Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 26

$7$

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $S=\int_{0}^{\sqrt 3} \sqrt{(3t^2)^2+(3t)^2} dt=\int_{0}^{\sqrt 3} \sqrt{9t^4+9t^2} dt$ or, $(3t) \int_{0}^{\sqrt 3} \sqrt{t^2+1} dt=(\dfrac{3}{2}) \int_{0}^{\sqrt 3} \sqrt{t^2+1} (2t) dt$ Plug $t^2+1 =p \implies dp=2tdt$ Thus, $S=(\dfrac{3}{2}) \int_{1}^{4} p^{1/2} dp$ or, $S=[u^{3/2}]_{1}^{4} =7$