# Chapter 6: Applications of Definite Integrals - Section 6.4 - Areas of Surfaces of Revolution - Exercises 6.4 - Page 340: 20

$\dfrac{16 \sqrt 2-5 \sqrt 5 }{12}\pi$

#### 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, $=\dfrac{16 \sqrt 2-5 \sqrt 5 }{12}\pi$

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