Q: The center of mass of the quarter circle given by $y=\sqrt{r^2−x^2}$, $x\in[0,r]$ is the point $(x, y)$.
My first thought was that the circle was bounded by the points $(0,0)$, $(0,r)$, $(r,0)$ and the graph $\sqrt{r^2−x^2}$ but even with that I'm not sure how to proceed.
HINT
The coordinates of the center of mass are given by $x_{cm}$ and $y_{cm}$, where \begin{align*} x_{cm} = \frac{\displaystyle\int_{D}x\mathrm{d}m}{\displaystyle\int_{D}1\mathrm{d}m} = \frac{\displaystyle\int_{0}^{r}\int_{0}^{\sqrt{r^{2} - x^{2}}}x\mathrm{d}y\mathrm{d}x}{\displaystyle\int_{0}^{r}\int_{0}^{\sqrt{r^{2} - x^{2}}}1\mathrm{d}y\mathrm{d}x} \end{align*} Similarly, one has that
\begin{align*} y_{cm} = \frac{\displaystyle\int_{D}y\mathrm{d}m}{\displaystyle\int_{D}1\mathrm{d}m} = \frac{\displaystyle\int_{0}^{r}\int_{0}^{\sqrt{r^{2} - x^{2}}}y\mathrm{d}y\mathrm{d}x}{\displaystyle\int_{0}^{r}\int_{0}^{\sqrt{r^{2} - x^{2}}}1\mathrm{d}y\mathrm{d}x} \end{align*}
Can you take it from here?