University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.6 - Cylinders and Quadric Surfaces - Exercises - Page 636: 26

Answer

Elliptical cone. See image: .

Work Step by Step

Divide with 36; the equation is of the form $\displaystyle \frac{x^{2}}{3^{2}}+\frac{y^{2}}{2^{2}}=\frac{z^{2}}{4^{2}}$ and, comparing to Table 11.1, this is an elliptical cone. The traces in planes parallel to the xy plane are ellipses. For $z=\pm 2$, the ellipses have semiaxes 3 and 2 (use this to sketch).
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