Answer
$$128\pi $$
Work Step by Step
$$\eqalign{
& \int_0^4 {\int_{ - 2\sqrt {16 - {y^2}} }^{2\sqrt {16 - {y^2}} } {\int_0^{16 - {x^2}/4 - {y^2}} {dzdydx} } } \cr
& {\text{Integrate with respecto to }}z \cr
& \int_0^4 {\int_{ - 2\sqrt {16 - {y^2}} }^{2\sqrt {16 - {y^2}} } {\left[ z \right]_0^{16 - {x^2}/4 - {y^2}}dx} dy} \cr
& \int_0^4 {\int_{ - 2\sqrt {16 - {y^2}} }^{2\sqrt {16 - {y^2}} } {\left( {16 - \frac{{{x^2}}}{4} - {y^2}} \right)dx} dy} \cr
& {\text{Integrate with respecto to }}x \cr
& = \int_0^4 {\left[ {16x - \frac{{{x^3}}}{{12}} - x{y^2}} \right]_{ - 2\sqrt {16 - {y^2}} }^{2\sqrt {16 - {y^2}} }dy} \cr
& = \int_0^4 {\left[ {64\sqrt {16 - {y^2}} - \frac{{4{{\left( {\sqrt {16 - {y^2}} } \right)}^3}}}{3} - 4{y^2}\sqrt {16 - {y^2}} } \right]dy} \cr
& {\text{Let }} \cr
& y = 4\sin \theta ,{\text{ }}\sqrt {16 - {y^2}} = 4\cos \theta ,{\text{ }}dy = 4\cos \theta d\theta \cr
& = \int_0^{\pi /2} {\left( {256\cos \theta - \frac{{256{{\cos }^3}\theta }}{3} - 256{{\sin }^2}\theta \cos \theta } \right)\left( {4\cos \theta } \right)d\theta } \cr
& = 1024\int_0^{\pi /2} {\left( {{{\cos }^2}\theta - \frac{1}{3}{{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)d\theta } \cr
& = 1024\int_0^{\pi /2} {\left( {{{\cos }^2}\theta - \frac{1}{3}\left( {1 - {{\sin }^2}\theta } \right){{\cos }^2}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)d\theta } \cr
& = 1024\int_0^{\pi /2} {\left( {{{\cos }^2}\theta - \frac{1}{3}{{\cos }^2}\theta + \frac{1}{3}{{\sin }^2}\theta {{\cos }^2}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)d\theta } \cr
& = 1024\int_0^{\pi /2} {\left( {\frac{2}{3}{{\cos }^2}\theta - \frac{2}{3}{{\sin }^2}\theta {{\cos }^2}\theta } \right)d\theta } \cr
& = \frac{{2048}}{3}\int_0^{\pi /2} {\left( {{{\cos }^2}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)d\theta } \cr
& = \frac{{2048}}{3}\int_0^{\pi /2} {{{\cos }^2}\theta \left( {1 - {{\sin }^2}\theta } \right)d\theta } \cr
& = \frac{{2048}}{3}\int_0^{\pi /2} {{{\cos }^4}\theta d\theta } \cr
& {\text{Integrating}} \cr
& = \frac{{2048}}{3}\left( {\frac{3}{{16}}\pi } \right) \cr
& = \frac{{2048}}{{16}}\pi \cr
& = 128\pi \cr} $$
