Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.5 - Surface Area - 15.5 Exercise - Page 1028: 7

Answer

$\dfrac{2 \pi}{8} [\dfrac{2}{3} \sqrt {(17)^3} -\dfrac{2}{3} \sqrt {(5)^3}]$

Work Step by Step

The surface area of the part $z=f(x,y)$ can be written as: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$ and, $\iint_{D} dA$ is the projection of the surface on the xy-plane. Now, the area of the given surface is: $A(S)=\iint_{D} \sqrt {1+(-2x)^2+(2y)^2} dA= \iint_{D} \sqrt {1+4(x^2+y^2)} dA$ We can use the polar co-ordinates because of the part $x^2+y^2$ $A(S)=\int_{0}^{2 \pi} \int_{1}^2 \sqrt {1+4r^2} r dr d\theta $ Let us suppose that $a =1+4r^2 \implies da = 8 r dr$ Thus, $A(S)=(1/8) \int_{0}^{2 \pi} [\dfrac{2 a^{3/2}}{3}da]_1^2 d\theta \\ =\int_{0}^{2 \pi} d\theta \times \dfrac{}{8} [\dfrac{2}{3} (1+4(2)^2)^{3/2} -\dfrac{2}{3} (1+4(1)^2)^{3/2}] \\ =\dfrac{2 \pi}{8} [\dfrac{2}{3} \sqrt {(17)^3} -\dfrac{2}{3} \sqrt {(5)^3}]$
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