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