A lamina occupies the part of the disk 
x2 + y2 ≤ 64
 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis.
(x, y) =

To find the center of mass \((x, y)\) of a lamina occupying the region \( x^2 + y^2 \leq 64 \) in the first quadrant, with density proportional to the distance from the \( x \)-axis, we can follow these steps:

1. **Define the Region and Density Function**:
   - The lamina occupies the quarter disk in the first quadrant, where \( x \geq 0 \), \( y \geq 0 \), and \( x^2 + y^2 \leq 64 \).
   - The density function \( \rho(x, y) \) is proportional to the distance from the \( x \)-axis. Let \( \rho(x, y) = k \cdot y \), where \( k \) is a constant of proportionality.

2. **Set Up the Center of Mass Formulas**:
   - The coordinates of the center of mass \((x, y)\) are given by:
     \[
     x = \frac{\iint_R x \rho(x, y) \, dA}{\iint_R \rho(x, y) \, dA}, \quad y = \frac{\iint_R y \rho(x, y) \, dA}{\iint_R \rho(x, y) \, dA}
     \]
   - Since \( \rho(x, y) = k \cdot y \), the expressions become:
     \[
     x = \frac{\iint_R x \cdot (k \cdot y) \, dA}{\iint_R k \cdot y \, dA}, \quad y = \frac{\iint_R y \cdot (k \cdot y) \, dA}{\iint_R k \cdot y \, dA}
     \]
   - The factor \( k \) will cancel out in both expressions, simplifying the calculations.

3. **Convert to Polar Coordinates**:
   - Using polar coordinates \( x = r \cos \theta \) and \( y = r \sin \theta \), we have \( dA = r \, dr \, d\theta \).
   - The region \( R \) corresponds to \( 0 \leq r \leq 8 \) and \( 0 \leq \theta \leq \frac{\pi}{2} \) (first quadrant).
   - Substitute \( \rho(x, y) = k \cdot y = k \cdot r \sin \theta \).

4. **Calculate \( \iint_R k \cdot y \, dA \) (Total Mass)**:
   \[
   \iint_R k \cdot y \, dA = \iint_R k \cdot r \sin \theta \cdot r \, dr \, d\theta = k \int_0^{\frac{\pi}{2}} \int_0^8 r^2 \sin \theta \, dr \, d\theta
   \]
   - Integrating with respect to \( r \):
     \[
     = k \int_0^{\frac{\pi}{2}} \sin \theta \int_0^8 r^2 \, dr \, d\theta = k \int_0^{\frac{\pi}{2}} \sin \theta \cdot \frac{8^3}{3} \, d\theta = \frac{512k}{3} \int_0^{\frac{\pi}{2}} \sin \theta \, d\theta
     \]
   - Integrating with respect to \( \theta \):
     \[
     = \frac{512k}{3} \cdot \left[ -\cos \theta \right]_0^{\frac{\pi}{2}} = \frac{512k}{3} \cdot (1) = \frac{512k}{3}
     \]

5. **Calculate \( x \) Coordinate of the Center of Mass**:
   \[
   x = \frac{\iint_R x \rho(x, y) \, dA}{\iint_R \rho(x, y) \, dA} = \frac{\iint_R x \cdot k \cdot y \, dA}{\frac{512k}{3}} = \frac{\iint_R k \cdot r \cos \theta \cdot r \sin \theta \cdot r \, dr \, d\theta}{\frac{512k}{3}}
   \]
   - Simplifying:
     \[
     = \frac{k \int_0^{\frac{\pi}{2}} \int_0^8 r^3 \cos \theta \sin \theta \, dr \, d\theta}{\frac{512k}{3}}
     \]
   - Integrating with respect to \( r \):
     \[
     = \frac{k \int_0^{\frac{\pi}{2}} \cos \theta \sin \theta \cdot \frac{8^4}{4} \, d\theta}{\frac{512k}{3}} = \frac{4096k}{4} \int_0^{\frac{\pi}{2}} \cos \theta \sin \theta \, d\theta
     \]
     - Use the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \):


A lamina occupies the part of the disk x^2 + y^2 ≤ 64 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis.

Learn to calculate the center of mass for a lamina occupying a quarter disk (x^2 + y^2 ≤ 64) in the first quadrant, with density proportional to the distance from the x-axis. This solution covers key multivariable calculus concepts including polar coordinates, density functions, and center of mass formulas.

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