Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 6: Applications of Definite Integrals - Section 6.4 - Areas of Surfaces of Revolution - Exercises 6.4 - Page 340: 18

Answer

$\dfrac{16\pi }{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_{1}^{3} (\dfrac{-y^{3/2}}{3} +y^{1/2}) \times (\dfrac{y^{1/2}}{2} +y^{-1/2}) dy$ or, $=\dfrac{ \pi}{3} [\dfrac{-y^{3}}{3} +y^2+3y]_{1}^{3}$ or, $=\dfrac{16\pi }{9} $
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