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 16.6 - Surface Integrals - Exercises 16.6 - Page 1001: 24

Answer

$2 \pi a$

Work Step by Step

We have: $Normal \ Vector ; n=\vec{r_u} \times \vec{r_v}=\cos u i+ \sin u j+0 k$ Now, $F \cdot n = \cos^2 u+\sin^2 u+0=1$ and $\iint_{S} F \cdot N \ d S=\int_{0}^{2 \pi} \int_{0}^{a} (1) \ dv \ du$ or, $= \int_{0}^{2 \pi}[v]_{0}^{a} \ du$ or, $=a \int_{0}^{2 \pi} \ du$ or, $=a [u]_{0}^{2\pi}$ or, $=a (2 \pi-0)$ or, $=2 \pi a$
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