Answer
$M_x=M_y=0$
Work Step by Step
$M=\int_{-1}^1 \int_{-1} (x^2+y^2+\dfrac{1}{3}) \ dy \ dx=\int_{-1}^1 (2x^2 +\dfrac{4}{3}) \ dx=4$
Now, $M_x=\int_{-1}^1 \int_{-1} y(x^2+y^2+\dfrac{1}{3}) \ dy \ dx=0$
and $M_y=\int_{-1}^1 \int_{-1} x(x^2+y^2+\dfrac{1}{3}) \ dy \ dx=\int_{-1}^1 (2x^3+\dfrac{4}{3} x ) \ dx =0$
