Answer
$$- \pi $$
Work Step by Step
Here, we have: $ \dfrac{dr}{dt}=\cos t i -\sin t j+k$
The work done can be computed as follows:
$W=\int_a^b F[r(t)] \dfrac{dr}{dt}(dt) $
or, $=\int_0^{2 \pi } ( t i +\sin t j+\cos t k) \cdot (\cos t -\sin t +k) dt $
or, $=\int_0^{2 \pi} (t+1) \cos t \ dt -\int_0^{2 \pi} \sin^2 t dt $
or, $= [\int (t+1) \cos t \ dt ]_0^{2 \pi} -\int_0^{2 \pi} \sin^2 t dt $
or, $=[(t+1) \sin t -\int \sin t dt ]_0^{2 \pi} - \int_0^{2 \pi} \dfrac{1}{2} - \dfrac{\cos 2t}{2} dt $
or, $=[ (t+1) \sin t +\cos t ]_0^{2 \pi} - \pi $
or, $= 0-\pi $
or, $=- \pi $
