University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Section 6.4 - Areas of Surfaces of Revolution - Exercises - Page 376: 26

Answer

$\pi r \sqrt {r^2+h^2} $

Work Step by Step

The formula to determine the surface area is as follows: $S= \int_{m}^{n} 2 \pi y \sqrt {1+(\dfrac{dy}{dx})^2} $ Now, $ S=(2 \pi)\int_{0}^{h} \dfrac{r \space x}{h}\times \sqrt {\dfrac{r^2+h^2}{h^2} } dx \\=|\dfrac{\pi r \sqrt {r^2+h^2} (x^2) }{h^2} ]_0^h \\=\pi r \sqrt {r^2+h^2} $
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