## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 6 - Section 6.4 - Areas of Surfaces of Revolution - Exercises - Page 375: 12

#### Answer

$4 \pi \sqrt 5$

#### 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, $S=\int_{1}^{2} 2 \pi [(2y-1) \sqrt {1+4}] dy = 2\sqrt 5 \pi [y^2-y] _{1}^{2}=4 \pi \sqrt 5$

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