How this instrument works
A hemisphere's curved surface is exactly half of a full sphere's: 2πr² next to the sphere's 4πr². Slice a ball straight through its center and each half carries an equal share of the outer skin, since the two caps are congruent by symmetry — no calculus is needed to see why the factor of two survives from the sphere formula down to a single dome.
Archimedes reached that sphere formula by comparison rather than by summing infinitesimal strips. Wrap a sphere in the tightest cylinder that touches it at the equator — radius r, height 2r — and that cylinder's own curved wall works out to 2πr times 2r, or 4πr², matching the sphere exactly. Halve the sphere and the matching curved area halves too, which is where the plain 2πr² for one dome comes from.
Add a flat base and the total climbs to 3πr², because a hemisphere is a genuinely different solid from an open dome shell — a mixing bowl has no base, a dome-shaped roof or a solid paperweight does. At r = 0 both figures collapse to zero, and the ratio between total and curved stays fixed at 3 to 2 for any radius at all, since both scale together with r².
- Enter the hemisphere's radius into the Radius field, in any length unit you like.
- Read Curved surface area for the domed cap alone — 2πr² in that unit, squared.
- Read Total surface area (curved + base) once the flat circular base is folded in — 3πr².
- Subtract the two outputs to isolate the base on its own: the gap always equals πr².
- Keep one unit throughout — both outputs inherit whatever length unit you entered the radius in, squared.
Worked example — a radius-5 dome
Picture a hemispherical shell fabrication job: a radius-5 dome needs a metal skin over just its curved surface. A_curved = 2π(5)² = 2π × 25 = 50π ≈ 157.079633 square units — the curved area alone, the figure you would send to a supplier quoting only the dome itself, with no base included.
Add the flat base disk the dome sits on and the reading changes: A_total = 3π(5)² = 3π × 25 = 75π ≈ 235.619449 square units. The difference between the two figures, 235.619449 minus 157.079633, equals 78.539816 — exactly π × 5², the base circle's own area, recovered here as a built-in check on both numbers at once.
Questions
Why is a hemisphere's curved area exactly half a sphere's 4πr²?
Because a plane through a sphere's center splits it into two congruent domes, and each carries an equal share of the outer skin by symmetry. Half of 4πr² is 2πr², the curved-surface formula this sheet uses — no separate derivation is needed once the sphere is accepted as cut evenly in two.
Where does the extra πr² in the total-area formula come from?
From the flat circular base a solid hemisphere sits on — an actual disk of radius r, with the ordinary circle area πr². Add that to the curved 2πr² and the total is 3πr². Leave the base out, as you would for an open bowl with no bottom, and 2πr² is the figure you want instead.
How does Archimedes' cylinder result explain the 2π in 2πr²?
Archimedes showed a sphere's curved surface equals the lateral surface of the tightest cylinder wrapped around it: circumference 2πr times height 2r gives 4πr², matching the sphere exactly. Split the sphere in half and the matching curved area splits too, leaving 2πr² for one dome — the constant falls out of that cylinder comparison rather than being assumed.
What's the most common mistake when computing hemisphere surface area?
Reporting 4πr² by mistake — the formula for a whole sphere, not a half — or the opposite error of forgetting the base disk and calling 2πr² the 'total' area when a base is actually present. The two separate outputs on this sheet exist precisely so the right one gets read for whichever object is in hand.
Does a mixing bowl use the curved figure or the total figure?
The curved figure, 2πr² — an open bowl or shell has no flat base disk at all, so its whole surface is the dome. Reach for the total, 3πr², only when the object is genuinely closed off flat on one side, like a solid paperweight or a dome resting on a foundation slab.
How does surface area relate to a hemisphere's volume?
They are separate quantities that scale with the radius at different rates: volume runs (2/3)πr³, growing with the cube of r, while both surface areas here grow with the square. Doubling the radius scales either area by four and the volume by eight, so area and volume are never proportional to one another.