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: 3


$\dfrac{4 \pi}{3}$

Work Step by Step

Consider $\vec{r} (u,v) =\cos u \sin v i+\sin u \sin v j+\cos v k$ So, $\vec{r_u} \times \vec{r_v}=-\cos u \sin^2 v i-\sin u \sin^2 v j -\sin v \cos v k$ and $|\vec{r_x} \times \vec{r_y}| =\sqrt {(-\cos u \sin^2 v)^2 +(-\sin u \sin^2 v)^2 +( -\sin v \cos v )^2}=\sqrt {\sin^2 v(\sin^2 v +\cos^2 v) }=\sqrt {\sin^2 v}=\sin v$ Consider, $I=\int_{0}^{\pi} \int_{0}^{2 \pi} \cos^2 u \sin^3 v \ du \ dv $ or, $=(1/2)\int_{0}^{\pi} \int_{0}^{2 \pi} \cos^2 u \sin^3 v \ du \ dv $ or, $=\dfrac{1}{2} \int_{0}^{\pi} [u+\dfrac{\sin 2u}{2} ]_{0}^{2 \pi} \sin^3 v \ dv $ or, $=\pi \int_{0}^{\pi} \sin^3 v \ dv$ Since, $\sin^2 v+\cos^2 v=1 \implies \sin^2 v=1-\cos^2 v$ So, $I= \pi \int_{0}^{\pi} \sin v \times \sin^2 v \ dv=\pi \int_{0}^{\pi} \sin v (1-\cos^2 v) \ dv$ or, $=\pi \int_0^{\pi} (\sin v -\sin v \cos^2 v ) \ dv$ or, $=\pi [-\cos v +\dfrac{\cos^3 v}{3}]_0^{\pi}$ or, $=\dfrac{4 \pi}{3}$
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