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We are given a lamina that occupies the part of the disk \( x^2 + y^2 \leq 1 \) in the first quadrant. We need to find the moments of inertia \( I_x \), \( I_y \), and \( I_0 \), with a density proportional to the square of its distance from the origin.

**Key Information:**
1. The disk is in the first quadrant, so \( x \geq 0 \) and \( y \geq 0 \).
2. The equation of the disk is \( x^2 + y^2 \leq 1 \).
3. The density \( \rho \) at any point is proportional to the square of its distance from the origin, i.e., \( \rho(x, y) = k(x^2 + y^2) \), where \( k \) is the proportionality constant.

### Step 1: Define the Moments of Inertia
- **Moment of inertia about the \( x \)-axis**:
  \[
  I_x = \iint_{R} y^2 \, \rho(x, y) \, dA
  \]
  where \( R \) is the region in the first quadrant defined by \( x^2 + y^2 \leq 1 \). Substituting \( \rho(x, y) \):
  \[
  I_x = \iint_{R} y^2 \cdot k(x^2 + y^2) \, dA
  \]

- **Moment of inertia about the \( y \)-axis**:
  \[
  I_y = \iint_{R} x^2 \, \rho(x, y) \, dA = \iint_{R} x^2 \cdot k(x^2 + y^2) \, dA
  \]

- **Moment of inertia about the origin**:
  \[
  I_0 = \iint_{R} (x^2 + y^2) \, \rho(x, y) \, dA = \iint_{R} (x^2 + y^2) \cdot k(x^2 + y^2) \, dA
  \]

### Step 2: Convert to Polar Coordinates
Since \( R \) is a circular region, it is convenient to switch to polar coordinates:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
- \( dA = r \, dr \, d\theta \)

The given region corresponds to \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq \frac{\pi}{2} \).

#### Moments in Polar Coordinates:
1. **For \( I_x \)**:
   \[
   I_x = k \int_0^{\frac{\pi}{2}} \int_0^1 (r \sin \theta)^2 \cdot r^2 \, r \, dr \, d\theta = k \int_0^{\frac{\pi}{2}} \sin^2 \theta \, d\theta \int_0^1 r^5 \, dr
   \]

2. **For \( I_y \)**:
   \[
   I_y = k \int_0^{\frac{\pi}{2}} \int_0^1 (r \cos \theta)^2 \cdot r^2 \, r \, dr \, d\theta = k \int_0^{\frac{\pi}{2}} \cos^2 \theta \, d\theta \int_0^1 r^5 \, dr
   \]

3. **For \( I_0 \)**:
   \[
   I_0 = k \int_0^{\frac{\pi}{2}} \int_0^1 r^2 \cdot r^2 \, r \, dr \, d\theta = k \int_0^{\frac{\pi}{2}} d\theta \int_0^1 r^5 \, dr
   \]

### Step 3: Evaluate the Integrals
1. **Integral over \( r \)**:
   \[
   \int_0^1 r^5 \, dr = \frac{r^6}{6} \Bigg|_0^1 = \frac{1}{6}
   \]

2. **Integrals over \( \theta \)**:
   - \( \int_0^{\frac{\pi}{2}} \sin^2 \theta \, d\theta = \frac{\pi}{4} \)
   - \( \int_0^{\frac{\pi}{2}} \cos^2 \theta \, d\theta = \frac{\pi}{4} \)
   - \( \int_0^{\frac{\pi}{2}} d\theta = \frac{\pi}{2} \)

### Step 4: Calculate the Moments
1. \( I_x = k \cdot \frac{\pi}{4} \cdot \frac{1}{6} = \frac{k\pi}{24} \)
2. \( I_y = k \cdot \frac{\pi}{4} \cdot \frac{1}{6} = \frac{k\pi}{24} \)
3. \( I_0 = k \cdot \frac{\pi}{2} \cdot \frac{1}{6} = \frac{k\pi}{12} \)

### Final Answers
- \( I_x = \frac{k\pi}{24} \)
- \( I_y = \frac{k\pi}{24} \)
- \( I_0 = \frac{k\pi}{12} \)

Would you like more details or have any questions?

### Related Questions:
1. How would the moments of inertia change if the lamina occupied the entire disk instead of just the first quadrant?
2. What would be the impact on \( I_x \), \( I_y \), and \( I_0 \) if the density function had a different proportionality?
3. Can the moments of inertia be derived if the region were an ellipse instead of a disk?
4. What physical significance do these moments of inertia hold for the given lamina?
5. How do the values of \( I_x \), \( I_y \), and \( I_0 \) compare in higher dimensions?

**Tip:** When solving problems in circular regions, always consider converting to polar coordinates for simplification.

A lamina occupies the part of the disk x^2 + y^2 ≤ 1 in the first quadrant. Find the moments of inertia I_x, I_y, and I_0 for the lamina if the density at any point is proportional to the square of its distance from the origin. Assume that the coefficient of proportionality is k.

Find the moments of inertia I_x, I_y, and I_0 for a lamina occupying the part of the disk x^2 + y^2 ≤ 1 in the first quadrant, with a density proportional to the square of its distance from the origin. This solution involves converting to polar coordinates, using double integration, and applying the concept of density proportionality. Ideal for undergraduate physics and engineering students.

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