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The problem asks to find the volume of the region bounded below by the plane \( z = 1 \) and above by the sphere \( x^2 + y^2 + z^2 = 4 \), using cylindrical coordinates.

### Step-by-Step Solution:

1. **Convert the equation of the sphere to cylindrical coordinates:**
   - In cylindrical coordinates, the transformations are:
     \[
     x = r \cos \theta, \quad y = r \sin \theta, \quad z = z, \quad r^2 = x^2 + y^2
     \]
   - The equation of the sphere \( x^2 + y^2 + z^2 = 4 \) becomes:
     \[
     r^2 + z^2 = 4
     \]
     which simplifies to \( z = \sqrt{4 - r^2} \).

2. **Set up the limits for \( r \), \( \theta \), and \( z \):**
   - The sphere intersects the plane \( z = 1 \), so at \( z = 1 \), the radius \( r \) must satisfy:
     \[
     1^2 + r^2 = 4 \quad \Rightarrow \quad r^2 = 3 \quad \Rightarrow \quad r = \sqrt{3}
     \]
   - The limits for \( r \) are from 0 to \( \sqrt{3} \).
   - The limits for \( \theta \) (since the region is symmetric) are from 0 to \( 2\pi \).
   - The limits for \( z \) are from the plane \( z = 1 \) to the surface of the sphere, given by \( z = \sqrt{4 - r^2} \).

3. **Volume element in cylindrical coordinates:**
   The volume element in cylindrical coordinates is:
   \[
   dV = r \, dz \, dr \, d\theta
   \]

4. **Set up the integral:**
   The volume \( V \) is given by:
   \[
   V = \int_0^{2\pi} \int_0^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}} r \, dz \, dr \, d\theta
   \]

5. **Evaluate the integral:**

   First, integrate with respect to \( z \):
   \[
   \int_1^{\sqrt{4 - r^2}} dz = \sqrt{4 - r^2} - 1
   \]

   Now, the volume integral becomes:
   \[
   V = \int_0^{2\pi} \int_0^{\sqrt{3}} r \left( \sqrt{4 - r^2} - 1 \right) dr \, d\theta
   \]

6. **Simplify the radial integral:**

   Distribute the \( r \):
   \[
   V = \int_0^{2\pi} \int_0^{\sqrt{3}} \left( r \sqrt{4 - r^2} - r \right) dr \, d\theta
   \]

   Separate the integrals:
   \[
   V = \int_0^{2\pi} \left[ \int_0^{\sqrt{3}} r \sqrt{4 - r^2} \, dr - \int_0^{\sqrt{3}} r \, dr \right] d\theta
   \]

   **First Integral:**
   Use the substitution \( u = 4 - r^2 \), \( du = -2r \, dr \):
   \[
   \int_0^{\sqrt{3}} r \sqrt{4 - r^2} \, dr = -\frac{1}{3} \left(4 - r^2\right)^{3/2} \Bigg|_0^{\sqrt{3}} = \frac{16}{3} - 4
   \]

   **Second Integral:**
   \[
   \int_0^{\sqrt{3}} r \, dr = \frac{1}{2} r^2 \Bigg|_0^{\sqrt{3}} = \frac{3}{2}
   \]

   Combine the two:
   \[
   V = \int_0^{2\pi} \left( \frac{16}{3} - 4 - \frac{3}{2} \right) d\theta
   \]

7. **Final result:**
   Simplify the constants:
   \[
   V = \int_0^{2\pi} \frac{5}{6} \, d\theta = \frac{5}{6} \times 2\pi = \frac{5\pi}{3}
   \]

Thus, the volume of the region is \( \boxed{\frac{5\pi}{3}} \).

Would you like more details on any step or have any questions?

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Here are 5 related questions:

1. How do you convert between Cartesian and cylindrical coordinates?
2. What is the volume of a sphere using spherical coordinates?
3. How do you compute triple integrals over cylindrical regions?
4. What is the Jacobian determinant in cylindrical coordinates?
5. How do you find the bounds of integration for complex 3D regions?

**Tip:** Always check the limits of integration carefully when setting up problems involving volumes in multiple coordinate systems.

Use cylindrical coordinates to find the volume of the region bounded below by the plane z = 1 and above by the sphere x^2 + y^2 + z^2 = 4.

Find the volume of the region bounded below by the plane z = 1 and above by the sphere x^2 + y^2 + z^2 = 4 using cylindrical coordinates. This problem involves key concepts like triple integrals, cylindrical coordinates, and volume calculation. Suitable for college-level students.

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