University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 15 - Section 15.7 - Stokes' Theorem - Exercises - Page 896: 16


$25 \pi$

Work Step by Step

Applying Stoke's Theorem, we have $\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$ $r_u=\lt \cos v,\sin v,1 \gt$ and $r_u=\lt -u\sin v,u\cos v,0 \gt$ Then, we have $\iint _S (\nabla \times F) \cdot n d\sigma=\int _{0}^{2 \pi}\int_{0}^{5} (u \cos v+u \sin v+u) du dv $ This implies that $\int _{0}^{2 \pi}(\dfrac{25}{2}) [ \cos v+\sin v+1]dv=25 \pi$
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