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 341: 25

Answer

$4 \pi a^2$

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_{-a}^{a} \sqrt {a^2-x^2} \times \sqrt {\dfrac{a^2}{a^2-x^2} } dx $ or, $ =(2 \pi)\int_{-a}^{a} a dx $ or, $=2 \pi a (x)_{-a}^a$ or, $=4 \pi a^2$
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