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 600: 58


Center: $(0,-\dfrac{1}{3},\dfrac{1}{3})$ and radius: $\dfrac{\sqrt {29}}{3}$

Work Step by Step

We know that the standard equation of a sphere is written as: $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2\ \ \ $ (1) Here, $(x_0,y_0,z_0)$ represents the center and $r$ is the radius of the sphere. Since, we have $3x^2+3y^2+3z^2+2y-2z=9$ or, $2(x^2+\dfrac{1}{2}x)+2(y^2+\dfrac{1}{2}y)+2(z^2+\dfrac{1}{2}z)=9$ or, $3x^2+3(y+\dfrac{1}{3})^2+3(z-\dfrac{1}{3})^2=\dfrac{29}{3}$ or, $x^2+(y+\dfrac{1}{3})^2+(z-\dfrac{1}{3})^2=\dfrac{29}{9}$ or, $(x-0)^2+(y -(-\dfrac{1}{3}))^2+(z-\dfrac{1}{3})^2=(\sqrt{\dfrac{29}{9}})^2$ or, $(x-0)^2+(y -(-\dfrac{1}{3}))^2+(z-\dfrac{1}{3})^2=(\dfrac{\sqrt {29}}{3})^2$ Compare the above equation with equation (1), we get Center :$(0,-\dfrac{1}{3},\dfrac{1}{3})$ and radius: $\dfrac{\sqrt {29}}{3}$
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