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: 28

Answer

$2\pi rh$

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+h} \sqrt {R^2-x^2} \times \dfrac{r}{\sqrt {r^2-x^2}} dx $ or, $=(2 \pi ) \int_{a}^{a+h} r dx$ or, $= 2 \pi r (a+h-a)$ or, $=2\pi rh$
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