Such a Bullet

The volume can be written as two times the volume of a half (for positive values of $x$ ). This means that the volume equals the sum of all cylinders with height $dx$ and base $\pi\cdot y^2(x)$ . So the volume becomes: $V=2\displaystyle \int_{0}^{4}\pi\cdot 9(1-\frac{x^2}{16})dx=18\pi \displaystyle \int_{0}^{4} (1-\frac{x^2}{16})dx$ $V=18\pi\cdot 4 - 18\pi\cdot \frac{1}{16}\cdot\frac{4^3}{3}=48\pi$

The semi-axes of the ellipsoid will be 4,3, and 3, and the volume is $V=4\times3\times3\times\frac{4\pi}{3}=\boxed{48}\pi$ : Consider a unit sphere and stretch it out in the directions of the coordinate axes.