Calculate Iz of the homogenous 3D object occupying the top half of the unit ball.

To calculate the moment of inertia \( I_z \) of the homogeneous 3D object occupying the top half of the unit ball about the z-axis, we follow these steps:

1. **Define the region**: The top half of the unit ball is defined by:
   \[
   x^2 + y^2 + z^2 \leq 1 \quad \text{and} \quad z \geq 0
   \]

2. **Moment of inertia definition**: The moment of inertia \( I_z \) about the z-axis for a density \(\rho\) is given by:
   \[
   I_z = \iiint_V \rho (x^2 + y^2) \, dV
   \]
   For a homogeneous object with constant density, we can set \(\rho = 1\).

3. **Convert to spherical coordinates**: To simplify the integral, we convert to spherical coordinates where:
   \[
   x = r \sin \theta \cos \phi, \quad y = r \sin \theta \sin \phi, \quad z = r \cos \theta
   \]
   The volume element in spherical coordinates is:
   \[
   dV = r^2 \sin \theta \, dr \, d\theta \, d\phi
   \]

4. **Set the limits of integration**: For the top half of the unit ball, the limits are:
   \[
   0 \leq r \leq 1, \quad 0 \leq \theta \leq \frac{\pi}{2}, \quad 0 \leq \phi \leq 2\pi
   \]

5. **Express the integrand in spherical coordinates**: Substituting \(x\) and \(y\) into the integrand \(x^2 + y^2\), we get:
   \[
   x^2 + y^2 = r^2 \sin^2 \theta (\cos^2 \phi + \sin^2 \phi) = r^2 \sin^2 \theta
   \]

6. **Set up the integral**:
   \[
   I_z = \int_0^{2\pi} \int_0^{\pi/2} \int_0^1 r^2 \sin^2 \theta \cdot r^2 \sin \theta \, dr \, d\theta \, d\phi
   \]

7. **Simplify the integrand**:
   \[
   I_z = \int_0^{2\pi} \int_0^{\pi/2} \int_0^1 r^4 \sin^3 \theta \, dr \, d\theta \, d\phi
   \]

8. **Evaluate the integral**:
   \[
   I_z = \int_0^{2\pi} d\phi \int_0^{\pi/2} \sin^3 \theta \, d\theta \int_0^1 r^4 \, dr
   \]
   First, compute the radial integral:
   \[
   \int_0^1 r^4 \, dr = \left[ \frac{r^5}{5} \right]_0^1 = \frac{1}{5}
   \]
   Next, compute the angular integrals:
   \[
   \int_0^{2\pi} d\phi = 2\pi
   \]
   \[
   \int_0^{\pi/2} \sin^3 \theta \, d\theta
   \]
   Use the substitution \(u = \cos \theta\), \(du = -\sin \theta \, d\theta\):
   \[
   \int_0^{\pi/2} \sin^3 \theta \, d\theta = \int_1^0 (1 - u^2) \, (-du) = \int_0^1 (1 - u^2) \, du = \left[ u - \frac{u^3}{3} \right]_0^1 = 1 - \frac{1}{3} = \frac{2}{3}
   \]

9. **Combine all results**:
   \[
   I_z = 2\pi \cdot \frac{2}{3} \cdot \frac{1}{5} = \frac{4\pi}{15}
   \]

Thus, the moment of inertia \( I_z \) of the homogeneous 3D object occupying the top half of the unit ball about the z-axis is:
\[
I_z = \frac{4\pi}{15}
\]

Would you like details or have any questions?

Here are 5 questions you might ask next:
1. How is the moment of inertia calculated for different shapes?
2. Can you explain how to convert Cartesian coordinates to spherical coordinates?
3. What are the applications of moment of inertia in physics?
4. How does the density distribution affect the moment of inertia?
5. Can you show an example of calculating the moment of inertia for another 3D object?

**Tip**: When dealing with integrals in spherical coordinates, always remember to include the Jacobian determinant (\( r^2 \sin \theta \)) when changing variables.

Learn how to calculate the moment of inertia Iz of a homogeneous 3D object occupying the top half of a unit ball using integral calculus and spherical coordinates. This problem involves detailed steps for setting up the integral in spherical coordinates and evaluating it step-by-step. Suitable for advanced college-level students and anyone interested in physics and mathematical applications.

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