Chapter 14 - Section 14.1 - Double and Iterated Integrals over Rectangles - Exercises - Page 760: 26

$\dfrac{160}{3}$

Re-arrange the given integral as follows: $V=\int_{0}^{2} \int_{0}^{2} (16-x^2-y^2) dy dx$ This implies that $\int_{0}^{2} (16y-x^2y-\dfrac{y^3}{y}]_{0}^{2} dx=\int_{0}^{2} [32-2x^2-\dfrac{8}{3}] dx$ $\int_0^2 (\dfrac{88}{3}-2x^2)dx=[\dfrac{88}{3}x-\dfrac{2x^3}{3}]_0^2$ Hence, $V=\dfrac{88}{3}(2)-\dfrac{2(2)^3}{3}=\dfrac{160}{3}$