Chapter 16: Integrals and Vector Fields - Section 16.2 - Vector Fields and Line Integrals: Work, Circulation, and Flux - Exercises 16.2 - Page 955: 22

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Work done is given as: $W=\int_a^b F(r(t)) \dfrac{dr}{dt}(dt)$ Here, $ \dfrac{dr}{dt}=(\cos t) i-(\sin t) j +(\dfrac{1}{6}) k$ Now, $W=\int_0^{2 \pi} \cos t -\cos^2 t\sin t+2 \sin t dt$ Thus, $[\sin t+(1/3) cos^3 t-2 \cos t ]_0^{2 \pi}=0$