<s>[INST] 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?


 [/INST] 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
$$\int_{-\sqrt{2}}^{\sqrt{2}}\int_x^{\sqrt{4-x^2}}dydx=\int_{-\sqrt{2}}^{\sqrt{2}}\sqrt{4-x^2}-x~dx=2+\pi-2=\pi.$$
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