Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - 15.7 Triple-Integrals in Cylindrical Coordinates - 15.7 Exercises - Page 1083: 6


plane with $y=\dfrac{\sqrt 3x}{3}$

Work Step by Step

In the cylindrical coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$ Conversion of rectangular to cylindrical coordinate system, we have $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$ Here, the given angle is $\theta=\dfrac{\pi}{6}$ that means that surface will be a plane passing through the origin with an angle $\dfrac{\pi}{6}$ above the $x$-axis in the $xy$ plane. Now, $\tan (\dfrac{\pi}{6})=\dfrac{y}{x}$ $\implies \dfrac{y}{x}=\dfrac{\sqrt 3}{3}$ Thus, $y=\dfrac{\sqrt 3x}{3}$ Hence, we have an equation of a plane with $y=\dfrac{\sqrt 3x}{3}$.
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