Answer
$9\pi$
Work Step by Step
Here, $F(r(t)) \cdot \dfrac{dr}{dt}=\lt 3\cos t, 3 \sin t \gt \cdot \lt 6\sin t,9 \cos t \gt =27 \cos ^2 t-18\sin t^2 t$
Now, $\int _C F(r(t)) \cdot dr =\int _0^{2\pi} 27 \cos ^2 t-18\sin t^2 tdt$
or, $\int _0^{2\pi} \dfrac{45}{2} \cos 2t +\dfrac{9}{2} =[\dfrac{45}{2} \sin 2t +\dfrac{9}{2}t] _0^{2\pi}$
or, $[\dfrac{45}{2} ( \sin 4\pi-\sin 0) +\dfrac{9}{2}(2\pi-0)]=9\pi$
