## Thomas' Calculus 13th Edition

$M_x=M_y=0$
$$M=\int_{-1}^1 \int_{-1} (x^2+y^2+\dfrac{1}{3}) \ dy \ dx \\=\int_{-1}^1 (2x^2 +\dfrac{4}{3}) \ dx \\=4$$ Now, $$M_x=\int_{-1}^1 \int_{-1} y \times (x^2+y^2+\dfrac{1}{3}) \ dy \ dx \\=0$$ $$M_y=\int_{-1}^1 \int_{-1} x \times (x^2+y^2+\dfrac{1}{3}) \ dy \ dx \\=\int_{-1}^1 (2x^3+\dfrac{4}{3} x ) \space dx \\ =0$$