Find the volume of the solid defined by the inequality $ x^2 \plus{} y^2 \plus{} \ln (1 \plus{} z^2)\leq \ln 2$.
Note that you may not directively use double integral here for Japanese high school students who don't study it.
1. **Convert to cylindrical coordinates:**
   The given inequality is \( x^2 + y^2 + \ln(1 + z^2) \leq \ln 2 \). In cylindrical coordinates, \( x^2 + y^2 = r^2 \), where \( r \) is the radial distance from the \( z \)-axis. Thus, the inequality becomes:
   \[
   r^2 + \ln(1 + z^2) \leq \ln 2
   \]

2. **Solve for \( r \):**
   To find the range of \( r \), we isolate \( r \):
   \[
   r^2 \leq \ln 2 - \ln(1 + z^2)
   \]
   Using the properties of logarithms, we can rewrite the right-hand side:
   \[
   r^2 \leq \ln \left( \frac{2}{1 + z^2} \right)
   \]
   Since \( \ln \left( \frac{2}{1 + z^2} \right) \) must be non-negative for \( r^2 \) to be non-negative, we have:
   \[
   \frac{2}{1 + z^2} \geq 1 \implies 2 \geq 1 + z^2 \implies 1 \geq z^2 \implies -1 \leq z \leq 1
   \]

3. **Determine the range of \( r \):**
   For \( -1 \leq z \leq 1 \), the maximum value of \( \ln \left( \frac{2}{1 + z^2} \right) \) occurs when \( z = 0 \):
   \[
   r^2 \leq \ln 2
   \]
   Therefore, the range of \( r \) is:
   \[
   0 \leq r \leq \sqrt{\ln 2}
   \]

4. **Calculate the volume:**
   The volume element in cylindrical coordinates is \( dV = r \, dr \, d\theta \, dz \). The limits for \( r \), \( \theta \), and \( z \) are:
   \[
   0 \leq r \leq \sqrt{\ln 2}, \quad 0 \leq \theta \leq 2\pi, \quad -1 \leq z \leq 1
   \]
   The volume \( V \) is given by the triple integral:
   \[
   V = \int_{-1}^{1} \int_{0}^{2\pi} \int_{0}^{\sqrt{\ln 2}} r \, dr \, d\theta \, dz
   \]

5. **Evaluate the integral:**
   First, integrate with respect to \( r \):
   \[
   \int_{0}^{\sqrt{\ln 2}} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{\sqrt{\ln 2}} = \frac{(\sqrt{\ln 2})^2}{2} = \frac{\ln 2}{2}
   \]
   Next, integrate with respect to \( \theta \):
   \[
   \int_{0}^{2\pi} d\theta = 2\pi
   \]
   Finally, integrate with respect to \( z \):
   \[
   \int_{-1}^{1} dz = 2
   \]
   Combining these results, the volume is:
   \[
   V = \left( \frac{\ln 2}{2} \right) \cdot 2\pi \cdot 2 = 2\pi \ln 2
   \]

The final answer is \( \boxed{2\pi \ln 2} \).