Triple Integral

For a fixed $-1\leq z\leq 1$ , the cross section is an annulus bounded by concentric circles with radii $3\pm \sqrt{1-z^2}$ . The area of this annulus is $12\pi\sqrt{1-z^2}$ . Thus the volume of the solid region is $V=\int_{-1}^{1}12\pi\sqrt{1-z^2}dz=12\pi\times\frac{\pi}{2}=6\pi^2$ . The answer is $\boxed{6}$ .