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, we have the mass of the wire is given as:
$m=\int_C \rho(x,y) ds=(k a^3) \int_{0}^{\pi/2} \cos (t) \sin (t) dt=\dfrac{ka^3}{2}$
Thus, the mass of the wire is: $m=\dfrac{ka^3}{2}$
Now, $\overline {x}=\dfrac{1}{m}\int_{C} y \times \rho(x,y) ds=\dfrac{2}{ka^3}(k a^3)\int_{0}^{\pi/2} (a \sin t) \times (\cos t) (\sin t ) dt$
or, $=2a \int_{0}^{\pi/2} (\sin^2 t) (\cos t) dt$
or, $=\dfrac{2a}{3}$
Since, the quarter-circle is symmenrtic about the line $y=x$, we have $\overline {x}=\overline {y}=\dfrac{2a}{3}$
Thus, the center of mass is: $(\overline {x}, \overline {y})=(\dfrac{2a}{3},\dfrac{2a}{3})$
