Compute multiple integral of function $\frac{xy}{2}$ within a domain D that is area formed by following curves: $L_1: x=0, L_2: x^2+y^2=4, L_3:y=-x$ So I need to do a multiple integral and first I wanted to know does it matter which way I do it, first integrate by $y$ and then  by $x$ or vice verse. I gather that it would be any different, because the integrals $f_{xy}$ and $f_{yx}$ are equal. I first wanted to integrate it by $y$, because it's simpler that way.
$$\int^{\sqrt{2}}_0 \int^{\sqrt{4-x^2}}_{-x} \frac{xy}{2}dydx+\int^2_{\sqrt{2}} \int^{-\sqrt{4-x^2}}_{\sqrt{4-x^2}}\frac{xy}{2}dydx=\int^{\sqrt{2}}_0 \left. \frac{xy^2}{4}\right|^{\sqrt{4-x^2}}_{-x}dx + 0=\frac{4x-x^3}{4}-\frac{x^3}{4}=\frac{1}{4}\int^{\sqrt{2}}_0 4x-2x^3 dx=\left. \frac{x^2}{2}\right|^{\sqrt{2}}_0=\frac{1}{4} (2x^2-\frac{x^4}{2}\left. \right|^{\sqrt{2}}_0)=\frac{7}{8}$$
I assume that this answer is incorrect as it would be too simple. So, I would like to know here my error is. Also, I don't know if I get noting right, because it looks like too integral multiplication in the beginning, but I heard that it doesn't matter where do you write that $dx$, I may be awfully wrong. The second question is about the domain that I try to calculate. I sketched the domain and from there it looks like the function $x^2+y^2=4$ is above the $y=-x$ so I made the $-x$ the lower bound of the integral and $2-x$ the upper. The same goes, where I calculate the other integral by $y$.
 A: The domain is a portion of the circle centered at the origin with radius equal to $2$ and therefore the set up should be in polar coordinates
$$\int_{\theta_1}^{\theta_2}\int_0^2\frac{\cos \theta \sin \theta}{2} r^3drd\theta=$$
where the limit for $\theta$ depends upon which part we are considering for the domain.

A: Since OP specifies in the comments that the integral must be done using Cartesian coordinates, here is an indication of how to set that up for one of the possible domains.
The "blue" domain is fairly straightforward but the "orange" domain must be broken into two parts depending upon whether $0\le x\le\sqrt{2}$ or $\sqrt{2}\le x\le2$
when integrating in the order $dx\,dy$. If integrating in the order $dy\,dx$ the integral will have to be broken into two parts depending upon whether $-\sqrt{2}\le y\le0$ or $0\le y\le2$.

$$ \int_D\int\frac{xy}{2}\,dA=\int_0^\sqrt{2}\int_{-x}^\sqrt{4-x^2}\frac{xy}{2}\,dy\,dx +\int_\sqrt{2}^2\int_{-\sqrt{4-x^2}}^\sqrt{4-x^2}\frac{xy}{2}\,dy\,dx$$
You should find that the first integral has a value of $\frac{1}{2}$ and the second, a value of $0$.
