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: 30


a) $6075 \pi $ and b) $19085 \space ft^2$

Work Step by Step

Our aim is to integrate the integral to compute the surface area. In order to solve the integral, we have: a) $Surface \space Area(S_A)= (2 \pi)\int_{a}^{b} y \sqrt {1+(\dfrac{dy}{dx})^2}$ or, $ =(2 \pi)\int_{-22.5}^{45} \sqrt {R^2-x^2} \times \dfrac{R}{\sqrt {R^2-x^2}} dx =(2 \pi ) \int_{-22.5}^{45} R dx= 6075 \pi$ b) The required nearest square foot $6075 \pi \approx 19085 \space ft^2$
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