Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 11 - Three-Dimensional Space; Vectors - 11.1 Rectangular Coordinates In 3-Space; Spheres; Cylindrical Surfaces - Exercises Set 11.1 - Page 772: 25

Answer

The surface is a sphere with a radius $r=\frac{3 \sqrt{6}}{4}$ and centered at $C(1 / 2,-3 / 4,-5 / 4)$

Work Step by Step

We need to describe the surface whose equation is \[ 0=-2+2 z^{2}+2 y^{2}+2 x^{2}-2 x-3 y+5 z \] If we multiply this equation by $\left(\frac{1}{2}\right)$, then the result will be: \[ 1=-x-3 / 2 y+5 / 2 z+z^{2}+x^{2}+y^{2} \] This is the equation of a sphere. We complete squares to find its center and radius. \[ \begin{array}{l} 1=\left(-3 / 2 y+y^{2}\right)+\left(\frac{5}{2} z+z^{2}\right)+\left(-x+x^{2}\right) \\ \Rightarrow \left(-\frac{3}{2} y+y^{2}+\frac{9}{16}\right)+\left(\frac{5}{2} z+z^{2}+\frac{25}{10}\right)+\left(1 / 4-x+x^{2}\right)=\frac{9}{16}+\frac{25}{16}+1+\frac{1}{4}+\frac{9}{16} \\ \Rightarrow (5 / 4+z)^{2}+(3 / 4+y)^{2}+(-1 / 2+x)^{2}=\frac{54}{16}=\frac{27}{8} \\ \Rightarrow \quad (5 / 4+z)^{2}+(3 / 4+y)^{2}+(-1 / 2+x)^{2}=\left(\frac{3 \sqrt{3}}{2 \sqrt{2}}\right)^{2}=\left(\frac{3 \sqrt{6}}{9}\right)^{2} \end{array} \] And therefore, the surface is a sphere of radius $r=\frac{3 \sqrt{6}}{4}$ and is concentrated at $C(1 / 2,-3 / 4,-5 / 4)$
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