University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 11 - Section 11.1 - Three-Dimensional Coordinate Systems - Exercises - Page 599: 12

Answer

A circle of radius $\sqrt{3}$, centered at the origin, in the xz-plane.

Work Step by Step

$x^{2}+(y-1)^{2}+z^{2}=4$ is a sphere centered at $(0,1,0)$, with radius $2$. $y=0$ is the xz-plane. The equation of their intersection in the xz-plane is obtained by substituting $y=0$ into the equation of the sphere: $x^{2}+1+z^{2}=4$ $x^{2}+z^{2}=3$, in the xz-plane. A circle of radius $\sqrt{3}$, centered at the origin, in the xz-plane.
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