Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 10 - Section 10.6 - The Three-Dimensional Coordinate System - Exercises - Page 482: 40

Answer

(a) $0.5625$ (b) $0.4219$

Work Step by Step

We can write the general equation of a sphere: $(x-a)^2+(y-b)^2+(z-c)^2 = r^2$ where $(a,b,c)$ is the center of the sphere and $r$ is the radius The equation of the first sphere is: $(x-1)^2+(y+2)^2+(z-4)^2 = 36$ The radius of the first sphere is $r_1 = 6$ The equation of the second sphere is: $x^2+y^2+z^2 = 64$ The radius of the second sphere is $r_2 = 8$ (a) We can find the surface area of the first sphere: $A_1 = 4~\pi~r_1^2$ $A_1 = (4~\pi)~(6)^2$ $A_1 = 144~\pi$ We can find the surface area of the second sphere: $A_2 = 4~\pi~r_2^2$ $A_2 = (4~\pi)~(8)^2$ $A_2 = 256~\pi$ We can find the ratio of the surface areas: $\frac{A_1}{A_2} = \frac{144~\pi}{256~\pi} = 0.5625$ (b) We can find the volume of the first sphere: $V_1 = \frac{4}{3}~\pi~r_1^3$ $V_1 = \frac{4}{3}~\pi~(6)^3$ $V_1 = 288~\pi$ We can find the volume of the second sphere: $V_2 = \frac{4}{3}~\pi~r_2^3$ $V_2 = \frac{4}{3}~\pi~(8)^3$ $V_2 = 682.7~\pi$ We can find the ratio of the volumes: $\frac{V_1}{V_2} = \frac{288~\pi}{682.7~\pi} = 0.4219$
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