Answer
$I_x = 0, I_y = 0$
Work Step by Step
The wire lies along $\textbf r (t) = a\cos t \textbf i + a\sin t \textbf j = x(t) \textbf i + y(t) \textbf j$
It follows that $x(t) a \cos t , y(t) = a\sin t$
$\textbf r' = -a\sin t \textbf i + a\cos t \textbf j \Rightarrow \vert \textbf r' \vert = a$
$\rho (x,y) = y = a\sin t$
$I_x = \int_C y^2\rho(x,y) ds = \int_{t=0}^{2\pi}a^2\sin ^2 t.a\sin t.\vert \textbf r' \vert dt = a^4[-\cos t + \frac{\cos ^3 t}{3}]_0^{2\pi}=0$
$I_y = \int_C x^2\rho(x,y) ds = \int_{t=0}^{2\pi}a^2\cos ^2 t.a\sin t.\vert \textbf r' \vert dt = a^3[-\frac{\cos^3t}{3}]_0^{2\pi} = 0$
