Answer
$$20$$
Work Step by Step
We must integrate the integral as follows:
$$ \iint_{R} f(x,y) dA = \int_{0}^{2} \int_{0}^{2-x} (12-3y^2) dy dx \\= \int_{0}^{2}[12y-y^3]_0^{2-x} dx \\=\int_{0}^{2} (24-12x-(2-x)^3) dx \\=[24 x-6x^2+ \dfrac{(2-x)^4}{4}]_0^2 \\ = 24(2-0) -6[4-0]+[\dfrac{(2-2)^4}{4}-\dfrac{(2-0)^4}{4}] \\=20$$
