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 - Section 16.7 - Stoke's Theorem - Exercises 16.7 - Page 1013: 2



Work Step by Step

$F(r(t)) \cdot \dfrac{dr}{dt}=\lt 3\cos t, 3 \sin t \gt \cdot \lt 6\sin t,9 \cos t \gt \\ \implies F(r(t)) \cdot \dfrac{dr}{dt}=(27 \cos ^2 t-18\sin^2 t)$ The, we have $\int _C F(r(t)) \cdot dr =\int _0^{2\pi} [27 \cos ^2 t-18\sin^2 (t)]dt=\int _0^{2\pi} (\dfrac{45}{2}) \cos (2t) +(\dfrac{9}{2}) =[\dfrac{45}{2} \sin 2t +(\dfrac{9}{2})(t)] _0^{2\pi}$ Thus, $[\dfrac{45}{2} ( \sin 4\pi-\sin 0) +\dfrac{9}{2}(2\pi-0)]=0+9\pi=9\pi$
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