The centre of mass of a thin rectangular plate (fig - x) with sides of length a and b, whose mass per unit area  $\left(\sigma \right)$ varies as $\sigma =\frac{{\sigma }_0\mathrm{x}}{\mathrm{ab}}$ (where ${\sigma }_0$ is a constant), would be ________

$\sigma$ is constant in y-direction
$\text{So,}{\mathrm{y}}_{\mathrm{cm}}=\mathrm{b}/2$

$=\frac{{\int }_0^{a}x{\sigma }_{x}dA}{{\int }_0^a{\sigma }_{x}dA}$

$=\frac{{\int }_0^{a}x\frac{{\sigma }_{\circ }x}{ab}bdx}{{\int }_0^a\frac{{\sigma }_{0}x}{ab}bdx}$

$\begin{array}{rl} & x_{\mathrm{cm}}=\frac{{\int }_0^{\mathrm{a}}{\mathrm{x}}^{2}dx}{{\int }_0^{\mathrm{a}}\mathrm{xdx}}\\ & =\frac{{\left(\frac{{\mathrm{x}}^3}{3}\right)}_0^{\mathrm{a}}}{{\left(\frac{{\mathrm{x}}^2}{2}\right)}_0^a}=\frac{{\mathrm{a}}^3/3}{{\mathrm{a}}^2/2}\\ & =\frac{2\mathrm{a}}{3}\end{array}$
Since $\sigma(x,y)$ depends only on $x$ and not on $y$, the mass is symmetrically distributed along the $y$-axis.

Thus, the center of mass along $y$ remains at the midpoint:

$$Y_{\text{CM}} = \frac{b}{2}$$

$$(X_{\text{CM}}, Y_{\text{CM}}) = \left(\frac{2a}{3}, \frac{b}{2} \right)$$