Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.2 Vectors in Three Dimensions - 11.2 Exercises: 25

Answer

$(x+2)^2 + y^2 + (z-4)^2 \leq 1^2$

Work Step by Step

Equation of sphere with radius $r$ and centered at $(a,b,c)$: $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$ Plug in center and radius: $(x+2)^2 + (y-0)^2 + (z-4)^2 = 1^2$ Because it's a ball, the equation must account for the fact that the sphere is not hollow. $(x+2)^2 + (y-0)^2 + (z-4)^2 \leq 1^2$
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