(1 point)
Evaluate the triple integral of $f(x,y,z)=z{(x^2+y^2+z^2)}^{-3/2}$ over the part of the ball $x^2+y^2+z^2\le 64$ defined by $z\ge 4$.
$${\iiint }_{\mathcal{W}}f(x,y,z)dV=$$

Step 1 ：First, we convert the given integral into spherical coordinates. The conversion from Cartesian coordinates to spherical coordinates is given by $x=r\sin (\theta )\cos (\varphi )$, $y=r\sin (\theta )\sin (\varphi )$, and $z=r\cos (\theta )$. The volume element $dV$ in spherical coordinates is $r^2\sin (\theta )drd\theta d\varphi$.

Step 2 ：The function $f(x,y,z)$ in spherical coordinates becomes $f(r,\theta ,\varphi )=r\cos (\theta )(r^2{)}^{-3/2}=\cos (\theta )r^{-2}$.

Step 3 ：The region of integration is defined by $0\le r\le 8$, $0\le \varphi \le 2\pi$, and ${\cos }^{-1}(4/8)=\pi /3\le \theta \le \pi$.

Step 4 ：Substituting these into the triple integral, we get ${\iiint }_{\mathcal{W}}f(r,\theta ,\varphi )r^2\sin (\theta )drd\theta d\varphi ={\int }_0^{2\pi }d\varphi {\int }_{\pi /3}^{\pi }\sin (\theta )d\theta {\int }_0^8\cos (\theta )r^{-2}{r}^{2}dr$.

Step 5 ：Solving the integral with respect to $r$ first, we get ${\int }_0^{8}dr=8$.

Step 6 ：Next, we solve the integral with respect to $\theta$, which gives ${\int }_{\pi /3}^{\pi }\sin (\theta )\cos (\theta )d\theta =\frac{1}{2}{[{\sin }^2(\theta )]}_{\pi /3}^{\pi }=\frac{1}{2}(0-{\sin }^2(\pi /3))=-\frac{1}{4}$.

Step 7 ：Finally, we solve the integral with respect to $\varphi$, which gives ${\int }_0^{2\pi }d\varphi =2\pi$.

Step 8 ：Multiplying these results together, we get $8\times -\frac{1}{4}\times 2\pi =-4\pi$.

Step 9 ：Therefore, the value of the triple integral is $\boxed{{\displaystyle -4\pi }}$.