Physics - Center of Mass Question with Solution | TestHub

The total mass of the plate is $M={\int }_0^a{\int }_0^a\sigma (x,y)dxdy={\int }_0^a{\int }_0^{a}kxydxdy=k{\left[\frac{x^2}{2}\right]}_0^a{\left[\frac{y^2}{2}\right]}_0^a=k\frac{a^2}{2}\frac{a^2}{2}=\frac{k{a}^4}{4}$. The x-coordinate of the center of mass is $x_{CM}=\frac{1}{M}{\int }_0^a{\int }_0^{a}x\sigma (x,y)dxdy=\frac{1}{M}{\int }_0^a{\int }_0^{a}k{x}^{2}ydxdy=\frac{k}{M}{\left[\frac{x^3}{3}\right]}_0^a{\left[\frac{y^2}{2}\right]}_0^a=\frac{k}{M}\frac{a^3}{3}\frac{a^2}{2}=\frac{k{a}^5}{6}$. Substituting $M=\frac{k{a}^4}{4}$, we get $x_{CM}=\frac{k{a}^5/6}{k{a}^4/4}=\frac{4a}{6}=\frac{2a}{3}$.