1. Interpret the figure and given expression

   The shaded region is a quarter disc of radius such that its area is given as $a$.

   For a quarter disc, the centroid coordinates measured from the two straight edges are known to be $\bar x = \bar y = \frac{4R}{3\pi}.$

   The question writes the centre of mass as $\left(\frac{x}{3}\frac{a}{\pi},\; \frac{x}{3}\frac{a}{\pi}\right).$

   This is almost certainly a notation issue from the scanned problem: the standard result is of the form $\left(\frac{x}{3}\frac{r}{\pi},\; \frac{x}{3}\frac{r}{\pi}\right),$ where $r$ is the radius, and then $x=4$.

2. Derive the centroid of a quarter disc

   Let the quarter disc lie in the first quadrant with centre at the origin, radius $R$.

   Using polar coordinates:

   • $0 \le r \le R$
   • $0 \le \theta \le \frac{\pi}{2}$
   • $dA = r\,dr\,d\theta$

   Total area of the quarter disc: $A = \frac{\pi R^2}{4}.$

3. Find $\bar x$

   $\bar x = \frac{1}{A}\iint x\,dA$ with $x = r\cos\theta.$

   So, $\bar x = \frac{1}{A}\int_0^{\pi/2}\int_0^R (r\cos\theta)(r\,dr\,d\theta).$

   $\bar x = \frac{1}{A}\left(\int_0^{\pi/2}\cos\theta\,d\theta\right)\left(\int_0^R r^2\,dr\right).$

   Evaluate: $\int_0^{\pi/2}\cos\theta\,d\theta = 1,$ $\int_0^R r^2\,dr = \frac{R^3}{3}.$

   Therefore,

   = \frac{4R}{3\pi}.$$

4. By symmetry

   $\bar y = \frac{4R}{3\pi}.$

   Hence the centroid is $\left(\frac{4R}{3\pi},\frac{4R}{3\pi}\right).$

5. Compare with the given form

   Given form: $\left(\frac{x}{3}\frac{R}{\pi},\frac{x}{3}\frac{R}{\pi}\right).$

   Matching with $\left(\frac{4R}{3\pi},\frac{4R}{3\pi}\right),$ we get $x=4.$

6. Final answer

   $\boxed{4}$

The stored answer is also $4$, so it agrees.