Thomas' Calculus 13th Edition

$2 \pi -2$
Given: $\delta =2-z$ As we know that $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$ or, $ds=\sqrt{(0)^2+( -\sin t )^2+(\cos t)^2} dt$ and $ds=(1)dt= dt$ Now, $I_x=\int_C (y^2+z^2) \delta ds=\int_C (y^2+z^2) (2-z) ds$ or, $I_x=\int_0^{\pi} (cos^2 t+\sin^2 t)( 2-\sin t) dt$ So, $I_x= \int_0^{\pi} (2-\sin t dt)=2 \pi -2$