Question #111583
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}.
1
Expert's answer
2020-04-23T17:47:15-0400

We shall show that the closed sphere with centre (2,3,7) and radius 10 in R3R^3  is contained in the

open cube P={(x,y,z):x2<11,y3<11,z7<11}.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 R3R^3 is

(x2)2+(y3)2+(z7)2=102(x2)20,xR(y3)20,yR(z7)20,zR(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(x2)2102x2100(y3)2102y3100(z7)2102z7100≤(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

x2<11,y3<11,z7<11,x,y,zRx-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 R3R^3 is contained in the

open cube P={(x,y,z):x2<11,y3<11,z7<11}.P = \{(x, y,z): x − 2 <11, y − 3 <11, z − 7 <11\}.



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