Answer
$\dfrac{253 \pi}{20}$
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}^{2} (x) \sqrt {1+\dfrac{(4y^6-1)^2}{16y^6}} dy $
or, $ =(2 \pi)\int_{1}^{2} (y) (\dfrac{4y^6+1}{4y^3} ) dy $
or, $= (\pi/2) \int_{1}^{2} [4y^6+\dfrac{1}{y^2}] dy$
or, $= (\dfrac{\pi}{2}) [\dfrac{4y^3}{3}-y^{-1}]_{1}^{2}$
or, $=\dfrac{253 \pi}{20}$
