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: 28


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

Work Step by Step

a) The formula to determine the surface area is as follows: $S= \int_{m}^{n} 2 \pi y \sqrt {1+(\dfrac{dy}{dx})^2} \\= 2 \pi \times \int_{-22.5}^{45} \sqrt {R^2-x^2} \times \dfrac{R}{\sqrt {R^2-x^2}} dx \\=(2 \pi ) \int_{-22.5}^{45} \space R \space dx \\= 6075 \pi$ b) We need the required nearest square foot $6075 \pi \approx 19085 \space ft^2$
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