The disc of mass $$M$$ with uniform surface mass density $$\sigma$$ is shown in the figure. The center of mass of the quarter disc (the shaded area) is at the position $$\left(\frac{xa}{3\pi}, \frac{xa}{3\pi}\right)$$ where $$x$$ is ________. (Round off to the Nearest Integer) [$$a$$ is an area as shown in the figure]

For a uniform quarter disc of radius $$a$$ lying in the first quadrant (from $$\theta = 0$$ to $$\theta = \frac{\pi}{2}$$), we find the center of mass coordinates by integration.

Using polar coordinates, the $$x$$-coordinate of the center of mass is $$\bar{x} = \frac{\int_0^{\pi/2}\int_0^a r\cos\theta \cdot r \, dr \, d\theta}{\int_0^{\pi/2}\int_0^a r \, dr \, d\theta}$$. The denominator (total mass divided by $$\sigma$$) is $$\int_0^{\pi/2}\int_0^a r \, dr \, d\theta = \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}$$.

The numerator is $$\int_0^{\pi/2}\int_0^a r^2 \cos\theta \, dr \, d\theta = \frac{a^3}{3} \cdot [\sin\theta]_0^{\pi/2} = \frac{a^3}{3} \cdot 1 = \frac{a^3}{3}$$.

Therefore $$\bar{x} = \frac{a^3/3}{\pi a^2/4} = \frac{4a}{3\pi}$$. By symmetry, $$\bar{y} = \frac{4a}{3\pi}$$ as well.

The center of mass is at $$\left(\frac{4a}{3\pi}, \frac{4a}{3\pi}\right)$$. Comparing with the given form $$\left(\frac{xa}{3\pi}, \frac{xa}{3\pi}\right)$$, we get $$x = 4$$.