Answer
$\pi^2$
Work Step by Step
$dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$
This implies that
$S=\int_{0}^{\pi/2} \sqrt{(-8(\sin t)+(8) \sin t (8t) (\cos t))^2+(8 (\cos t)-8 (\cos t)+8(t) (\sin (t))^2}$
or, $\implies S=\int_{0}^{\pi/2} \sqrt{(8t \cos t)^2+(8t \sin t)^2} dt=\int_{0}^{\pi/2} \sqrt {64t^2(\cos^2 t+\sin^2 t)} dt$
$\implies \int_{0}^{\pi/2} \sqrt {64t^2} dt=[4t^2]_{0}^{\pi/2} $
Hence, $S=4(\dfrac{\pi}{2})^2=\dfrac{4\pi^2}{4}=\pi^2$
