Answer
$\dfrac{16 \sqrt 2-5 \sqrt 5 }{12}\pi $
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_{5/8}^{1} ( \sqrt {2y-1}) \sqrt {\dfrac{2y}{2y-1} } dy$
or, $= 2\sqrt 2 \pi \int_{5/8}^{1}\sqrt y dy$
or, $=( 2 \sqrt 2 \pi) \int_{5/8}^{1} (2/3) y^{3/2} dy$
or, $ =\dfrac{ 4 \sqrt 2 \pi}{3} (1- \dfrac{5 \sqrt 5} {16 \sqrt 2}) $
or, $=\dfrac{16 \sqrt 2-5 \sqrt 5 }{12}\pi $
