Answer
$4\pi$
Work Step by Step
$F(r(t)) \cdot \dfrac{dr}{dt}=\lt \cos (t), 2 \sin (t) \gt \cdot \lt -\sin (t),2 \cos (t) ,0 \gt \\ \implies F(r(t)) \cdot \dfrac{dr}{dt}=(4) \cos ^2 t-\sin (t) \cos^2 (t)$
Then, we have
$\implies \int _C F(r(t)) \cdot dr =\int _0^{2\pi} 4 \cos ^2 t-\sin t \cos^2 t dt$
This implies that $(\sin 4\pi-\sin 0) +2 (2 \pi-0)+\dfrac{1}{3} [\cos ^3 (2 \pi)-\cos^3 (0)=(\sin 4\pi-0 +4 \pi+\dfrac{1}{3} [\cos ^3 (2 \pi)-\cos^3 (0)=4\pi$
