Physics - Center of Mass Question with Solution | TestHub

$\begin{array}{rl}& dm=\sigma dA\\ & \begin{array}{rl}d& =\sigma (dx)(dy)=\frac{{\sigma }_{0}x}{ab}(dx)(dy)\\ x_{com}& =\frac{\int xdm}{\int dm}=\frac{\int x\frac{\left({\sigma }_{0}x\right)}{ab}(dx)(dy)}{{\int }_0\frac{{\sigma }_0}{ab}(dx)(dy)}\\ & =\frac{{\int }_0^a{x}^{2}dx{\int }_0^{b}dy}{{\int }_0^{b}xdx{\int }_0^{b}dy}=\frac{2a}{3}\\ y_{com}& =\frac{\int ydm}{\int dm}=\frac{\int y\left(\frac{{\sigma }_{0}x}{ab}\right)(dx)(dy)}{\int \frac{{\sigma }_{0}x}{ab}(dx)(dy)}\end{array}\end{array}$ $=\frac{{\int }_0^{a}xdx{\int }_0^{b}ydy}{{\int }_0^{a}xdx{\int }_0^{b}dy}=\frac{b}{2}$ i.e., ${\vec{r}}_{\mathrm{com}}\equiv \left(\frac{2a}{3},\frac{b}{2}\right)$