(1 point)
Evaluate the triple integral of $f(x, y, z)=z\left(x^{2}+y^{2}+z^{2}\right)^{-3 / 2}$ over the part of the ball $x^{2}+y^{2}+z^{2} \leq 64$ defined by $z \geq 4$.
\[
\iiint_{\mathcal{W}} f(x, y, z) d V=
\]

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

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

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

Step 4 ：Substituting these into the triple integral, we get \(\iiint_{\mathcal{W}} f(r, \theta, \phi) r^2 \sin(\theta) dr d\theta d\phi = \int_{0}^{2\pi} d\phi \int_{\pi/3}^{\pi} \sin(\theta) d\theta \int_{0}^{8} \cos(\theta) r^{-2} r^2 dr\).

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 \(\int_{\pi/3}^{\pi} \sin(\theta) \cos(\theta) d\theta = \frac{1}{2} \left[ \sin^2(\theta) \right]_{\pi/3}^{\pi} = \frac{1}{2} (0 - \sin^2(\pi/3)) = -\frac{1}{4}\).

Step 7 ：Finally, we solve the integral with respect to \(\phi\), which gives \(\int_{0}^{2\pi} d\phi = 2\pi\).

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 \(\boxed{-4\pi}\).