Answer
The center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( {\frac{1}{6},\frac{1}{6}} \right)$.
Work Step by Step
We have the region ${\cal D}:\left| x \right| + \left| y \right| \le 1$ and the mass density $\delta \left( {x,y} \right) = \left( {x + 1} \right)\left( {y + 1} \right)$.
We sketch the graph and notice that we can divide ${\cal D}$ into two horizontally simple region: ${{\cal D}_1}$ and ${{\cal D}_2}$ such that $D = {{\cal D}_1}\bigcup {{\cal D}_2}$.
1. Region ${{\cal D}_1}$
The region ${{\cal D}_1}$ consists of the part of ${\cal D}$ in the first and the second quadrant, bounded left by the line $y=1+x$ and bounded right by the line $y=1-x$. We describe ${{\cal D}_1}$ as a horizontally simple region, thus the description:
${{\cal D}_1} = \left\{ {\left( {x,y} \right)|0 \le y \le 1,y - 1 \le x \le - \left( {y - 1} \right)} \right\}$
2. Region ${{\cal D}_2}$
The region ${{\cal D}_2}$ consists of the part of ${\cal D}$ in the third and the fourth quadrant, bounded left by the line $y=-1-x$ and bounded right by the line $y=-1+x$. We describe ${{\cal D}_2}$ as a horizontally simple region, thus the description:
${{\cal D}_2} = \left\{ {\left( {x,y} \right)| - 1 \le y \le 0, - \left( {y + 1} \right) \le x \le y + 1} \right\}$
Step 1. Evaluate the mass of ${\cal D}$:
Using the linearity property of the double integral:
$M = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} \delta \left( {x,y} \right){\rm{d}}A + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} \delta \left( {x,y} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = y - 1}^{ - \left( {y - 1} \right)} \left( {x + 1} \right)\left( {y + 1} \right){\rm{d}}x{\rm{d}}y + \mathop \smallint \limits_{y = - 1}^0 \mathop \smallint \limits_{x = - \left( {y + 1} \right)}^{y + 1} \left( {x + 1} \right)\left( {y + 1} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^1 \left( {\left( {\frac{1}{2}{x^2} + x} \right)|_{y - 1}^{ - \left( {y - 1} \right)}} \right)\left( {y + 1} \right){\rm{d}}y + \mathop \smallint \limits_{y = - 1}^0 \left( {\left( {\frac{1}{2}{x^2} + x} \right)|_{ - \left( {y + 1} \right)}^{y + 1}} \right)\left( {y + 1} \right){\rm{d}}y$
Since the terms in even exponents cancel, so
$M = \mathop \smallint \limits_{y = 0}^1 \left( { - y + 1 - y + 1} \right)\left( {y + 1} \right){\rm{d}}y + \mathop \smallint \limits_{y = - 1}^0 \left( {y + 1 + y + 1} \right)\left( {y + 1} \right){\rm{d}}y$
$ = 2\mathop \smallint \limits_{y = 0}^1 \left( { - y + 1} \right)\left( {y + 1} \right){\rm{d}}y + 2\mathop \smallint \limits_{y = - 1}^0 \left( {y + 1} \right)\left( {y + 1} \right){\rm{d}}y$
$ = 2\mathop \smallint \limits_{y = 0}^1 \left( { - {y^2} + 1} \right){\rm{d}}y + 2\mathop \smallint \limits_{y = - 1}^0 \left( {{y^2} + 2y + 1} \right){\rm{d}}y$
$ = 2\left( { - \frac{1}{3}{y^3} + y} \right)|_0^1 + 2\left( {\frac{1}{3}{y^3} + {y^2} + y} \right)|_{ - 1}^0$
$ = 2\left( { - \frac{1}{3} + 1} \right) - 2\left( { - \frac{1}{3} + 1 - 1} \right) = 2$
So, $M=2$.
Step 2. Evaluate the $x$-coordinate of the center of mass:
${x_{CM}} = \frac{{{M_y}}}{M} = \frac{1}{2}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\delta \left( {x,y} \right){\rm{d}}A$
$ = \frac{1}{2}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} x\delta \left( {x,y} \right){\rm{d}}A + \frac{1}{2}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} x\delta \left( {x,y} \right){\rm{d}}A$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = y - 1}^{ - \left( {y - 1} \right)} x\left( {x + 1} \right)\left( {y + 1} \right){\rm{d}}x{\rm{d}}y + \frac{1}{2}\mathop \smallint \limits_{y = - 1}^0 \mathop \smallint \limits_{x = - \left( {y + 1} \right)}^{y + 1} x\left( {x + 1} \right)\left( {y + 1} \right){\rm{d}}x{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = y - 1}^{ - \left( {y - 1} \right)} \left( {{x^2} + x} \right)\left( {y + 1} \right){\rm{d}}x{\rm{d}}y + \frac{1}{2}\mathop \smallint \limits_{y = - 1}^0 \mathop \smallint \limits_{x = - \left( {y + 1} \right)}^{y + 1} \left( {{x^2} + x} \right)\left( {y + 1} \right){\rm{d}}x{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 \left( {\left( {\frac{1}{3}{x^3} + \frac{1}{2}{x^2}} \right)|_{y - 1}^{ - \left( {y - 1} \right)}} \right)\left( {y + 1} \right){\rm{d}}y + \frac{1}{2}\mathop \smallint \limits_{y = - 1}^0 \left( {\left( {\frac{1}{3}{x^3} + \frac{1}{2}{x^2}} \right)|_{ - \left( {y + 1} \right)}^{y + 1}} \right)\left( {y + 1} \right){\rm{d}}y$
Since the terms in even exponents cancel, so
${x_{CM}} = - \frac{1}{3}\mathop \smallint \limits_{y = 0}^1 {\left( {y - 1} \right)^3}\left( {y + 1} \right){\rm{d}}y + \frac{1}{3}\mathop \smallint \limits_{y = - 1}^0 {\left( {y + 1} \right)^3}\left( {y + 1} \right){\rm{d}}y$
$ = - \frac{1}{3}\mathop \smallint \limits_{y = 0}^1 \left( {{y^4} - 2{y^3} + 2y - 1} \right){\rm{d}}y + \frac{1}{3}\mathop \smallint \limits_{y = - 1}^0 \left( {{y^4} + 4{y^3} + 6{y^2} + 4y + 1} \right){\rm{d}}y$
$ = - \frac{1}{3}\left( {\left( {\frac{1}{5}{y^5} - \frac{1}{2}{y^4} + {y^2} - y} \right)|_0^1} \right) + \frac{1}{3}\left( {\left( {\frac{1}{5}{y^5} + {y^4} + 2{y^3} + 2{y^2} + y} \right)|_{ - 1}^0} \right)$
$ = - \frac{1}{3}\left( {\frac{1}{5} - \frac{1}{2} + 1 - 1} \right) - \frac{1}{3}\left( { - \frac{1}{5} + 1 - 2 + 2 - 1} \right) = \frac{1}{6}$
Step 3. Evaluate the $y$-coordinate of the center of mass:
${y_{CM}} = \frac{{{M_x}}}{M} = \frac{1}{2}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y\delta \left( {x,y} \right){\rm{d}}A$
$ = \frac{1}{2}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} y\delta \left( {x,y} \right){\rm{d}}A + \frac{1}{2}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} y\delta \left( {x,y} \right){\rm{d}}A$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = y - 1}^{ - \left( {y - 1} \right)} y\left( {x + 1} \right)\left( {y + 1} \right){\rm{d}}x{\rm{d}}y + \frac{1}{2}\mathop \smallint \limits_{y = - 1}^0 \mathop \smallint \limits_{x = - \left( {y + 1} \right)}^{y + 1} y\left( {x + 1} \right)\left( {y + 1} \right){\rm{d}}x{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 y\left( {y + 1} \right)\left( {\left( {\frac{1}{2}{x^2} + x} \right)|_{y - 1}^{ - \left( {y - 1} \right)}} \right){\rm{d}}y + \frac{1}{2}\mathop \smallint \limits_{y = - 1}^0 y\left( {y + 1} \right)\left( {\left( {\frac{1}{2}{x^2} + x} \right)|_{ - \left( {y + 1} \right)}^{y + 1}} \right){\rm{d}}y$
Since the terms in even exponents cancel, so
${y_{CM}} = - \mathop \smallint \limits_{y = 0}^1 y\left( {y + 1} \right)\left( {y - 1} \right){\rm{d}}y + \mathop \smallint \limits_{y = - 1}^0 y\left( {y + 1} \right)\left( {y + 1} \right){\rm{d}}y$
$ = - \mathop \smallint \limits_{y = 0}^1 \left( {{y^3} - y} \right){\rm{d}}y + \mathop \smallint \limits_{y = - 1}^0 \left( {{y^3} + 2{y^2} + y} \right){\rm{d}}y$
$ = - \left( {\left( {\frac{1}{4}{y^4} - \frac{1}{2}{y^2}} \right)|_0^1} \right) + \left( {\frac{1}{4}{y^4} + \frac{2}{3}{y^3} + \frac{1}{2}{y^2}} \right)|_{ - 1}^0$
$ = - \left( {\frac{1}{4} - \frac{1}{2}} \right) - \left( {\frac{1}{4} - \frac{2}{3} + \frac{1}{2}} \right) = \frac{1}{6}$
Thus, the center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( {\frac{1}{6},\frac{1}{6}} \right)$.
