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 1000: 7


$\dfrac{11 \pi}{12}$

Work Step by Step

Consider $\vec{r} (r,\theta) =r \cos \theta i+r \sin \theta j+(1-r^2) k$ So, $\vec{r_{r}} \times \vec{r_{\theta}}=-r \sin \theta+r \cos \theta+0$ and $|\vec{r_{r}} \times \vec{r_{\theta}}| =\sqrt {4r^4+r^2}=r \sqrt {4r^2 +1}$ Now, $S=\int_{0}^{ 2 \pi} \int_{0}^{1} (\cos^2 \theta )(4r^5 +r^3) \ dr \ d \theta $ or, $= \int_{0}^{2 \pi} [(2/3) r^6+\dfrac{1}{4} r^4 \cos^2 \theta ]_0^1 \ dx$ or, $=\int_{0}^{2 \pi} \dfrac{11}{12} \cos^2 \theta d \theta$ By using a calculator, we evaluate the integral as: $\int_{0}^{ 2 \pi} \int_{0}^{1} (\cos^2 \theta )(4r^5 +r^3) \ dr \ d \theta=\dfrac{11 \pi}{12}$
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