Answer
The mass of the wire is: $m=\dfrac{ka^3}{2}$ and the center of mass is: $(\overline {x}, \overline {y})=(\dfrac{2a}{3},\dfrac{2a}{3})$
Work Step by Step
Here, $m=\int_C \rho(x,y) ds=k a^3 \int_{0}^{\pi/2} \cos t \sin t dt=\dfrac{ka^3}{2}$
Now, $\overline {x}=\dfrac{1}{m}\int_{C} y \rho(x,y) ds=\dfrac{2}{ka^3}(k a^3)\int_{0}^{\pi/2} (a \sin t)(\cos t \sin t ) dt=2a \int_{0}^{\pi/2} \sin^2 t \cos t dt=\dfrac{2a}{3}$
Since, the quarter-circle is symmenrtic about the line $y=x$, $\overline {x}=\overline {y}=\dfrac{2a}{3}$
Hence, the mass of the wire is: $m=\dfrac{ka^3}{2}$ and the center of mass is: $(\overline {x}, \overline {y})=(\dfrac{2a}{3},\dfrac{2a}{3})$
