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: 21


$\dfrac{1}{6} \pi a^3$

Work Step by Step

We have: $Normal \ Vector ; n=\vec{r_u} \times \vec{r_v}=a^2 \cos u \sin^2 v i+a^2 \sin u \sin^2 v j+a^2 \sin v \cos v$ Now, $F \cdot n \ d \theta =a^3 \sin v \cos^2 v$ and $\iint_{S} F (u,v) \ d S=\int_{0}^{\pi/2} \int_{0}^{\pi/2} a^3 \sin v \cos^2 v \ dv \ du$ or, $=(\dfrac{-1}{3} ) a^3 \int_{0}^{\pi/2}[\cos^3 v]_{0}^{\pi/2} \ du$ or, $=(\dfrac{-1}{3} ) a^3 \int_{0}^{\pi/2}[\cos^3 (\pi/2) -\cos^3 (0) ] \ du$ or, $=\dfrac{1}{3} a^3 [u]_{0}^{\pi/2}$ We evaluate the integral as: $\int_{0}^{\pi/2} \int_{0}^{\pi/2} a^3 \sin v \cos^2 v \ dv \ du=\dfrac{1}{6} \pi a^3$
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