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The total 3D space enclosed within the sphere.
USING FORMULA:
V = (4/3) · π · r³
V = 4/3 * π * r³.Why is the coefficient $4/3$? This specific number comes from the method of integration, often called the "Disk Method" in calculus.
Archimedes famously proved in his work "The Sphere and the Cylinder" that a sphere occupies exactly two-thirds of the volume of the smallest cylinder that can contain it. If a cylinder has a height of $2r$ and a radius of $r$, its volume is $V_{cyl} = \pi r^2 (2r) = 2\pi r^3$.
Taking two-thirds of that cylinder's volume gives us:
$V_{sphere} = \frac{2}{3} \times 2\pi r^3 = \frac{4}{3}\pi r^3$
| Object | Approx. Size | Context |
|---|---|---|
| Planet Earth | Radius ~6,371 km | Slightly oblate, but treated as a sphere for general calc. |
| Basketball (Size 7) | Diameter ~24 cm | Volume is approx 7.2 liters of air. |
| Soap Bubble | Varies | Forms a sphere to minimize surface area (tension). |
The formula is V = (π * d³) / 6. This is derived by substituting radius ($d/2$) into the standard formula.
Since the radius is cubed ($r^3$), doubling the radius increases the volume by a factor of 8 ($2^3 = 8$). A small increase in width creates a massive increase in volume.