Answer
$W=\dfrac{m(b^2-a^2)}{2} $
Work Step by Step
Work done: $W=\int_C F\cdot dr=\int_C F\cdot r'(t) dt=\int_0^{\pi/2} (-ma \sin t i-mb \cos t j) \cdot (a \cos t i-b \sin t j+ck) dt=m (b^2-a^2) \int_0^{\pi/2} \sin t \cos t dt$
or, $=\dfrac{m(b^2-a^2)}{2} \int_0^{\pi/2} \sin 2 t dt $
Thus, work done: $W=\dfrac{m(b^2-a^2)}{2} [\dfrac{-\cos 2t}{2}]_0^{\pi/2}$
or, work done: $W=\dfrac{m(b^2-a^2)}{2} [\dfrac{1-(-1)}{2}=\dfrac{m(b^2-a^2)}{2} $
