Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 14 - Vector Calculus - 14.6 Surface Integrals - 14.6 Exercises - Page 1123: 2

Answer

$x =\dfrac{az}{h} \cos \theta; y=\dfrac{az}{h} \sin \theta$ and $z=Z$

Work Step by Step

Let us consider $Z$ be the axis parameter and the radial parameters can be written as: $\sin \theta$ and $\cos \theta$. Assume that $a$ be the radius of the base of the cone with height $h$. Here, $\theta \in [0, 2\pi]$ and $0 \leq Z \leq h$. Thus, the parametric description for a cone with radius $a$ and at the height $z$ of the cone can be expressed as: $x =\dfrac{az}{h} \cos \theta; y=\dfrac{az}{h} \sin \theta$ and $z=Z$
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