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