Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 883: 62

$\dfrac{128}{15}$

Consider $I= \iint_{R} f(x,y) dA$ or, $= \int_{0}^{2} \int_{0}^{4-x^2} (4-x^2-y) dy dx$ or, $= \int_{0}^{2}[4y-\dfrac{y^2}{2}- x^2y]_0^{4-x^2} dx$ or, $=\int_{0}^{2} (8-4x^2+\dfrac{x^4}{2}) dx$ Thus, we have $I=\dfrac{32}{10} -\dfrac{32}{3}+16-0=\dfrac{128}{15}$