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