Answer
$\dfrac{16\pi }{9} $
Work Step by Step
Our aim is to integrate the integral to compute the surface area. In order to solve the integral, we have:
$Surface \space Area(S_A)= (2 \pi)\int_{a}^{b} y \sqrt {1+(\dfrac{dy}{dx})^2}$
or, $ =(2 \pi)\int_{1}^{3} (\dfrac{-y^{3/2}}{3} +y^{1/2}) \times (\dfrac{y^{1/2}}{2} +y^{-1/2}) dy$
or, $=\dfrac{ \pi}{3} [\dfrac{-y^{3}}{3} +y^2+3y]_{1}^{3}$
or, $=\dfrac{16\pi }{9} $
