Thomas' Calculus 13th Edition

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

Chapter 16: Integrals and Vector Fields - Practice Exercises - Page 1028: 10



Work Step by Step

Write the parametric equations as follows: $$ x=2 \cos \space t\\ y=2 \sin \space t $$ and $$ dx= -2 \sin \space t dt \\ dy=2 \cos\space t \space dt $$ We need to plug the above values in the given integral. $$\int_{0}^{2 \pi}( -2 \sin t )^2 (-2 \sin t dt)+(2 \cos t)^2 (2 \cos t dt)= \int_{0}^{2 \pi} (-8\sin^2 t+8 \cos^3 \space t) \space dt\\=\int_{0}^{2 \pi} -8 sin t +8 \sin t\cos^2 t +8 \cos \space t -8 \cos (t) \sin^2 (t) \\=[8\cos (2 \pi)-\dfrac{8}{3}\cos^3 (2 \pi)+8 \sin (2 \pi)-\dfrac{8}{3}\sin^3 (2 \pi))-(8\cos (0)-\dfrac{8}{3}\cos^3 (0)+8 \sin (0)-\dfrac{8}{3}\sin^3 (0)] \\= 0$$
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