Volume

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A measuring cup can be used to measure volumes of liquids. This cup measures volume in units of cups, fluid ounces, and millilitres.

Volume is the quantity of three-dimensional space enclosed by some closed boundary, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains.[1] Volume is often quantified numerically using the SI derived unit, the cubic metre. The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.

Three dimensional mathematical shapes are also assigned volumes. Volumes of some simple shapes, such as regular, straight-edged, and circular shapes can be easily calculated using arithmetic formulas. The volumes of more complicated shapes can be calculated by integral calculus if a formula exists for the shape's boundary. One-dimensional figures (such as lines) and two-dimensional shapes (such as squares) are assigned zero volume in the three-dimensional space.

The volume of a solid (whether regularly or irregularly shaped) can be determined by fluid displacement. Displacement of liquid can also be used to determine the volume of a gas. The combined volume of two substances is usually greater than the volume of one of the substances. However, sometimes one substance dissolves in the other and the combined volume is not additive.[2]

In differential geometry, volume is expressed by means of the volume form, and is an important global Riemannian invariant. In thermodynamics, volume is a fundamental parameter, and is a conjugate variable to pressure.

Units[edit]

Volume measurements from the 1914 The New Student's Reference Work.
Approximate conversion to millilitres:[3]
Imperial U.S. liquid U.S. dry
Gill 142 ml 118 ml 138 ml
Pint 568 ml 473 ml 551 ml
Quart 1137 ml 946 ml 1101 ml
Gallon 4546 ml 3785 ml 4405 ml

Any unit of length gives a corresponding unit of volume, namely the volume of a cube whose side has the given length. For example, a cubic centimetre (cm3) would be the volume of a cube whose sides are one centimetre (1 cm) in length.

In the International System of Units (SI), the standard unit of volume is the cubic metre (m3). The metric system also includes the litre (L) as a unit of volume, where one litre is the volume of a 10-centimetre cube. Thus

1 litre = (10 cm)3 = 1000 cubic centimetres = 0.001 cubic metres,



so

1 cubic metre = 1000 litres.

Small amounts of liquid are often measured in millilitres, where

1 millilitre = 0.001 litres = 1 cubic centimetre.

Various other traditional units of volume are also in use, including the cubic inch, the cubic foot, the cubic mile, the teaspoon, the tablespoon, the fluid ounce, the fluid dram, the gill, the pint, the quart, the gallon, the minim, the barrel, the cord, the peck, the bushel, and the hogshead.

Related terms[edit]

Volume and capacity are the 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 capacity management, where capacity is defined as volume over a specified time period. However in this context the term volume may be more loosely interpreted to mean quantity.

The density of an object is defined as mass per unit volume. The inverse of density is specific volume which is defined as volume divided by mass. Specific volume is a concept important in thermodynamics where the volume of a working fluid is often an important parameter of a system being studied.

The volumetric flow rate in fluid dynamics is the volume of fluid which passes through a given surface per unit time (for example cubic meters per second [m3 s−1]).

Volume formulas[edit]

Shape Volume formula Variables
Cube a^3\; a = length of any side (or edge)
Cylinder \pi r^2 h\; r = radius of circular face, h = height
Prism B \cdot h B = area of the base, h = height
Rectangular prism l \cdot w \cdot h l = length, w = width, h = height
Triangular prism \frac{1}{2}bhl b = base length of triangle, h = height of triangle, l = length of prism or distance between the triangular bases
Sphere \frac{4}{3} \pi r^3 r = radius of sphere
which is the integral of the surface area of a sphere
Oval \frac{4}{3} \pi abc a, b, c = semi-axes of ellipsoid
Torus (\pi r^2)(2\pi R) = 2\pi^2 Rr^2 r = minor radius, R = major radius
Pyramid \frac{1}{3}Bh B = area of the base, h = height of pyramid
Square pyramid \frac{1}{3} s^2 h\; s = side length of base, h = height
Rectangular pyramid \frac{1}{3} lwh l = length, w = width, h = height
Cone \frac{1}{3} \pi r^2 h r = radius of circle at base, h = distance from base to tip or height
Tetrahedron[4] {\sqrt{2}\over12}a^3 \, edge length a
Parallelepiped 
a b c  \sqrt{K}



\begin{align}
 K =& 1+2\cos(\alpha)\cos(\beta)\cos(\gamma) \\
 & - \cos^2(\alpha)-\cos^2(\beta)-\cos^2(\gamma)
\end{align}

a, b, and c are the parallelepiped edge lengths, and α, β, and γ are the internal angles between the edges
Any volumetric sweep
(calculus required)
\int_a^b A(h) \,\mathrm{d}h 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. a and b are the limits of integration for the volumetric sweep.
(This will work for any figure if its cross-sectional area can be determined from h).
Any rotated figure (washer method)
(calculus required)
\pi \int_a^b \left({\left[R_O(x)\right]}^2 - {\left[R_I(x)\right]}^2\right) \mathrm{d}x R_O and R_I are functions expressing the outer and inner radii of the function, respectively.

Volume ratios for a cone, sphere and cylinder of the same radius and height[edit]

A cone, sphere and cylinder of radius r and height h

The above formulas can be used to show that the volumes of a cone, sphere and cylinder of the same radius and height are in the ratio 1 : 2 : 3, as follows.

Let the radius be r and the height be h (which is 2r for the sphere), then the volume of cone is

\tfrac{1}{3} \pi r^2 h = \tfrac{1}{3} \pi r^2 (2r) = (\tfrac{2}{3} \pi r^3) \times 1,

the volume of the sphere is

\tfrac{4}{3} \pi r^3 = (\tfrac{2}{3} \pi r^3) \times 2,

while the volume of the cylinder is

\pi r^2 h = \pi r^2 (2r) = (\tfrac{2}{3} \pi r^3) \times 3.

The discovery of the 2 : 3 ratio of the volumes of the sphere and cylinder is credited to Archimedes.[5]

Volume formula derivations[edit]

Sphere[edit]

The volume of a sphere is the integral of an infinite number of infinitesimally small circular disks of thickness dx. The calculation for the volume of a sphere with center 0 and radius r is as follows.

The surface area of the circular disk is \pi r^2 .

The radius of the circular disks, defined such that the x-axis cuts perpendicularly through them, is;

y = \sqrt{r^2-x^2}

or

z = \sqrt{r^2-x^2}

where y or z can be taken to represent the radius of a disk at a particular x value.

Using y as the disk radius, the volume of the sphere can be calculated as  \int_{-r}^r \pi y^2 \,dx = \int_{-r}^r \pi(r^2 - x^2) \,dx.

Now \int_{-r}^r \pi r^2\,dx - \int_{-r}^r \pi x^2\,dx = \pi (r^3 + r^3) - \frac{\pi}{3}(r^3 + r^3) =  2\pi r^3 - \frac{2\pi r^3}{3}.

Combining yields V = \frac{4}{3}\pi r^3.

This formula can be derived more quickly using the formula for the sphere's surface area, which is 4\pi r^2. The volume of the sphere consists of layers of infinitesimally thin spherical shells, and the sphere volume is equal to

 \int_0^r 4\pi u^2 \,du =   \frac{4}{3}\pi r^3.

Cone[edit]

The cone is a type of pyramidal shape. The fundamental equation for pyramids, one-third times base times altitude, applies to cones as well.

However, using calculus, the volume of a cone is the integral of an infinite number of infinitesimally small circular disks of thickness dx. The calculation for the volume of a cone of height h, whose base is centered at (0,0,0) with radius r, is as follows.

The radius of each circular disk is r if x = 0 and 0 if x = h, and varying linearly in between—that is, r\frac{(h-x)}{h}.

The surface area of the circular disk is then  \pi \left(r\frac{(h-x)}{h}\right)^2 =  \pi r^2\frac{(h-x)^2}{h^2}.

The volume of the cone can then be calculated as  \int_{0}^h \pi r^2\frac{(h-x)^2}{h^2} dx,

and after extraction of the constants: \frac{\pi r^2}{h^2} \int_{0}^h (h-x)^2 dx

Integrating gives us \frac{\pi r^2}{h^2}\left(\frac{h^3}{3}\right) = \frac{1}{3}\pi r^2 h.

See also[edit]

References[edit]

  1. ^ "Your Dictionary entry for "volume"". Retrieved 2010-05-01. 
  2. ^ One litre of sugar (about 970 grams) can dissolve in 0.6 litres of hot water, producing a total volume of less than one litre. "Solubility". Retrieved 2010-05-01. "Up to 1800 grams of sucrose can dissolve in a liter of water." 
  3. ^ "General Tables of Units of Measurement". NIST Weights and Measures Division. Retrieved 2011-01-12. 
  4. ^ Coxeter, H. S. M.: Regular Polytopes (Methuen and Co., 1948). Table I(i).
  5. ^ Rorres, Chris. "Tomb of Archimedes: Sources". Courant Institute of Mathematical Sciences. Retrieved 2007-01-02. 

External links[edit]