The volume of any solid, liquid, plasma, vacuum or theoretical object is how much three-dimensional 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. Volume is commonly presented in units such as mL or cm3 (milliliters or cubic centimeters).
Volumes of some simple shapes, such as regular, straight-edged and circular shapes can be easily calculated using arithmetic formulas. More complicated shapes can be calculated by integral calculus if a formula exists for its boundary. The volume of any shape can be determined by displacement.
In differential geometry, volume is expressed by means of the volume form, and is an important global Riemannian invariant.
Volume is a fundamental parameter in thermodynamics and it is conjugate to pressure.
Template:Conjugate variables (thermodynamics)
Traditional cooking measures
| measure | US | Imperial | metric |
|---|---|---|---|
| teaspoon | 1/6 U.S. fluid ounce (about 4.929 mL) | 1/6 Imperial fluid ounce (about 4.736 mL) | 5 mL |
| tablespoon = 3 teaspoons | ½ U.S. fluid ounce (about 14.79 mL) | ½ Imperial fluid ounce (about 14.21 mL) | 15 mL |
| cup | 8 U.S. fluid ounces or ½ U.S. liquid pint (about 237 mL) | 8 Imperial fluid ounces or 2/5 fluid pint (about 227 mL) | 250 mL |
In the UK, a tablespoon can also be five fluidrams (about 17.76 mL).
Volume formulas
| Shape | Equation | Variables |
|---|---|---|
| A cube | $ a^3 $ | a = length of any side (or edge) |
| A rectangular prism: | $ l\cdot w\cdot h $ | l = length, w = width, h = height |
| A cylinder: | $ \pi r^2h $ | r = radius of circular face, h = height |
| A general prism: | $ B\cdot h $ | B = area of the base, h = height |
| A sphere: | $ \frac{4\pi}{3}r^3 $ | r = radius of sphere which is the integral of the Surface Area of a sphere |
| An ellipsoid: | $ \frac{4\pi}{3}abc $ | a, b, c = semi-axes of ellipsoid |
| A pyramid: | $ \frac{1}{3}Ah $ | A = area of the base, h = height of pyramid |
| A cone (circular-based pyramid): | $ \frac{\pi}{3}r^2h $ | 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)</small>. |
The units of volume depend on the units of length. If the lengths are in meters, the volume will be in cubic meters.
- For their volume formulas, see the articles on tetrahedron and parallelepiped.
Volume formula derivation
Sphere
The volume of a sphere is the integral of infinitesimal circular slabs of thickness $ dx $ . 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 area of the circular slab is $ \pi y^2 $ .
The volume of the sphere can be calculated as $ \int\limits_{-r}^r\pi(r^2-x^2)\,dx=\int\limits_{-r}^r\pi r^2\,dx-\int\limits_{-r}^r\pi x^2\,dx $
- Now $ \int\limits_{-r}^r\pi r^2\,dx=2\pi r^3 $
- and $ \int\limits_{-r}^r\pi x^2\,dx=\frac{2\pi}{3}r^3 $
Combining yields $ \left(2-\frac{2}{3}\right)\pi r^3=\frac{4\pi}{3}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 infinitesimal spherical slabs, and the sphere volume is equal to
$ \int\limits_0^r 4\pi r^2\,dr=\frac{4\pi}{3}r^3 $
Cone
The volume of a cone is the integral of infinitesimal circular slabs of thickness $ dx $ . The calculation for the volume of a cone of height $ h $ , whose base is centered at (0,0) with radius $ r $ is as follows.
The radius of each circular slab is $ \begin{cases}r&x=0\\0&x=h\end{cases} $ , and varying linearly in between—that is, $ r\left(1-\frac{x}{h}\right) $
The surface area of the circular slab is then $ \pi\left(r(1-\frac{x}{h})\right)^2=\pi r^2\frac{(h-x)^2}{h^2} $
The volume of the cone can then be calculated as $ \int\limits_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\limits_0^h (h-x)^2dx $
Integrating gives us $ \frac{\pi r^2}{h^2}\left(\frac{h^3}{3}\right)=\frac{\pi}{3}r^2h $
See also
 |
The Wikibook Calculus has a page on the topic of: Volume |
|
The Wikibook Geometry has a page on the topic of: Perimeters, Areas, Volumes |
- Area
- Conversion of units
- Density
- Dimensional weight
- Dimensioning
- Length
- Mass
- Orders of magnitude (volume)
- Volume form
- Weight
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