Answer
The mass of the wire is: $m=2k \pi$ and $(\overline {x}, \overline {y})=(\dfrac{4}{\pi},0)$
Work Step by Step
Here, we have $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}=\sqrt{(-2\sin t)^2+(2\cos t)^2}dt=2 dt$
The mass of the wire is: $m=\int_C k ds=2k\int_{-\pi/2}^{\pi/2} dt=2k \pi$
Now, $\overline {x}=(\dfrac{1}{2k \pi}) \int_{C} xk ds=(1/2 \pi) \int_{C} (2 \cos t) 2dt=\dfrac{4}{\pi}$
and $\overline {y}=(\dfrac{1}{2k \pi}) \int_{C} yk ds=(1/2 \pi) \int_{C} (2 \sin t) 2dt=0$
Thus, $(\overline {x}, \overline {y})=(\dfrac{4}{\pi},0)$
