Double integral over square shaped region $$\iint_R(y-2x^2)dxdy$$ where $R$ is the region inside the square $|x|+|y|=1$.
So the area is:
\begin{align}
4×\left[\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}(y-2x^2)dydx\right]
&=4×\left[\int_0^1\int_{y=0}^{y=1-x}ydydx-2\int_0^1\int_{y=0}^{y=1-x}x^2dydx\right]\\
&=4×\left[\frac{1}{2}\int_0^1(1-x)^2dx-2\int_0^1x^2(1-x)dx\right]\\
&=4×\left[\int_0^1(1-x)^2dx-2\int_0^1x^2dx+2\int_0^1x^3dx\right]\\
&=4×\left[-\frac{1}{6}[(1-x)^3]_0^1-\frac{2}{3}[x^3]_0^1+\frac{2}{4}[x^4]_0^1\right]\\
&=4×\left[\frac{1}{6}-\frac{2}{3}+\frac{1}{2}\right]\\
&=0
\end{align}
I can't find my fault please check this..
 A: Here is the region $R$ of integration

It is given by the equations $$y=x+1,\quad y=x-1,\quad y=-x+1,\quad y=-x-1$$
As Nick has noted in the comments, you cannot split the integral into 4 symmetric pieces, because the function you're integrating on this region is not symmetric itself.
We have $$\iint_Ry-2x^2\text{ d}x\text{ d}y$$
Let's make a change of variables so we don't have to break this area up into two integrals. We have $$x=u+v,\phantom{x}y=u-v,\quad J_f=\left[\begin{array}{cc}
     1& 1 \\
     1& -1 
\end{array}\right],\qquad \left|\det\left(J_f\right)\right|=2$$
Plugging in our transformation into the equations defining our region, we have
$$y=x+1\implies v=-\frac12$$
$$y=x-1\implies v=\frac12$$
$$y=-x+1\implies u=\frac12$$
$$y=-x-1\implies u=-\frac12$$

Our region is now the square inside the 4 constant lines.
Our integral on the other hand transforms into
\begin{align*}
\iint_Ry-2x^2\text{ d}x\text{ d}y&=\int_{-\frac12}^{\frac12}\int_{-\frac12}^{\frac12}\left((u-v)-2(u+v)^2\right)\cdot 2\text{ d}u\text{ d}v\\
&=\int_{-\frac12}^{\frac12}\int_{-\frac12}^{\frac12}-4 u^2 - 8 u v + 2 u - 4 v^2 - 2 v\text{ d}u\text{ d}v\\
&=\int_{-\frac12}^{\frac12}-\frac13 - 2 v - 4 v^2\text{ d}v\\
&=\boxed{-\frac23}
\end{align*}
If you want to do it the regular two integral way, it would be slightly more complicated. Since the region does not pass the vertical line test, we must split it into regions that do. Here i will split it by slicing it across the $y$ axis.
You'd do something like this

It's annoying though and I won't latex it.
A: HINT
The proposed integral can be expressed as follows:
\begin{align*}
I = \int_{-1}^{0}\int_{-x-1}^{x+1}(y - 2x^{2})\mathrm{d}y\mathrm{d}x + \int_{0}^{1}\int_{x - 1}^{-x + 1}(y - 2x^{2})\mathrm{d}y\mathrm{d}x
\end{align*}
Can you take it from here?
A: First, let's break it into two integrals:
$$
\underbrace{\iint_R y - 2x^2 \ \ dxdy}_{I} = \underbrace{\iint_R y \ \ dxdy}_{I_1} -2 \underbrace{\iint x^2\ \ dxdy}_{I_2} 
$$

*

*As $y$ is an odd function, then the integral over the upper part is the negative of lower part:

\begin{align*}
I_{1} & = \iint_{R} y \  \  dxdy  \\
& = \iint_{lower} y \ dx dy + \iint_{upper} y \ \ dx dy \\
& = \iint_{lower} y \ dx \ dy + \left(-\iint_{lower} y \ \ dx dy\right) \\
& = 0
\end{align*}


*As $x^2$ is an even function, then the integral over the left part is the same as the right part:

\begin{align*}
I_{2} & = \iint_{R} x^2 \  \  dxdy  \\
& = \iint_{left} x^2 \ dx dy + \iint_{right} x^2 \ \ dx dy \\
& = \iint_{right} x^2 \ dx \ dy + \iint_{right} x^2 \ \ dx dy \\
& = 2\iint_{right} x^2 \ dx \ dy  \\
& = 2 \int_{0}^{1} \left( \int_{-1+x}^{1-x} x^2 \ dy \right) dx \\
& = 2 \int_{0}^{1} x^2 \cdot \left([1-x] - [-1+x]\right) \ dx \\
& = 4 \int_{0}^{1} x^2 (1-x) \ dx \\
& = 4 \int_{0}^{1} x^2 -x^3 \ dx \\
& = 4 \left[\dfrac{x^3}{3} - \dfrac{x^4}{4}\right]_{0}^{1}  \\
& = 4 \left[\dfrac{1}{3} - \dfrac{1}{4}\right] = \dfrac{1}{3}
\end{align*}
Then
$$
\boxed{I = I_1 - 2I_2 = 0 - 2\cdot \dfrac{1}{3} = \dfrac{-2}{3}}
$$
