Volume of a cylinder with a hemisphere on top
The volume of a cylinder with a hemisphere on top is the cylinder plus half a sphere of the same radius: V = πr²h + ⅔πr³. With r = 3 in and a straight part h = 8 in, that is 226.19 + 56.55 ≈ 282.74 in³, and the solid stands 11 in tall. A capsule with a hemisphere on each end adds a whole sphere instead: V = πr²h + (4 ÷ 3)πr³.
Contents
V = π × r² × h + (2/3) × π × r³. With r = 3 in and h = 8 in: 226.19 + 56.55 = 282.74 in³; overall height 11 in. Two hemispheres: 339.29 in³. Cone 4 in tall on top: 263.89 in³.
The calculator above handles a hemisphere on one end, hemispherical ends on both sides (a capsule) or a cone on top, and shows the overall height.
Composite solids: add the parts
A hemisphere on top of a cylinder is a composite solid. The flat face of the hemisphere sits exactly on the circular top of the cylinder, so the two parts share the radius r and do not overlap. The volume is the sum of the two volumes:
| Part | Volume | r = 3 in, h = 8 in |
|---|---|---|
| Cylinder | πr²h | 226.19 in³ |
| Hemisphere | ⅔πr³ | 56.55 in³ |
| Sphere (two hemispheres) | (4 ÷ 3)πr³ | 113.10 in³ |
| Cone, height k = 4 in | ⅓πr²k | 37.70 in³ |
The cylinder part is the usual πr²h; if you need a refresher, see how to find the volume of a cylinder. The hemisphere is half of the sphere volume (4 ÷ 3)πr³. How the cylinder, the cone and the sphere compare when they share the same radius is the subject of the page on cylinder, cone and sphere volumes.
The three common shapes
| Shape | Volume | Overall height | r = 3, h = 8 (in) |
|---|---|---|---|
| Hemisphere on one end | πr²h + ⅔πr³ | h + r | 282.74 in³, 11 in |
| Hemispherical ends (capsule) | πr²h + (4 ÷ 3)πr³ | h + 2r | 339.29 in³, 14 in |
| Cone on top (height k = 4) | πr²h + ⅓πr²k | h + k | 263.89 in³, 12 in |
A handy factorization: πr²h + ⅔πr³ = πr²(h + ⅔r). A hemisphere on top adds as much volume as two-thirds of a radius of extra cylinder. In the example, 8 in of cylinder plus the dome hold exactly as much as a plain cylinder 10 in tall.
How to find the volume step by step
- Identify r and the straight height hIf you measured the total height H, subtract the dome: h = H − r (one hemisphere) or H − 2r (two).
- Cylinderπ × r² × h.
- Dome⅔ × π × r³ for one hemisphere, (4 ÷ 3) × π × r³ for two.
- AddThen convert: in³ ÷ 231 for US gallons, ft³ × 7.48052 for US gallons, cm³ ÷ 1,000 for liters.
Example 1: a silo with a domed roof (feet)
A round silo has a radius of 6 ft, a straight wall 30 ft high and a hemispherical roof. Cylinder: π × 36 × 30 ≈ 3,392.92 ft³. Dome: ⅔ × π × 216 ≈ 452.39 ft³. Total ≈ 3,845.31 ft³, and the silo stands 36 ft tall.
Example 2: a capsule tank (centimeters)
A tank with hemispherical ends has a radius of 50 cm and a straight section 200 cm long. Cylinder: π × 2,500 × 200 ≈ 1,570,796 cm³. Two ends = one sphere: (4 ÷ 3) × π × 125,000 ≈ 523,599 cm³. Total ≈ 2,094,395 cm³ ≈ 2,094 L. Overall length 300 cm.
Example 3: a cone on top (meters)
A hopper is a cylinder of radius 1 m and height 3 m with a cone 1.5 m tall on top. Cylinder: π × 1 × 3 ≈ 9.425 m³. Cone: ⅓ × π × 1 × 1.5 ≈ 1.571 m³. Total ≈ 10.996 m³.
Example 4: working backwards
A container with a hemisphere on top must hold 500 in³ with r = 3 in. Subtract the dome, 56.55 in³, then divide by the base area π × 9 ≈ 28.27 in²: h = (500 − 56.55) ÷ 28.27 ≈ 15.68 in of straight wall, for a total height of about 18.68 in.
Partially filled cylinder with hemispherical ends
Lying horizontally. At liquid depth y, the straight part holds the circular segment area times its length L, and the two ends together hold a spherical cap of height y:
V(y) = L × [r² arccos((r − y) ÷ r) − (r − y)√(2ry − y²)] + πy²(3r − y) ÷ 3
With r = 2 ft, L = 10 ft and y = 1 ft: the segment is about 2.4567 ft², so the straight part holds 24.567 ft³; the cap holds π × 1 × 5 ÷ 3 ≈ 5.236 ft³. Total ≈ 29.80 ft³, about 223 US gallons, which is 18.7 % of the full 159.17 ft³ (1,191 gal). The segment part alone is explained on the page about the volume of a horizontal cylinder.
Standing upright with the dome at the bottom: below the dome's top (y ≤ r), the liquid is a cap, πy²(3r − y) ÷ 3. Above it, add the full hemisphere and a cylinder slice: ⅔πr³ + πr²(y − r). For r = 2 ft, a depth of 1 ft holds about 5.24 ft³ (39.2 gal) and a depth of 5 ft holds about 54.45 ft³ (407.3 gal). More tank shapes are on the cylinder tank volume page, and shallower domed heads shaped like half an ellipsoid are handled with the elliptical cylinder volume formulas.
Surface area
The curved part of a hemisphere has area 2πr². The shared circle disappears inside the solid, so it is not counted.
- Hemisphere on top, flat bottom: 3πr² + 2πrh. For r = 3 in, h = 8 in: about 235.62 in².
- Capsule (two hemispheres): 4πr² + 2πrh. Same sizes: about 263.89 in².
- Cone on top, flat bottom: πr² + 2πrh + πr√(r² + k²). For k = 4 in, the slant is 5 in: about 226.19 in².
Common mistakes
- Using the total height as h. The dome is then counted twice. Subtract r (or 2r) first.
- Using the full sphere for one hemisphere. One dome is ⅔πr³, not (4 ÷ 3)πr³.
- Using the diameter in r³. The error is eight times on the dome, four times on the cylinder.
- Adding the shared circle to the surface area. It is inside the solid.
Shortcut: a hemisphere on top of a cylinder is worth ⅔r of extra cylinder height.
To check the cylinder part on its own, use the cylinder volume calculator.
FAQ
What is the formula for the volume of a cylinder with a hemispherical top?
V = πr²h + ⅔πr³, where r is the shared radius and h the height of the straight cylinder only. It can also be written V = πr²(h + ⅔r).
How do you find the volume of a cylinder with hemispherical ends?
Add the cylinder and one whole sphere, because two hemispheres of the same radius make a sphere: V = πr²h + (4 ÷ 3)πr³. With r = 3 in and a straight part of 8 in, V ≈ 226.19 + 113.10 = 339.29 in³.
Is the height in the formula the total height?
No. h is the straight cylindrical part. With a hemisphere on top, the total height is h + r; with hemispheres on both ends, it is h + 2r. If you only know the total height H, use h = H − r or h = H − 2r first.
How do you find the volume of a partially filled cylinder with hemispherical ends?
For a horizontal tank filled to depth y, add the liquid in the straight part, L × [r² arccos((r − y) ÷ r) − (r − y)√(2ry − y²)], and the liquid in the two ends, which together form a spherical cap: πy²(3r − y) ÷ 3.
What is the surface area of a cylinder with a hemisphere on top?
With a flat bottom: πr² (bottom) + 2πrh (side) + 2πr² (dome) = 3πr² + 2πrh. For r = 3 in and h = 8 in, that is about 235.62 in². A capsule with two domes has 4πr² + 2πrh.
Related
- How to find the volumeHow to find the volume of a cylinder: square the radius, multiply by π and the height. Worked examples in inches, feet, cm and m, plus common mistakes.
- Cylinder, cone, sphereA cone holds ⅓ of a cylinder with the same base and height; a sphere holds ⅔ of its circumscribed cylinder. Formulas, tables, surface areas, word problems.
- Cylinder tank volumeTank volume calculator for vertical and horizontal cylinder tanks, full or partly filled. Get the capacity in gallons and liters, with worked examples.
Updated October 12, 2026