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..

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
$$
\boxed{I = I_1 - 2I_2 = 0 - 2\cdot \dfrac{1}{3} = \dfrac{-2}{3}}
$$
