Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.6 - Cylinders and Quadric Surfaces - Exercises 12.6 - Page 732: 36

Answer

Elliptical cylinder. See image: .
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Work Step by Step

There is no z in the equation; thus we have a cylinder. In the xy-plane (z=0), the trace is an ellipse $\displaystyle \frac{x^{2}}{(1/4)^{2}}+\frac{y^{2}}{(1/2)^{2}}=1$ and likewise in all the planes parallel to $z=0$. The resulting surface is an elliptical cylinder, with the z-axis as its axis.
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