Answer
$\dfrac{ \pi (2\sqrt 2-1)}{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_{0}^{1} \dfrac{y^3}{3} \sqrt {y^4+1} dx$
or, $=[\dfrac{ \pi}{9} \times (y^4+1)^{3/2} ]_{0}^{1}$
or, $=\dfrac{ \pi (2\sqrt 2-1)}{9} $
