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 $\sin(2\theta) = 2 \sin \theta \cos \theta$: