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: 9

Answer

$4 \pi \sqrt 5$

Work Step by Step

Integrate the integral to calculate the surface area as follows: We have: $S_{A}= (2 \pi)\int_{a}^{b} y \sqrt {1+(\dfrac{dy}{dx})^2}$ or, $ =(2 \pi)\int_{1}^{4}(\dfrac{x}{2}) \cdot \sqrt {1+\dfrac{1}{4}} dx$ or, $=[\dfrac{\sqrt 5 \pi x^2}{4} ]_1^4$ Thus, $Surface \space Area=4 \pi \sqrt 5$
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