Definition D1
Take the plane α to be z = 0 and a point O on it as the origin. The part of the ball with center O and radius r where z ≥ 0 is called the hemisphere H.
Proof.
(1)The ball with center O and radius r is the set of all points at distance at most r from O. Its volume is called the volume of the sphere.convention
(2)Let C be the cylinder of height r whose base is the disk on α with center O and radius r. C lies between z = 0 and z = r.convention
(3)Let K be the cone with apex O whose base is the top face of C, and let W be C with K removed.convention
(4)H also lies between z = 0 and z = r.(1), convention
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Lemma L1
For 0 ≤ h ≤ r, the cross-section of the hemisphere H cut by the plane z = h has area π(r² − h²)
Proof.
(1)0 ≤ h ≤ r gives h² ≤ r², that is, r² − h² ≥ 0, so √(r² − h²) is defined.assumption, A11, A15
(2)Set M = (0, 0, h). The line OM is perpendicular to the plane z = h.D1, convention
(3)If a point P of the plane z = h differs from M, then OM ⊥ MP, so triangle OMP has a right angle at M and OP² = h² + MP²(2), A9, A6
(4)If P = M, then OP = h and MP = 0, so the same equation holds.(2)
(5)P belongs to H exactly when OP ≤ r, and since both numbers are nonnegative, that is OP² ≤ r², that is, MP² ≤ r² − h².(3), (4), D1, A11, A15
(6)MP ≥ 0 and (√(r²−h²))² = r² − h², so MP² ≤ r² − h² is the same as MP ≤ √(r² − h²).(1), (5), A10, A11
(7)The cross-section is the disk with center M and radius √(r² − h²), so its area is π(√(r²−h²))² = π(r² − h²)(6), A5, A10
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Lemma L2
For 0 ≤ h ≤ r, the cross-section of W cut by the plane z = h also has area π(r² − h²)
Proof.
(1)The cross-section of C is a disk congruent to the base, so its area is πr²D1, A12, A5
(2)The distance from the apex O of K to the base is r and to the plane z = h is h, so for h > 0 the cross-section has area (h/r)² πr² = πh²D1, A4, A5, A14
(3)For h = 0 the cross-section is the single apex, of area 0 = π × 0²D1, convention
(4)The two cross-sections are concentric disks and h ≤ r, so the cross-section of K is contained in the cross-section of C.(1), (2), (3), assumption, A11
(5)The cross-section of W is the cross-section of C minus the cross-section of K, so its area is πr² − πh² = π(r² − h²)(4), D1, A7, A13
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Lemma L3
The hemisphere H and the solid W have equal volume.
Proof.
(1)H and W both lie between the planes z = 0 and z = r.D1
(2)For every h with 0 ≤ h ≤ r, the two cross-sections have the same area π(r² − h²).L1, L2
(3)The two volumes are equal.(1), (2), A1
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Theorem T1
The volume of a hemisphere of radius r is ⅔πr³
Proof.
(1)The volume of C is πr² × r = πr³D1, A2, A5, A14
(2)The volume of K is ⅓ × πr² × r = ⅓πr³D1, A3, A5, A14
(3)C is divided without overlap into K and W, so the volume of W is πr³ − ⅓πr³ = ⅔πr³(1), (2), D1, A7, A13, A14
(4)The volume of H equals the volume of W, hence ⅔πr³(3), L3, E
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Theorem T2
The volume of a sphere of radius r is ⁴⁄₃πr³
Proof.
(1)The ball is divided by the plane z = 0 into the hemisphere with z ≥ 0 and the hemisphere with z ≤ 0.D1
(2)The lower hemisphere is the reflection of the upper hemisphere in the plane z = 0, so their volumes are equal.(1), A8
(3)The volume of the sphere is ⅔πr³ + ⅔πr³ = ⁴⁄₃πr³(1), (2), T1, A7, A14
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