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Instrument MI-01-673 · Mathematics

Volume of a Hemisphere Calculator

Slice a sphere exactly through its center, and each half is a hemisphere. Enter the radius, and this sheet returns that half's volume.

Instrument MI-01-673
Sheet 1 OF 1
Rev A
Verified
Type 05 — Geometry SER. 2026-01673

Volume

261.79938780

V = (2⁄3)πr³

The working Every figure verified twice
  1. volume = 2·π·5^3 ⁄ 3 = 261.79938780
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

A hemisphere is exactly half of a sphere, produced by slicing straight through the sphere's own center with a flat plane. Because that slice divides the sphere into two identical, mirror-image halves, a hemisphere's volume is simply half of a full sphere's volume: V = (2 ⁄ 3)πr³, exactly half of the familiar (4 ⁄ 3)πr³ formula for a complete sphere.

Like a full sphere, a hemisphere's volume scales with the CUBE of its radius, not with the radius directly — doubling the radius doesn't merely double the volume, it multiplies it by eight (2³). This cubic growth is easy to underestimate: a hemisphere twice as wide looks only modestly larger by eye, but holds eight times as much volume, a distinction that matters directly for anything sized by capacity, from a mixing bowl to a storage tank.

Hemispherical shapes appear often in engineering and architecture specifically because they enclose the maximum volume for a given amount of surface material among all possible dome shapes — the same efficiency that makes a full sphere the most volume-efficient enclosed shape overall, halved. Domes, bowl-shaped tanks, and some pressure-vessel end caps are all built as hemispheres for exactly this reason.

V=23πr3V = \frac{2}{3}\pi r^3
r — the hemisphere's radius; V — the resulting volume, exactly half of a full sphere with the same radius.
  • Enter the hemisphere's radius into the Radius field.
  • Read Volume: the sheet applies (2 ⁄ 3)πr³ directly.
  • For the equivalent full sphere's volume, simply double the result — the relationship is always exact.

Worked example — a hemisphere of radius 5

A hemisphere has a radius of 5 units. Its volume is V = (2 ⁄ 3) × π × 5³ = (2 ⁄ 3) × π × 125 ≈ 261.80 cubic units — exactly half of the roughly 523.60 cubic units a full sphere of the same radius would hold.

Compare a smaller hemisphere with radius 3: V = (2 ⁄ 3) × π × 27 ≈ 56.55 cubic units. Even though the radius only shrank from 5 to 3, a drop of less than half, the volume fell to roughly a fifth of the original — the cube in the formula amplifying any change in radius far more than a direct, linear relationship would.

Questions

What is the formula for the volume of a hemisphere?

V = (2 ⁄ 3)πr³, where r is the radius. This is exactly half of a full sphere's volume formula, (4 ⁄ 3)πr³, since a hemisphere is simply a sphere sliced precisely through its own center.

How is hemisphere volume different from hemisphere surface area?

They're separate figures for separate purposes: volume (this page) measures the space enclosed inside, while surface area measures the total exterior — the curved dome plus the flat circular base — needed for coating, painting, or material estimates. The two use entirely different formulas.

Does doubling the radius double the volume?

No — volume scales with the CUBE of the radius, so doubling the radius multiplies the volume by 2³ = 8, not by 2. This cubic growth applies identically to a hemisphere and to a full sphere, since a hemisphere's volume is always a fixed fraction of the matching sphere's.

Why are hemispheres common in tank and dome design?

Because a sphere (and therefore a hemisphere, exactly half of one) encloses the most volume for a given amount of surface material among all possible closed shapes — a genuine efficiency advantage that makes hemispherical domes and bowl-shaped tanks a practical choice wherever minimizing material for a given capacity matters.

What is the volume of a hemisphere with radius 0?

Exactly 0 — the hemisphere has collapsed to a single point with no space enclosed at all, the degenerate limit the formula handles cleanly without any special case needed.

References