Example 1
- r: 10
4188.79
What volume at r = 10? ≈ 4188.79 — not surface 1256.64.
Enter radius r — half the diameter, not the full diameter. The calculator computes volume V = (4/3)πr³. At r = 10 the result is about 4188.79; at r = 5 about 523.60 (eight times smaller). That is the interior, not surface area 4πr².
Enter a radius r greater than zero.
Domain: r > 0. This is not surface area 4πr² — the skin is on the sphere surface area calculator. The result uses two decimal places.
A sphere of radius r is the set of points in space at distance at most r from the center. Its volume is given by Archimedes’ formula V = (4/3)·π·r³. The calculator uses the number you enter as radius r: first it forms r³, then it multiplies by the constant 4/3 π from π. The result is in cm³ or in³, matching the unit switch.
The domain is simple: r > 0. Zero or a negative value does not describe a solid sphere with positive volume here — the calculator shows a message instead of inventing a result. An empty field leaves the result blank. A comma and a period in the field are equivalent. The unit switch changes unit labels only; it does not multiply or divide the numbers you entered.
At r = 10 you get V = 4188.79. At r = 5, half the radius, the volume is 523.60 — one eighth, because (1/2)³ = 1/8. The unit sphere r = 1 gives about 4.19. The result panel also shows r³, the factor 4/3 π, and diameter d = 2r as helpers.
The form field takes the radius, not the diameter. Diameter 10 means r = 5 and 523.60; entering 10 as r gives 4188.79. Paint on the surface is 4πr² — a sibling page. A cylinder and a cone need height; a sphere in this model has one parameter r.
Historically, Archimedes in On the Sphere and Cylinder proved that the sphere’s volume is ⅔ of the volume of the cylinder circumscribed about it (radius r, height 2r). Hence V = (2/3)·πr²·2r = (4/3)πr³. Euclid’s Elements (Book XII) treats spheres by exhaustion. This calculator does not compute surface area or other solids — only volume from Archimedes’ formula.
The sibling sphere surface area page takes the same r with formula 4πr². Cylinder and cube volume are separate tools. Here the only length field is always the radius.
V = (4/3) · π · r³, equivalently V = ⁴⁄₃ π r³.
Symbol r is the sphere radius (form field). The constant π comes from circle geometry. Domain: r > 0.
Mathematical model: Archimedes’ sphere volume. Sources: Archimedes, On the Sphere and Cylinder; Euclid, Elements XII; MathWorld — Sphere; NIST DLMF (π).
A sphere with center O and radius r is the set of points in space whose distance from O is at most r. The volume of that solid — the measure of the interior — is classically V = (4/3)πr³. The calculator takes one length parameter: the radius r. Units must be consistent: if r is in inches, the result is in cubic inches.
The surface area of the sphere is a separate quantity A = 4πr². This calculator does not mix the two formulas: the primary result is always volume. Diameter d = 2r appears in the panel as a helper, but the form field does not accept diameter directly.
Archimedes of Syracuse (c. 287–212 BC) gave the first rigorous proof of the sphere volume formula in On the Sphere and Cylinder (Περὶ σφαίρας καὶ κυλίνδρου). He showed that the sphere’s volume is ⅔ of the volume of the cylinder circumscribed about that sphere — a cylinder of base radius r and height 2r. Hence V_sphere = (2/3)·πr²·2r = (4/3)πr³. In parallel he showed that the sphere’s surface area is ⅔ of the total surface of that cylinder, which is 4πr².
Euclid’s Elements (Book XII) develops the method of exhaustion for solids of revolution, comparing spheres, cones and cylinders of common height. Modern calculus recovers the same result: V = ∫_{-r}^{r} π(r² − x²) dx = (4/3)πr³. Conceptually the calculator does not guess material or pressure — it computes only the geometric volume of an ideal sphere.
Input domain: r > 0 (in numerical practice also a finite floating-point number). For r ≤ 0 the geometric volume is not positive as a solid. Homogeneity: if all lengths are scaled by λ > 0, then r′ = λr and V′ = λ³V, matching the dimension of volume.
Equivalent forms: V = (4/3)πr³ = (1/6)πd³ when d = 2r. Relative sensitivity: ∂V/V = 3 ∂r/r, so a relative radius error enters with weight 3 — confusing diameter with radius overstates the result eightfold.
At fixed r the map is a single value. Scale: doubling r multiplies V by 8; halving r divides V by 8. The ratio V_sphere / V_circumscribed_cylinder = 2/3 is constant for every r.
Step 1. Consider a right circular cylinder of base radius r and height 2r — the cylinder circumscribed about a sphere of the same r. Its volume is V_c = πr²·2r = 2πr³.
Step 2. Archimedes shows that the sphere’s volume is ⅔ of that cylinder: V = (2/3)·2πr³ = (4/3)πr³.
Step 3 (modern integral). A plane cut of the sphere at distance x from the center is a disk of radius √(r² − x²). Volume is V = ∫_{-r}^{r} π(r² − x²) dx = 2π[r²x − x³/3]_0^r = 2π(r³ − r³/3) = (4/3)πr³.
Step 4 (consistency with diameter). With d = 2r: V = (1/6)πd³. Both routes (r and d) agree when d = 2r.
When r = 1, V = 4π/3 ≈ 4.188790 (the panel shows 4.19 to two places). When r = 10, V ≈ 4188.79. When r = 5, V ≈ 523.60 = 4188.79/8. A sphere inscribed in a cube of edge a = 2r has volume (4/3)π(a/2)³ = πa³/6, while the cube has volume a³.
The limit r → 0 gives V → 0. A hemisphere of the same r has half the volume: (2/3)πr³ — a different model, not this form. An ellipsoid with semi-axes a, b, c needs (4/3)πabc; this calculator assumes a = b = c = r.
The calculator uses π in floating-point arithmetic and the order (4/3)·π·r·r·r. Very large r can overflow the floating-point range; typical user sizes stay safe. The display is rounded to two decimal places, matching the worked examples on the page.
Comma and period in the field are equivalent. The calculator does not model wall thickness, meniscus or deformation.
In the lab, distinguish outer from inner diameter on spherical tanks: a tape on the outside measures d_outer, while liquid fills d_inner. Using d_outer/2 as r overstates volume. The formula assumes an ideal sphere.
Comparison with a cylinder: a cylinder of radius r and height 2r has volume 2πr³, which is 3/2 of the sphere’s volume. A cone of base r and height h has (1/3)πr²h — a different solid. The constant 4/3 π ≈ 4.188790 comes from geometry alone, not from an engineering approximation.
Related models on this site: sphere surface area (4πr²), cylinder volume (πr²h), cube volume (a³), circle area (πr²) and volume unit converters. This formula does not compute surface area or ellipsoid volume.
Sources: Archimedes, On the Sphere and Cylinder (Heath, The Works of Archimedes); Euclid, Elements, Book XII; NIST DLMF (π); MathWorld — Sphere; Wikipedia — Sphere. The calculator is an educational tool and does not replace an engineering standard for pressure vessels.
Sphere volume: V = (4/3)πr³. At r = 10 the result is 4188.79.
4188.79
What volume at r = 10? ≈ 4188.79 — not surface 1256.64.
4.19
What volume at r = 1? ≈ 4.19 (4π/3).
523.60
What volume at r = 5? ≈ 523.60 — 1/8 of the volume at r = 10.
33.51
What volume at r = 2? ≈ 33.51.
113.10
What volume at r = 3? ≈ 113.10 — not a cube of 27.
268.08
What volume at r = 4? ≈ 268.08.
904.78
What volume at r = 6? ≈ 904.78.
2144.66
What volume at r = 8? ≈ 2144.66 — between r = 5 and r = 10.
Radius r must be greater than zero. The calculator rejects zero and negative values — they do not define a solid sphere with positive volume in this model. An empty field leaves the result blank, not zero.
The volume is 4188.79 to two decimal places. That is (4/3)·π·10³. The same figure appears in example 1 and in the press-seed paragraph. Surface area at that r is 1256.64 on a sibling page.
It is Archimedes’ classical result from On the Sphere and Cylinder: the sphere’s volume is two thirds of the volume of the cylinder circumscribed about it (radius r, height 2r). Hence V = (2/3)·πr²·2r = (4/3)πr³. The calculator evaluates that formula with π.
No. Diameter 10 means r = 5 and volume 523.60. Entering 10 as r gives 4188.79 — eight times more, because volume scales as r³. Divide the diameter by two before you enter the radius.
No. 4πr² is surface area (the skin). Volume is (4/3)πr³. At r = 10 the skin is 1256.64 and the interior is 4188.79. Surface area is on the sibling page.
Because r is cubed. If r₂ = r₁/2, then V₂/V₁ = (1/2)³ = 1/8. At r = 10 the volume is 4188.79; at r = 5 it is 523.60. Surface area would drop by four, because of r².
Radius and result must share one length unit. With Metric the labels are cm and cm³; with US they are in and in³. The header switch changes labels only — it does not convert the numbers you entered.
No. The currency and unit settings in the page header do not affect this result. Only r and π matter.
Yes. 2.5 and 2,5 are the same radius. Spaces inside a number are ignored.
On the sphere surface area page: the formula there is 4πr². This page computes only volume (4/3)πr³.
No. The form field takes radius r. If you know the diameter, divide by two before entering the value.
No. Cube volume a³ is on a separate page. A sphere at r = 10 gives 4188.79; a cube of edge 10 gives 1000 — a different solid and a different formula.
Cube the radius, multiply by π, then by 4/3. At r = 10: 10³ = 1000, V ≈ 4188.79. At r = 3: 27 × 4π/3 ≈ 113.10. At r = 5 the volume is ≈ 523.60.
First r = d / 2, then V = (4/3)πr³. Diameter 10 means r = 5 and volume ≈ 523.60. On this page enter r already — do not put d in the radius field.
Volume V = (4/3)πr³ is the interior. Surface area 4πr² is the skin. At r = 10 volume ≈ 4188.79 and surface ≈ 1256.64 — a sibling page computes surface alone.
Because the formula uses r³. At r = 5 volume ≈ 523.60; at r = 10 ≈ 4188.79 — about eight times larger. Surface area grows only with r² (fourfold).
In theory yes: r = ∛(3V / (4π)). This page goes the other way — from r to V. If you have V and need r, rearrange on paper or use a tool that accepts volume as input.
The formula, derivation and historical context below come from established geometry sources.
Page Updated November 2, 2026.