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Volume

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The volume of a solid object is the three-dimensional concept of how much space it occupies, often quantified numerically. One-dimensional figures (such as lines) and two-dimensional shapes (such as squares) are assigned zero volume in the three-dimensional space.

Volumes of straight-edged and circular shapes are calculated using arithmetic formulae. Volumes of other curved shapes are calculated using integral calculus, by approximating the given body with a large amount of small cubes or concentric cylindrical shells, and adding the individual volumes of those shapes. The volume of irregularly shaped objects can be determined by displacement. If an irregularly shaped object is less dense than the fluid, you will need a weight to attach to the floating object. A sufficient weight will cause the object to sink. The final volume of the unknown object can be found by subtracting the volume of the attached heavy object and the total fluid volume displaced.

The generalization of volume to arbitrarily many dimensions is called content.^{[citation needed]} In differential geometry, volume is expressed by means of the volume form.

Volume and Capacity are sometimes distinguished, with capacity being used for how much a container can hold (with contents measured commonly in litres or its derived units), and volume being how much space an object displaces (commonly measured in cubic metres or its derived units).

Volume and capacity are also distinguished in a capacity management setting, where capacity is defined as volume over a specified time period.

Volume is a fundamental parameter in thermodynamics and it is conjugate to pressure.

Conjugate variables of thermodynamics
Pressure	Volume
(Stress)	(Strain)
Temperature	Entropy
Chem. potential	Particle no.

1 Volume formulas
2 Volume measures: USA
3 Volume measures: UK
4 Volume measures: cooking
5 Relationship to density
6 Volume formula derivation
7 See also
8 External links

Common equations for volume:
Shape	Equation	Variables
A cube:	$s^3 = s \cdot s \cdot s$	s = length of a side
A rectangular prism:	$l \cdot w \cdot h$	l = length, w = width, h = height
A cylinder (circular prism):	$\pi r^2 \cdot h$	r = radius of circular face, h = height
Any prism that has a constant cross sectional area along the height**:	$A \cdot h$	A = area of the base, h = height
A sphere:	$\frac{4}{3} \pi r^3$	r = radius of sphere which is the integral of the Surface Area of a sphere
An ellipsoid:	$\frac{4}{3} \pi abc$	a, b, c = semi-axes of ellipsoid
A pyramid:	$\frac{1}{3} lwh$	l = length, w = width, h = height
A cone (circular-based pyramid):	$\frac{1}{3} \pi r^2 h$	r = radius of circle at base, h = distance from base to tip
Any figure (calculus required)	$\int A(h) \,dh$	h = any dimension of the figure, A(h) = area of the cross-sections perpendicular to h described as a function of the position along h. This will work for any figure if its cross-sectional area can be determined from h (no matter if the prism is slanted or the cross-sections change shape).

(The units of volume depend on the units of length - if the lengths are in meters, the volume will be in cubic meters, etc)

The volume of a parallelepiped is the absolute value of the scalar triple product of the subtending vectors, or equivalently the absolute value of the determinant of the corresponding matrix.

The volume of any tetrahedron, given its vertices a, b, c and d, is (1/6)·|det(a−b, b−c, c−d)|, or any other combination of pairs of vertices that form a simply connected graph.

U.S. fluid ounce, about 29.6 mL
U.S. liquid pint = 16 fluid ounces, or about 473 mL
U.S. dry pint = 1/64 U.S. bushel, or about 551 mL (used for things such as blueberries)
U.S. liquid quart = 32 fluid ounces, or about 946 mL
U.S. dry quart = 1/32 U.S. bushel, or about 1.101 L
U.S. liquid gallon = 128 fluid ounces or four U.S. quarts, about 3.785 L
U.S. dry gallon = 1/8 U.S. bushel, or about 4.405 L
U.S. (dry level) bushel = 2150.42 cubic inches, or about 35.239 L

The acre foot is often used in measuring the volume of water in a reservoir or an aquifer. It is the volume of water that would cover an area of one acre to a depth of one foot. It is equivalent to 43,560 cubic feet or exactly 1233.481 837 547 52 m³.

cubic inch = 16.387064 cm³
cubic foot = 1,728 in³ ≈ 28.317 dm³
cubic yard = 27 ft³ ≈ 0.7646 m³
cubic mile = 5,451,776,000 yd³ = 3,379,200 acre-feet ≈ 4.168 km³

The UK is undergoing metrication and is increasingly using the SI metric system's units of volume, i.e. cubic meter and litre. However, some former units of volume are still in varying degrees of usage:

Imperial units of volume:

UK fluid ounce, about 28.4 mL (this equals the volume of an avoirdupois ounce of water under certain conditions)
UK pint = 20 fluid ounces, or about 568 mL
UK quart = 40 ounces or two pints1.137 L
UK gallon = 4 quarts, or exactly 4.546 09 L

The quart is now obsolete and the fluid ounce extremely rare. The gallon is only used for transportation uses, (it is illegal for petrol and diesel to be sold by the gallon). The pint is the only Imperial unit that is in everyday use, for the sale of draught beer and cider (bottled and canned beer is mainly sold in SI units) and for milk (this too is increasingly being sold in SI units, mainly Liters).

Traditional cooking measures for volume also include:

teaspoon = 1/6 U.S. fluid ounce (about 4.929 mL)
teaspoon = 1/6 Imperial fluid ounce (about 4.736 mL)
teaspoon = 5 mL (metric)
tablespoon = ½ U.S. fluid ounce or 3 teaspoons (about 14.79 mL)
tablespoon = ½ Imperial fluid ounce or 3 teaspoons (about 14.21 mL)
tablespoon = 15 mL or 3 teaspoons (metric)
tablespoon = 5 fluidrams (about 17.76 mL) (British)
cup = 8 U.S. fluid ounces or ½ U.S. liquid pint (about 237 mL)
cup = 8 Imperial fluid ounces or ½ fluid pint (about 227 mL)
cup = 250 mL (metric)

The volume of an object is equal to its mass divided by its average density. This is a rearrangement of the calculation of density as mass per unit volume.

The term specific volume is used for volume divided by mass. This is the reciprocal of the mass density, expressed in units such as cubic meters per kilogram (m³·kg^-1).

Shape	Volume formula derivation
Sphere	The volume of a sphere is the integral of infinitesimal circular slabs of width $d x$ . The calculation for the volume of a sphere with center 0 and radius r is as follows. The radius of the circular slabs is $y = \sqrt(r^2-x^2)$ The surface of the circular slab is $\pi \cdot y^2$ The volume of the sphere can be calculated as $\int_{-r}^r \pi(r^2-x^2) \,dx$ Replacing $x$ by $\frac{x}{r}$ , so that the integral boundaries become -1 and +1, we get $\pi r^3 \int_{-1}^1 (1-x^2) \,dx$ The antiderivative needed can be determined very easily as $x-\frac{x^3}{3}$ Thus, the sphere volume amounts to V_sphere = $\pi r^3 \cdot[1-1/3-(-1+1/3)]$ = $\frac{4}{3}\pi r^3$ This formula can be derived more quickly using the formula for the sphere surface area, which is $4π r 2$ . The volume of the sphere consists of layers of infinitesimal spherical slabs, and thus the sphere volume is equal to $\int_0^r 4\pi r^2 \,dr$ = $\frac{4}{3}\pi r^3$

Shape

Volume formula derivation

Sphere

The volume of a sphere is the integral of infinitesimal circular slabs of width

d x

The calculation for the volume of a sphere with center 0 and radius r is as follows.
The radius of the circular slabs is $y = \sqrt(r^2-x^2)$
The surface of the circular slab is $\pi \cdot y^2$
The volume of the sphere can be calculated as $\int_{-r}^r \pi(r^2-x^2) \,dx$
Replacing $x$ by $\frac{x}{r}$ , so that the integral boundaries become -1 and +1, we get $\pi r^3 \int_{-1}^1 (1-x^2) \,dx$
The antiderivative needed can be determined very easily as $x-\frac{x^3}{3}$
Thus, the sphere volume amounts to V_sphere = $\pi r^3 \cdot[1-1/3-(-1+1/3)]$ = $\frac{4}{3}\pi r^3$

This formula can be derived more quickly using the formula for the sphere surface area, which is $4π r 2$ . The volume of the sphere consists of layers of infinitesimal spherical slabs, and thus the sphere volume is equal to