Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 16 - Section 16.6 - Parametric Surfaces and Their Areas - 16.6 Exercise - Page 1182: 56

Answer

$$\approx 4.4506$$

Work Step by Step

$\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}=\dfrac{9 (\cos^3 v \sin 2v \sin 2u)}{4} (sin u \sin vi+\cos u \sin v j+\cos v \cos u \sin u k) $ and $|\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}|=\dfrac{9}{4} |(\cos^3 v \sin 2v \sin 2u)| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}}$ Therefore, $A(S)= \iint_{D} |\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}| \\=\iint_{D} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} dA \\=\int_0^{\pi} \int_{0}^{2 \pi} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} \ dv \ du$ Now, by using a calculator, we get: $$A(S) =\dfrac{9}{4} \times \int_0^{\pi} \int_{0}^{2 \pi} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} dv du \\ \approx 4.4506$$
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