Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 657: 28

$\dfrac{21}{2}$

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ This implies that $S=\int_{0}^{3} \sqrt{(\sqrt{2t+3})^2+(1+t)^2} dt=\int_{0}^{3} (t+2) dt$ Now, $S=[\dfrac{t^{2}}{2}+2t]_{0}^{3}=[\dfrac{9}{2}-0]+[(2)(3)-0]$ or, $S=\dfrac{9}{2}+(2)(3)=\dfrac{9}{2}+6=\dfrac{21}{2}$