Elliptic cylinder volume
An elliptic cylinder is a cylinder with an oval cross-section, and its volume is V = πabh: π times the two semi-axes a and b of the oval, times the length h. An oval tank 6 ft wide, 4 ft tall and 10 ft long has a = 3 ft and b = 2 ft, so it holds π × 3 × 2 × 10 ≈ 188.50 ft³, about 1,410 US gallons.
Contents
V = π × a × b × h. With a full width of 6 in (a = 3 in), a depth of 4 in (b = 2 in) and a length of 10 in: π × 3 × 2 × 10 = 188.50 in³ (0.816 US gal).
The calculator above accepts the full width and depth of the oval or the two semi-axes, and converts the elliptical cylinder volume to gallons and liters.
What is an elliptic cylinder?
Take an ellipse and push it straight along a line perpendicular to its plane: the solid you sweep is an elliptic (or elliptical) cylinder. Every slice cut square to the axis is the same ellipse. Oval fuel tanks lying under trucks, oval ducts, oval cake tins, and many tablets and soaps have this shape.
An ellipse has two axes. The major axis is its full width, the minor axis its full depth. The semi-axes a and b are half of each. When a = b the ellipse is a circle and the elliptic cylinder becomes an ordinary round cylinder of radius r = a.
Elliptical cylinder volume formula
The area of an ellipse is πab. A circle is the special case a = b = r, which gives the familiar πr². Since every slice is the same ellipse, the volume is that area times the length:
V = π × a × b × h or, with full width W and depth D, V = π × W × D × h ÷ 4.
This is the same "base area × length" rule used in how to find the volume of a cylinder. Formally, with the axis along z, the solid is the set of points with x²/a² + y²/b² ≤ 1 and 0 ≤ z ≤ h; the page on the equation of a cylinder covers the surface x²/a² + y²/b² = 1.
How to calculate the volume of an ellipse cylinder
- Measure the ovalFull width at the widest point, full depth at the deepest point, square to each other. Use inside dimensions for capacity.
- Halve botha = width ÷ 2, b = depth ÷ 2.
- Multiplyπ × a × b × length, all in the same unit.
- Convertft³ × 7.48052 for US gallons, in³ ÷ 231 for US gallons, cm³ ÷ 1,000 for liters.
Example 1: oval tank in feet
Width 6 ft, height 4 ft, length 10 ft. a = 3 ft, b = 2 ft. V = π × 3 × 2 × 10 ≈ 188.50 ft³. In gallons: 188.50 × 7.48052 ≈ 1,410 US gal.
Example 2: oval tank in centimeters
Inside width 120 cm, inside height 80 cm, length 200 cm. a = 60 cm, b = 40 cm. V = π × 60 × 40 × 200 ≈ 1,507,964 cm³ ≈ 1.508 m³, or about 1,508 L.
Example 3: a small oval container in inches
6 in wide, 4 in deep, 10 in tall: V = π × 3 × 2 × 10 ≈ 188.50 in³, which is 188.50 ÷ 231 ≈ 0.816 US gal.
| Oval (width × depth) | Length | Volume | Capacity |
|---|---|---|---|
| 6 in × 4 in | 10 in | 188.50 in³ | 0.816 US gal |
| 6 ft × 4 ft | 10 ft | 188.50 ft³ | 1,410 US gal |
| 120 cm × 80 cm | 200 cm | 1.508 m³ | 1,508 L |
An oval of 6 × 4 has the same area as a circle of radius √(3 × 2) ≈ 2.449. That equivalent radius lets you reuse any round-cylinder tool, including the cylinder volume calculator.
Partially filled elliptical tank
Oval tanks usually lie flat, wider than tall. The trick is to see the ellipse as a circle of radius b stretched sideways by a ÷ b. A stretch multiplies every area by the same factor, so the wet area at depth y is the circular-segment area times a ÷ b:
A(y) = (a ÷ b) × [b² arccos((b − y) ÷ b) − (b − y)√(2by − y²)], and V = A(y) × length.
Here b is the vertical semi-axis and arccos is in radians. For the 6 ft × 4 ft × 10 ft tank (a = 3, b = 2):
| Liquid depth | Fraction full | Volume | US gallons |
|---|---|---|---|
| 0.5 ft | 7.2 % | 13.60 ft³ | 101.7 |
| 1 ft | 19.6 % | 36.85 ft³ | 275.7 |
| 2 ft | 50 % | 94.25 ft³ | 705.0 |
| 3 ft | 80.4 % | 151.64 ft³ | 1,134.4 |
| 4 ft | 100 % | 188.50 ft³ | 1,410.0 |
The fraction full depends only on y ÷ (2b), the same as for a round tank. The round case is worked in detail on the page about the volume of a horizontal cylinder, and the full tank method on the cylinder tank volume page.
Tanks with ellipsoidal ends
Some round tanks have domed ends shaped like half an ellipsoid instead of flat ends. Each such head, with the tank radius r and a dome depth c, holds ⅔πr²c. Two heads hold (4 ÷ 3)πr²c. A tank of radius 2 ft with a 10 ft straight section and two 0.5 ft deep heads holds π × 4 × 10 + (4 ÷ 3) × π × 4 × 0.5 ≈ 125.66 + 8.38 = 134.04 ft³, about 1,003 US gallons. Hemispherical ends are the case c = r, treated on the cylinder with a hemisphere page.
Surface area: why the perimeter is only approximate
The volume of an elliptic cylinder is exact and simple. Its side area is not: it equals the perimeter of the ellipse times the length, and the perimeter of an ellipse has no exact formula in elementary functions. A good approximation, due to Ramanujan, is:
P ≈ π × [3(a + b) − √((3a + b)(a + 3b))]
For a = 3 and b = 2 it gives about 15.865, which matches a numerical computation to four decimals. The side of the 10 ft tank is then about 158.65 ft², plus two oval ends of π × 3 × 2 ≈ 18.85 ft² each. Treat any perimeter or side area as an approximation; the volume πabh is exact.
Common mistakes
- Using full width and depth as a and b. The volume comes out four times too big: 754 ft³ instead of 188.5 ft³ for the example tank.
- Averaging the axes into a radius. Using r = (a + b) ÷ 2 gives π × 2.5² = 19.63 instead of 18.85 for the oval area: about 4 % too high here, more for flatter ovals. Use √(ab) if you need an equivalent radius.
- Measuring outside dimensions for capacity. The wall thickness reduces both axes.
- Treating a racetrack shape as an ellipse. A rectangle with two half circles on its ends is not an ellipse; its area is the rectangle plus one full circle.
Check your answer with a = b: the formula must give the round cylinder πr²h.
FAQ
What is the formula for the volume of an elliptical cylinder?
V = πabh, where a and b are the semi-axes of the oval (half the width and half the depth) and h is the length. The oval area πab is multiplied by the length, just as πr² is for a round cylinder.
How do I find the volume of an oval cylinder from its width and height?
Halve both: a = width ÷ 2, b = depth ÷ 2. Then V = π × a × b × length. An oval tank 6 ft wide, 4 ft tall and 10 ft long holds π × 3 × 2 × 10 ≈ 188.50 ft³, about 1,410 US gallons.
Is an elliptic cylinder the same as an oblique cylinder?
No. An elliptic cylinder has oval cross-sections square to its axis. An oblique circular cylinder has circular bases and a tilted side. Slicing an oblique cylinder square to its axis does give an ellipse, which is why the two are often confused.
What is the equation of an elliptic cylinder?
With the axis along z, x²/a² + y²/b² = 1. The variable z is free, so the oval is repeated along the whole axis. When a = b the equation becomes a circular cylinder of radius a.
How do I find the volume of a partially filled elliptical tank?
For a horizontal oval tank with vertical semi-axis b, horizontal semi-axis a and liquid depth y, the wet area is (a ÷ b) × [b² arccos((b − y) ÷ b) − (b − y)√(2by − y²)]. Multiply by the length. At half depth the tank is exactly half full.
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 tank volumeTank volume calculator for vertical and horizontal cylinder tanks, full or partly filled. Get the capacity in gallons and liters, with worked examples.
- Horizontal cylinderHorizontal tank volume calculator for a partly filled cylinder: segment formula, depth to volume table, and a worked 48 in × 96 in example in gallons.
Updated October 12, 2026