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: 17


$-15 \pi$

Work Step by Step

Applying Stoke's Theorem, we have $\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$ $r_{\phi}=\lt\sqrt 3 \cos \phi \cos \theta, \sqrt 3 \cos \phi \sin \theta,-\sqrt 3 \sin \phi \gt$ and $r_{\theta}=\lt -\sqrt 3 \sin \phi \sin \theta, \sqrt 3 \sin \phi \cos \theta,0 \gt$ Then, we have $\iint _S (\nabla \times F) \cdot n d\sigma=\int _{0}^{\pi/2}\int_{0}^{2 \pi} (\dfrac{-15}{2} \sin 2 \phi )d\phi d\theta $ This implies that $(-15 \pi) \int _{0}^{\pi/2} \sin 2 \phi d \phi=-15 \pi$
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