Answer
$I_x=2 \pi -2$
Work Step by Step
Since, we have $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
or, $ds=\sqrt{(0)^2+( -\sin t )^2+(\cos t)^2} dt \implies ds= dt$
We are given that $\delta =2-z$
and $I_x=\int_C (y^2+z^2) \delta ds$
or, $I_x=\int_C (y^2+z^2) (2-z) ds$
or, $I_x=\int_0^{\pi} (cos^2 t+\sin^2 t)( 2-\sin t) dt= \int_0^{\pi} 2-\sin t dt$
Then $I_x=2 \pi -2$
