Volume
Contents 1 Units
2 Related terms
3
Volume
4.1
Volume
4.2.1 Sphere 4.2.2 Cone 4.2.3 Polyhedron 5
Volume
Units[edit]
Volume
Approximate conversion to metric (mL)[3] Imp. U.S. Liquid Dry Gill 142 118 138 Pint 568 473 551 Quart 1137 946 1101 Gallon 4546 3785 4405 Any unit of length gives a corresponding unit of volume: the volume of
a cube whose sides have the given length. For example, a cubic
centimetre (cm3) is the volume of a cube whose sides are one
centimetre (1 cm) in length.
In the
International System of Units
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]
Capacity is defined by the
Oxford English Dictionary
f ( x , y , z ) = 1 displaystyle f(x,y,z)=1 and is usually written as: ∭ D 1 d x d y d z . displaystyle iiint limits _ D 1,dx,dy,dz. The volume integral in cylindrical coordinates is ∭ D r d r d θ d z , displaystyle iiint limits _ D r,dr,dtheta ,dz, and the volume integral in spherical coordinates (using the convention for angles with θ displaystyle theta as the azimuth and ϕ displaystyle phi measured from the polar axis; see more on conventions) has the form ∭ D ρ 2 sin ϕ d ρ d θ d ϕ . displaystyle iiint limits _ D rho ^ 2 sin phi ,drho ,dtheta ,dphi .
Volume
Shape
Volume
Cube a 3 displaystyle a^ 3 ; a = length of any side (or edge) Circular Cylinder π r 2 h displaystyle pi r^ 2 h; r = radius of circular base, h = height Prism B h displaystyle Bh B = area of the base, h = height Cuboid l w h displaystyle lwh l = length, w = width, h = height Triangular prism 1 2 b h l displaystyle frac 1 2 bhl b = base length of triangle, h = height of triangle, l = length of prism or distance between the triangular bases
Triangular prism
1 4 h − a 4 + 2 ( a b ) 2 + 2 ( a c ) 2 − b 4 + 2 ( b c ) 2 − c 4 displaystyle frac 1 4 h sqrt a^ 4 +2(ab)^ 2 +2(ac)^ 2 b^ 4 +2(bc)^ 2 c^ 4 a, b, and c = lengths of sides h = height of the triangular prism Sphere 4 3 π r 3 = 1 6 π d 3 displaystyle frac 4 3 pi r^ 3 = frac 1 6 pi d^ 3 r = radius of sphere d = diameter of sphere which is the integral of the surface area of a sphere Ellipsoid 4 3 π a b c displaystyle frac 4 3 pi abc a, b, c = semiaxes of ellipsoid Torus ( π r 2 ) ( 2 π R ) = 2 π 2 R r 2 displaystyle left(pi r^ 2 right)left(2pi Rright)=2pi ^ 2 Rr^ 2 r = minor radius (radius of the tube), R = major radius (distance from center of tube to center of torus) Pyramid 1 3 B h displaystyle frac 1 3 Bh B = area of the base, h = height of pyramid Square pyramid 1 3 s 2 h displaystyle frac 1 3 s^ 2 h; s = side length of base, h = height
Rectangular
1 3 l w h displaystyle frac 1 3 lwh l = length, w = width, h = height Cone 1 3 π r 2 h displaystyle frac 1 3 pi r^ 2 h r = radius of circle at base, h = distance from base to tip or height Regular tetrahedron[6] 2 12 a 3 displaystyle sqrt 2 over 12 a^ 3 , Edge length, a displaystyle a Parallelepiped a b c K displaystyle abc sqrt K K = 1 + 2 cos ( α ) cos ( β ) cos ( γ ) − cos 2 ( α ) − cos 2 ( β ) − cos 2 ( γ ) displaystyle begin aligned K=1&+2cos(alpha )cos(beta )cos(gamma )\&cos ^ 2 (alpha )cos ^ 2 (beta )cos ^ 2 (gamma )end aligned 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) ∫ a b A ( h ) d h displaystyle int _ a ^ b A(h),mathrm d h h = any dimension of the figure, A(h) = area of the crosssections 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 crosssectional area can be determined from h). Any rotated figure (washer method; calculus required) π ∫ a b ( [ R O ( x ) ] 2 − [ R I ( x ) ] 2 ) d x displaystyle pi int _ a ^ b left( left[R_ O (x)right] ^ 2  left[R_ I (x)right] ^ 2 right)mathrm d x R O displaystyle R_ O and R I displaystyle R_ I are functions expressing the outer and inner radii of the function, respectively.
Volume
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 1 3 π r 2 h = 1 3 π r 2 ( 2 r ) = ( 2 3 π r 3 ) × 1 , displaystyle frac 1 3 pi r^ 2 h= frac 1 3 pi r^ 2 left(2rright)=left( frac 2 3 pi r^ 3 right)times 1, the volume of the sphere is 4 3 π r 3 = ( 2 3 π r 3 ) × 2 , displaystyle frac 4 3 pi r^ 3 =left( frac 2 3 pi r^ 3 right)times 2, while the volume of the cylinder is π r 2 h = π r 2 ( 2 r ) = ( 2 3 π r 3 ) × 3. displaystyle pi r^ 2 h=pi r^ 2 (2r)=left( frac 2 3 pi r^ 3 right)times 3. The discovery of the 2 : 3 ratio of the volumes of the
sphere and cylinder is credited to Archimedes.[7]
Volume
π r 2 displaystyle pi r^ 2 . The radius of the circular disks, defined such that the xaxis cuts perpendicularly through them, is y = r 2 − x 2 displaystyle y= sqrt r^ 2 x^ 2 or z = r 2 − x 2 displaystyle 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 ∫ − r r π y 2 d x = ∫ − r r π ( r 2 − x 2 ) d x . displaystyle int _ r ^ r pi y^ 2 ,dx=int _ r ^ r pi left(r^ 2 x^ 2 right),dx. Now ∫ − r r π r 2 d x − ∫ − r r π x 2 d x = π ( r 3 + r 3 ) − π 3 ( r 3 + r 3 ) = 2 π r 3 − 2 π r 3 3 . displaystyle int _ r ^ r pi r^ 2 ,dxint _ r ^ r pi x^ 2 ,dx=pi left(r^ 3 +r^ 3 right) frac pi 3 left(r^ 3 +r^ 3 right)=2pi r^ 3  frac 2pi r^ 3 3 . Combining yields V = 4 3 π r 3 . displaystyle 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 π r 2 displaystyle 4pi r^ 2 . The volume of the sphere consists of layers of infinitesimally thin spherical shells, and the sphere volume is equal to ∫ 0 r 4 π r 2 d r = 4 3 π r 3 . displaystyle int _ 0 ^ r 4pi r^ 2 ,dr= frac 4 3 pi r^ 3 . Cone[edit] The cone is a type of pyramidal shape. The fundamental equation for pyramids, onethird 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 thin 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 h − x h . displaystyle r frac hx h . The surface area of the circular disk is then π ( r h − x h ) 2 = π r 2 ( h − x ) 2 h 2 . displaystyle pi left(r frac hx h right)^ 2 =pi r^ 2 frac (hx)^ 2 h^ 2 . The volume of the cone can then be calculated as ∫ 0 h π r 2 ( h − x ) 2 h 2 d x , displaystyle int _ 0 ^ h pi r^ 2 frac (hx)^ 2 h^ 2 dx, and after extraction of the constants π r 2 h 2 ∫ 0 h ( h − x ) 2 d x displaystyle frac pi r^ 2 h^ 2 int _ 0 ^ h (hx)^ 2 dx Integrating gives us π r 2 h 2 ( h 3 3 ) = 1 3 π r 2 h . displaystyle frac pi r^ 2 h^ 2 left( frac h^ 3 3 right)= frac 1 3 pi r^ 2 h. Polyhedron[edit]
Main article:
Volume
ω =
g
d x 1 ∧ ⋯ ∧ d x n , displaystyle omega = sqrt g ,dx^ 1 wedge dots wedge dx^ n , where the d x i displaystyle dx^ i are 1forms that form a positively oriented basis for the cotangent bundle of the manifold, and g displaystyle g is the determinant of the matrix representation of the metric tensor
on the manifold in terms of the same basis.
Volume
Area
Banach–Tarski paradox
Conversion of units
Dimensional weight
Dimensioning
Length
Measure
Orders of magnitude (volume)
Perimeter
Volume
References[edit] ^ "Your Dictionary entry for "volume"". Retrieved 20100501.
^ 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 20100501. Up to 1800 grams of sucrose can
dissolve in a liter of water.
^ "General Tables of Units of Measurement". NIST Weights and Measures
Division. Archived from the original on 20111210. Retrieved
20110112.
^ "capacity".
Oxford English Dictionary
External links[edit] Wikimedia Commons has media related to Volume. Perimeters, Areas, Volumes at Wikibooks
Volume
