Integral over a 3-D domain
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Calculus |
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![{\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17d063dc86a53a2efb1fe86f4a5d47d498652766) |
- Rolle's theorem
- Mean value theorem
- Inverse function theorem
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Differential Definitions |
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- Differentiation notation
- Second derivative
- Implicit differentiation
- Logarithmic differentiation
- Related rates
- Taylor's theorem
| Rules and identities |
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| Definitions |
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| Integration by |
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In mathematics (particularly multivariable calculus), a volume integral (∭) is an integral over a 3-dimensional domain; that is, it is a special case of multiple integrals. Volume integrals are especially important in physics for many applications, for example, to calculate flux densities, or to calculate mass from a corresponding density function.
In coordinates
It can also mean a triple integral within a region
of a function
and is usually written as:
![{\displaystyle \iiint _{D}f(x,y,z)\,dx\,dy\,dz.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5fe06bc9a91420054921ca946e40ee29f2f7831)
A volume integral in cylindrical coordinates is
![{\displaystyle \iiint _{D}f(\rho ,\varphi ,z)\rho \,d\rho \,d\varphi \,dz,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ade33e992af29f94b9b8c494f6dcac99d89950f6)
and a volume integral in
spherical coordinates (using the ISO convention for angles with
![{\displaystyle \varphi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/33ee699558d09cf9d653f6351f9fda0b2f4aaa3e)
as the azimuth and
![{\displaystyle \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e5ab2664b422d53eb0c7df3b87e1360d75ad9af)
measured from the polar axis (see more on
conventions)) has the form
![{\displaystyle \iiint _{D}f(r,\theta ,\varphi )r^{2}\sin \theta \,dr\,d\theta \,d\varphi .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14e39eed164ac11b2fe1cbd828cf8517fcff497e)
Example
Integrating the equation
over a unit cube yields the following result:
![{\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}1\,dx\,dy\,dz=\int _{0}^{1}\int _{0}^{1}(1-0)\,dy\,dz=\int _{0}^{1}\left(1-0\right)dz=1-0=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75e991337f6946ea2456df716eab37d5500bed41)
So the volume of the unit cube is 1 as expected. This is rather trivial however, and a volume integral is far more powerful. For instance if we have a scalar density function on the unit cube then the volume integral will give the total mass of the cube. For example for density function:
![{\displaystyle {\begin{cases}f:\mathbb {R} ^{3}\to \mathbb {R} \\f:(x,y,z)\mapsto x+y+z\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbf1c7cbde9a1028d6c82cb9ffd374c649e632d6)
the total mass of the cube is:
![{\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}(x+y+z)\,dx\,dy\,dz=\int _{0}^{1}\int _{0}^{1}\left({\frac {1}{2}}+y+z\right)dy\,dz=\int _{0}^{1}(1+z)\,dz={\frac {3}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ce86f98f83ce069e0ff3173c1a92c29546dde8a)
See also
Mathematics portal
External links