## Thomas' Calculus 13th Edition

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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, $[\sin t+(1/3) cos^3 t-2 \cos t ]_0^{2 \pi}=0$