Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 275: 18

Answer

$4\pi $ square units

Work Step by Step

The Graph of the Circle is A=$\frac{1}{4}\pi r^2$ =$\frac{1}{4}\pi (4)^2$ =$4\pi$ =$\int^0_{-4}\sqrt{16-x^2}$ =$4\pi$ square Units
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