Answer
$$\dfrac{128}{15}$$
Work Step by Step
We must integrate the integral as follows:
$$ \iint_{R} f(x,y) dA = \int_{0}^{2} \int_{0}^{4-x^2} (4-x^2-y) dy dx \\= \int_{0}^{2}[4y-\dfrac{y^2}{2}- x^2y]_0^{4-x^2} dx \\=\int_{0}^{2} (8-4x^2+\dfrac{x^4}{2}) dx \\=[8x-\dfrac{4x^3}{3}+\dfrac{x^5}{10}]_0^2 \\=\dfrac{32}{10} -\dfrac{32}{3}+16-0 \\=\dfrac{128}{15}$$
