Definition D1
Write V(ρ) for the volume of the ball with center O and radius ρ, and S(ρ) for the area of the sphere bounding that ball.
Proof.
(1)S(ρ) is a fixed number for each ρ > 0.A1
(2)For 0 < ρ < σ, the set of all points whose distance from O is at least ρ and at most σ is called the shell of thickness σ − ρ.convention
(3)The ball of radius σ is divided without overlap into the ball of radius ρ and that shell, so the shell's volume is V(σ) − V(ρ)(2), A10
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Lemma L1
If t > 0, then S(r)·t ≤ V(r + t) − V(r) ≤ S(r + t)·t
Proof.
(1)The shell between the sphere of radius r and the sphere of radius r + t has thickness t, and its volume is V(r + t) − V(r).D1
(2)The volume of that shell is at least S(r)·t and at most S(r + t)·t.(1), A1, A2
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Lemma L2
If 0 < t < r, then S(r − t)·t ≤ V(r) − V(r − t) ≤ S(r)·t
Proof.
(1)r − t > 0, so the shell between the sphere of radius r − t and the sphere of radius r can be taken, and its thickness is t.assumption, D1
(2)The volume of that shell is V(r) − V(r − t).(1), D1
(3)Applying the squeeze with radius r − t and thickness t gives S(r − t)·t ≤ V(r) − V(r − t) ≤ S(r)·t(1), (2), A1, A2
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Lemma L3
V is differentiable and V′(r) = 4πr²
Proof.
(1)V(ρ) = ⁴⁄₃πρ³A3
(2)(ρ³)′ = 3ρ²A4
(3)V′(ρ) = ⁴⁄₃π × 3ρ² = 4πρ²(1), (2), A5, A11
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Theorem T
The area of a sphere of radius r is S(r) = 4πr²
Proof.
(1)For t > 0, dividing the left inequality of L1 by t gives S(r) ≤ (V(r + t) − V(r))/tL1, A9, A11
(2)As t → 0+ the right side tends to V′(r), and V′(r) = 4πr².A6, L3
(3)S(r) does not depend on t, so taking the limit gives S(r) ≤ 4πr²(1), (2), A7
(4)For 0 < t < r, dividing the right inequality of L2 by t gives (V(r) − V(r − t))/t ≤ S(r)L2, A9, A11
(5)Setting s = −t gives (V(r) − V(r − t))/t = (V(r + s) − V(r))/s, and t → 0+ is s → 0−, so this value tends to V′(r) = 4πr².A6, A11, L3, E
(6)Taking the limit gives 4πr² ≤ S(r)(4), (5), A7
(7)S(r) ≤ 4πr² and 4πr² ≤ S(r), so S(r) = 4πr²(3), (6), A8
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