Volume of region bounded by points satisfying the given inequality !

It's a torus with major radius of $3$ and minor radius of $1$ This can be seen by letting $y=0$ and solving for $z$ to get

$z=\sqrt { 1-{ (x\pm 3) }^{ 2 } }$

Symmetry takes care of the rest.

The volume of a torus is the area of the circular cross section multiplied by the length of the circumference of the circular centerline of the center of the circular cross section, or

$\left( 2\pi R \right) \left( \pi { r }^{ 2 } \right) =2{ \pi }^{ 2 }(3)({ 1 }^{ 2 })=6{ \pi }^{ 2 }=59.2176...$

Let's take a cylindrical coordinates approach here. Let $x^2+y^2=r^2$ which gives us:

$(r^2+z^2+8)^2 \le 36r^2 \Rightarrow r^2+z^2+8 \le 6r \Rightarrow (r^2-6r+9)+z^2 -1 \le 0 \Rightarrow z^2 \le 1 - (r-3)^2;$

or $-\sqrt{1-(r-3)^2} \le z \le \sqrt{1-(r-3)^2}$ . At $z=0$ we obtain the following $xy-plane$ curves: $r^2 + 8 \le 6r \Rightarrow r^2-6r+8 \le 0 \Rightarrow (r-2)(r-4) \le 0 \Rightarrow 2 \le r \le 4$ which sweep an entire circle ( $0 \le \theta \le 2\pi$ ). These parts finally converge into the following triple integral:

$X = \int_{0}^{2\pi} \int_{2}^{4} \int_{-\sqrt{1-(r-3)^2}}^{\sqrt{1-(r-3)^2}} r dz dr d\theta = 6\pi^{2} \approx 59.2176$

or $\lfloor 59.2176 \rfloor = \boxed{59}.$