y$.
At $y\ge\sqrt{2}$, $-\sqrt{4-y^2}\le x\le\sqrt{4-y^2}$.
So you can write your integral as
$$\int_0^{\sqrt{2}}\int_{-y}^ydxdy + \int_{\sqrt{2}}^2\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}dxdy=\int_0^{\sqrt{2}}2ydy+\int_{\sqrt{2}}^22\sqrt{4-y^2}dy=2+\pi-2=\pi.$$
EDIT (in response to comment): If you wish to evaluate the integral with respect to $y$ first... well, $x$ goes from $-\sqrt{2}$ to $\sqrt{2}$. For any given $x$, the lower bound on $y$ is $x$; the upper bound, $\sqrt{4-x^2}$. So then, your integral reads
$$\int_{-\sqrt{2}}^{\sqrt{2}}\int_x^{\sqrt{4-x^2}}dydx=\int_{-\sqrt{2}}^{\sqrt{2}}\sqrt{4-x^2}-x~dx=2+\pi-2=\pi.$$
A: Using Green's Theorem, we have 
$$\iint_D dA=\oint_C xdy \tag 1$$
Here, the contour $C$ in $(1)$ is comprised of $3$ segments; $C_1$, $C_2$, and $C_3$.  The integrals over these separate segments are for $C_1$, x=y
$$\begin{align}
\int_{C_1}xdy&=\int_0^{\sqrt{2}} ydy\\\\
&=1 \tag 2
\end{align}$$
for $C_2$, $x=-y$
$$\begin{align}
\int_{C_2}xdy&=\int_{\sqrt{2}}^0 -ydy\\\\
&=1\tag 3
\end{align}$$
and finally for $C_3$, $x=\pm \sqrt{4-y^2}$
$$\begin{align}
\int_{C_3}xdy&=2\int_{\sqrt{2}}^2\sqrt{4-y^2}dy\\\\
&=\pi-2\tag 4
\end{align}$$
Thus, adding the right-hand sides of $(2)-(4)$ reveals
$$\bbox[5px,border:2px solid #C0A000]{\iint_D dA=\pi}$$
as expected!
