Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 18 - Fundamental Theorems of Vector Analysis - 18.1 Green's Theorem - Exercises - Page 984: 16

Answer

$3 \pi$

Work Step by Step

The area of a given region can be calculated as: $Area; A=\dfrac{1}{2}\int_C xdy-y dx$ Here, we are given that $x=t-\sin t; y=1-\cos t$ Therefore, $A=\dfrac{1}{2}\int_C (t-\sin t) \sin t \ dt -(1-\cos t) (1-\cos t) \ dt \\=\dfrac{1}{2}\int_0^{2 \pi} (t \sin t -2+2 \cos t) \ dt \\=\dfrac{1}{2} [-t \cos t +\sin t -2t +2 \sin t]_0^{2 \pi} \\=-3 \pi$ But the area cannot be negative, so we have $Area=3 \pi$
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