The given problem is a triple integral over the region $\Omega$, where $\Omega$ is defined by $4 \leq x^2 + y^2 + z^2 \leq 16$, and the integral to evaluate is:

$\iiint_\Omega \frac{1}{\sqrt{x^2 + y^2 + z^2}} \, dx \, dy \, dz.$

Approach to Solve

The region $\Omega$ is a spherical shell bounded by two spheres with radii $2$ and $4$ (since $4 = 2^2$ and $16 = 4^2$). We switch to spherical coordinates, where:

• $x = \rho \sin\phi \cos\theta$,
• $y = \rho \sin\phi \sin\theta$,
• $z = \rho \cos\phi$,
• The volume element $dx \, dy \, dz = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$.

The given function $\frac{1}{\sqrt{x^2 + y^2 + z^2}}$ becomes $\frac{1}{\rho}$ in spherical coordinates. The integral now becomes:

$\iiint_\Omega \frac{1}{\sqrt{x^2 + y^2 + z^2}} \, dx \, dy \, dz = \int_0^{2\pi} \int_0^\pi \int_2^4 \frac{1}{\rho} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.$

Simplify the integrand: $\frac{1}{\rho} \rho^2 = \rho.$

Thus, the integral becomes: $\int_0^{2\pi} \int_0^\pi \int_2^4 \rho \sin\phi \, d\rho \, d\phi \, d\theta.$

Step 1: Evaluate the $\rho$-Integral

$\int_2^4 \rho \, d\rho = \left[ \frac{\rho^2}{2} \right]_2^4 = \frac{4^2}{2} - \frac{2^2}{2} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6.$

Step 2: Evaluate the $\phi$-Integral

$\int_0^\pi \sin\phi \, d\phi = \left[ -\cos\phi \right]_0^\pi = -\cos\pi + \cos0 = -(-1) + 1 = 2.$

Step 3: Evaluate the $\theta$-Integral

$\int_0^{2\pi} 1 \, d\theta = \left[ \theta \right]_0^{2\pi} = 2\pi.$

Combine the Results

Multiply the results from each step: $6 \cdot 2 \cdot 2\pi = 24\pi.$

Final Answer:

$\boxed{24\pi}$

Let me know if you need further clarifications or have questions!