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Chapter 9: Problem 40

Describe the solid satisfying the condition. \(x^{2}+y^{2}+z^{2}>-4 x+6 y-8 z-13\)

Short Answer

Expert verified
The inequality describes the region of 3D space outside a sphere centered at (-2,3,-4) with radius 6.

Step by step solution

01

Put Form Into Standard Equation of Sphere

Rewrite expression as \(x^{2}+y^{2}+z^{2}+4x-6y+8z=-13\). This comports to the equation of a sphere \(x^{2}+y^{2}+z^{2}+2ghx+2fky+2ilz+c=0\)
02

Determine Center and Radius of Sphere

By equating coefficients, we find \(h=-g=-2, k=-f=3, l=-i=-4\) and \(c=-13\). We then find \(r=\sqrt{h^{2}+k^{2}+l^{2}-c}\) which yields \(r=6\). So the center of the sphere is \(-2,3,-4\) with radius \(6\)
03

Determine the Zone the Inequality Represents

Given the inequality \(x^{2}+y^{2}+z^{2}>-4x+6y-8z-13\) which has transformed to \(x^{2}+y^{2}+z^{2}+4x-6y+8z>-13\) or \(r^{2}>0\), this represents the region outside the sphere

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