Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 657: 26

$7$

$dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ This implies that $S=\int_{0}^{\sqrt 3} \sqrt{(3t^2)^2+(3t)^2} dt$ or, $S=(3t) \int_{0}^{\sqrt 3} \sqrt{t^2+1} dt$ and $S=(\dfrac{3}{2}) \int_{0}^{\sqrt 3} \sqrt{t^2+1} (2t) dt$ Plug $t^2+1 =k$ or, $dk=(2t) dt$ Thus, $S=(\dfrac{3}{2}) \int_{1}^{4} k^{(1/2)} dp$ or, $S=[k^{(3/2)}]_{1}^{4} =7$