Answer to Question #111583 in Differential Equations for Rajeev

Question

Show that the closed sphere with centre (2,3,7) and radius 10 in 3 R is contained in the
open cube P = {(x, y, z) : x − 2 <11, y − 3 <11, z − 7 <11}.

Answer 1

We shall show that the closed sphere with centre (2,3,7) and radius 10 in $R^3$  is contained in the

open cube $P = \{(x, y,z): x − 2 <11, y − 3 <11, z − 7 <11\}.$

The equation of the sphere with centre (2, 3, 7) and radius 10 in $R^3$ is

$(x−2) ^2 +(y−3) ^2 +(z−7) ^2 =10^2\\ (x−2)^2 ≥0,x∈\R\\ (y-3)^2\geq0, y\in \R\\ (z-7)^2\geq0, z\in \R$

Then

$0≤(x−2)^2≤10^2\\ |x-2|\leq10\\ 0\leq(y-3)^2\leq 10^2\\ |y-3|\leq10\\ 0\leq(z-7)^2\leq10^2\\ |z-7|\leq10$

Hence

$x-2<11, y-3<11,z-7<11, x,y,z\in R$

This means that the closed sphere with centre (2,3,7) and radius 10 in $R^3$ is contained in the

open cube $P = \{(x, y,z): x − 2 <11, y − 3 <11, z − 7 <11\}.$