Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 29

$\pi^2$

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $S=\int_{0}^{\pi/2} \sqrt{(-8 \sin t+8 \sin t8t \cos t)^2+(8 \cos t-8 \cos t+8t \sin t)^2}$ or, $S=\int_{0}^{\pi/2} \sqrt{(8t \cos t)^2+(8t \sin t)^2} dt=\int_{0}^{\pi/2} \sqrt {64t^2(\cos^2 t+\sin^2 t)} dt$ Thus, $S=\int_{0}^{\pi/2} \sqrt {64t^2} dt$ or, $S=[4t^2]_{0}^{\pi/2} =4(\dfrac{\pi}{2})^2=\pi^2$