Answer
$(\dfrac{4a}{3 \pi}, \dfrac{4a}{3 \pi})$
Work Step by Step
The area of a quarter circle can be given by: $ \dfrac{1}{4} \times \ Area \ of \ circle =\dfrac{1}{4} \pi a^2$
The $x$ -coordinate of the center of mass can be given by: $\overline{x}=\dfrac{1}{2A} \oint_{C} x^2 dy \\= \dfrac{1}{\dfrac{2\pi a^2}{4}} (\dfrac{2 a^3}{3}) \\ =\dfrac{4a}{3 \pi}$
The $y$ -coordinate of the center of mass can be given by: $\overline{y}=\dfrac{1}{2A} \oint_{C} y^2 dx \\= -\dfrac{1}{\dfrac{2\pi a^2}{4}} (-\dfrac{2 a^3}{3}) \\ =\dfrac{4a}{3 \pi}$
So, the center of mass is: $(\dfrac{4a}{3 \pi}, \dfrac{4a}{3 \pi})$
