Answer
$0$
Work Step by Step
Here, we have $F(r(t)=\cos t i +\sin t j+\cos t \sin t k$ and $dr=(-\sin t i+\cos t j+k) dt$
$\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{\pi} (\cos t i +\sin t j+\cos t \sin t k) \cdot (-\sin t i+\cos t j+k) d t$
$= \int_0^{\pi} (\cos t) (\sin t) dt$
$= \int_0^{\pi} (\dfrac{1}{2}) [2 (\cos t) (\sin t)] dt$
$=(1/2) \int_0^{\pi} \sin (2 t) dt$
$=(1/2)[\dfrac{-\cos (2t) }{2}]_0^{\pi}$
$=0$
