Answer
$\dfrac{35\sqrt 5 \pi }{3} $
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}^{15/4} ( 2 \sqrt {4-y}) \sqrt {\dfrac{5-y}{4-y} } dy$
Suppose $a^2=5-y \implies -dy =2u du$
or, $ -8\pi \int_{\sqrt 5}^{\sqrt 5/2} a^2 da =(-8 \pi) \times [ \dfrac{ a^3}{3}]_{\sqrt 5}^{\sqrt 5/2}$
or, $=\dfrac{16\pi }{9} (\dfrac{ 5 \sqrt 5}{8}- 5 \sqrt 5)$
or, $=\dfrac{35\sqrt 5 \pi }{3} $
