Answer
$$16$$
Work Step by Step
We must integrate the integral as follows:
$$ \iint_{R} f(x,y) dA = \int_{0}^{2} \int_{0}^{3} (4-y^2) dx dy \\= \int_{0}^{2}[4x-y^2 x]_0^{3} dy \\=\int_{0}^{2} (12-3y^2) dy \\=[12y -\dfrac{3y^3}{3}]_{0}^{2} \\=12(2-0)-(2)^3 \\=16$$
