To evaluate the triple integral

$\iiint_{D} \frac{z}{(x^2 + y^2 + z^2)^{3/2}} \, dV,$

where $D$ is the region defined by $x^2 + y^2 + z^2 \leq 25$ and $z \geq 2.5$, it's advantageous to switch to spherical coordinates. In spherical coordinates, we have:

• $x = \rho \sin\theta \cos\phi$
• $y = \rho \sin\theta \sin\phi$
• $z = \rho \cos\theta$
• The Jacobian (volume element) $dV = \rho^2 \sin\theta \, d\rho \, d\theta \, d\phi$

With these substitutions, the function $f(x, y, z) = \frac{z}{(x^2 + y^2 + z^2)^{3/2}}$ becomes:

$f(\rho, \theta, \phi) = \frac{\rho \cos\theta}{\rho^3} = \frac{\cos\theta}{\rho^2}.$

Thus, the integral becomes:

$\iiint_{D} \frac{\cos\theta}{\rho^2} \cdot \rho^2 \sin\theta \, d\rho \, d\theta \, d\phi = \iiint_{D} \cos\theta \sin\theta \, d\rho \, d\theta \, d\phi.$

The limits for $\rho$, $\theta$, and $\phi$ are as follows:

1. $\rho$ ranges from $\frac{5}{2}$ (corresponding to $z = 2.5$) to 5 (the boundary of the sphere).
2. $\theta$ ranges from 0 to $\arccos\left(\frac{1}{2}\right)$, so that $z = \rho \cos \theta \geq 2.5$.
3. $\phi$ ranges from 0 to $2\pi$.

So the integral becomes:

$\int_0^{2\pi} \int_0^{\arccos\left(\frac{1}{2}\right)} \int_{\frac{5}{2}}^5 \cos\theta \sin\theta \, d\rho \, d\theta \, d\phi.$

Evaluating the Integral

1. Integrate with respect to $\rho$: $\int_{\frac{5}{2}}^5 \, d\rho = 5 - \frac{5}{2} = \frac{5}{2}.$

2. Integrate with respect to $\theta$ and $\phi$: