The problem asks to find the volume of a solid that is bounded by the cylinder $x^2 + y^2 = 4$, below the paraboloid $z = \frac{x^2 + y^2}{2}$, and above the plane $z = 0$. We will calculate this volume using triple integrals.

Step 1: Switching to cylindrical coordinates

Since the solid is defined by the cylinder $x^2 + y^2 = 4$, it's convenient to switch to cylindrical coordinates, where:

• $x = r \cos(\theta)$
• $y = r \sin(\theta)$
• $z = z$
• $x^2 + y^2 = r^2$

The bounds of the region are:

• The cylinder is $r = 2$ (from $x^2 + y^2 = 4$).
• The paraboloid equation becomes $z = \frac{r^2}{2}$.
• The plane is $z = 0$.

Step 2: Setting up the limits of integration

In cylindrical coordinates, the volume element $dV$ is $r \, dz \, dr \, d\theta$. The limits are:

• $r$ goes from 0 to 2 (from the cylinder's boundary).
• $z$ goes from 0 to $\frac{r^2}{2}$ (between the plane and the paraboloid).
• $\theta$ goes from 0 to $2\pi$ (for a full rotation around the z-axis).

Thus, the volume integral is: $V = \int_0^{2\pi} \int_0^2 \int_0^{\frac{r^2}{2}} r \, dz \, dr \, d\theta$

Step 3: Solving the triple integral

1. Integrating with respect to $z$: $\int_0^{\frac{r^2}{2}} dz = \frac{r^2}{2}$ This leaves: $V = \int_0^{2\pi} \int_0^2 \frac{r^3}{2} \, dr \, d\theta$

2. Integrating with respect to $r$: $\int_0^2 \frac{r^3}{2} \, dr = \frac{1}{2} \int_0^2 r^3 \, dr = \frac{1}{2} \left[\frac{r^4}{4}\right]_0^2 = \frac{1}{2} \cdot \frac{16}{4} = 2$

3. Integrating with respect to $\theta$: $\int_0^{2\pi} d\theta = 2\pi$

Step 4: Final result

Multiplying all the results together: $V = 2\pi \times 2 = 4\pi$

Thus, the volume of the solid is $4\pi$.

Would you like more details on any part of the process?

Here are some related questions to consider:

1. How do you apply triple integrals to find volumes in Cartesian coordinates?
2. Why are cylindrical coordinates preferred for this problem?
3. Can you solve the same problem using spherical coordinates?
4. What is the difference between cylindrical and spherical coordinates?
5. How would you solve this integral if the bounds were different?

Tip: Using symmetry in the problem can often simplify the integration process, especially when dealing with circular or rotationally symmetric shapes like cylinders or spheres.