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: 4


$\dfrac{2 \pi a^4}{3}$

Work Step by Step

$\vec{r_u} \times \vec{r_v}=a\cos u \cos v i+a \sin u \cos v j -a \sin v k$ and $|\vec{r_u} \times \vec{r_v}| =\sqrt {(a\cos u \cos v)^2 +(a \sin u \cos v)^2 +( -a \sin v )^2}=\sqrt {a^4 \sin^4 v(\cos^2 u +\sin^2 u)+a^4 \sin^2 v \cos^2 v }=\sqrt {a^4 \sin^2 v}=a^2 \sin v$ Consider, $I=\int_{0}^{\pi/2} \int_{0}^{2 \pi} (a^2 \cos^2 v) (a^2 \sin v) \ du \ dv $ or, $=a^4 \int_{0}^{\pi/2} \sin v \cos^2 v \times [u]_0^{2 \pi} dv $ or, $=-2 \pi a^4 \times \dfrac{1}{3} [\cos^2 v]_{0}^{\pi/2} $ or, $=\dfrac{-2 \pi a^4}{3} (\cos^3 \dfrac{\pi}{2}-\cos^3 0)$ or, $=\dfrac{2 \pi a^4}{3}$
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