"Solve 1≤x2+y2+z2≤4x2+y2+z2≤4, x≥0" was answered by our experts

High School
Algebra
Algebra I
Exponents and radicals

Dashawn Robbins

Answered question

2022-04-21

Solve

1 \leq x^{2} + y^{2} + z^{2} \leq 4

x^{2} + y^{2} + z^{2} \leq 4

x \geq 0

Answer & Explanation

Giovanny Howe

Beginner2022-04-22Added 18 answers

(a)
Given:

1 \leq x^{2} + y^{2} + z^{2} \leq 4

Compare the above inequality with the general equation of a sphere

x^{2} + y^{2} + z^{2} = r^{2}

implies.

x^{2} + y^{2} + z^{2} \leq 4

represents a solid ball with radius 2 centered at origin

(0, 0, 0)

.
And,

x^{2} + y^{2} + z^{2} \geq 1

represents a solid ball of radius 1 centered at origin.
Therefore, the region is, the solid ball of radius 1 removed from a solid ball of radius 2.
(b)
Given:

x^{2} + y^{2} + z^{2} \leq 4

Compare this inequality with the general equation of a sphere

x^{2} + y^{2} + z^{2} = r^{2}

implies.
It represents a solid ball of radius 2 centered at origin.
Thus, the required points are the points inside the solid ball.

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