Volume integral

Integral over a 3-D domain
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∫ a b f ′ ( t ) d t = f ( b ) − f ( a ) {\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}
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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 D ⊂ R 3 {\displaystyle D\subset \mathbb {R} ^{3}} of a function f ( x , y , z ) , {\displaystyle f(x,y,z),} and is usually written as:

∭ D f ( x , y , z ) d x d y d z . {\displaystyle \iiint _{D}f(x,y,z)\,dx\,dy\,dz.}

A volume integral in cylindrical coordinates is

∭ D f ( ρ , φ , z ) ρ d ρ d φ d z , {\displaystyle \iiint _{D}f(\rho ,\varphi ,z)\rho \,d\rho \,d\varphi \,dz,}
and a volume integral in spherical coordinates (using the ISO convention for angles with φ {\displaystyle \varphi } as the azimuth and θ {\displaystyle \theta } measured from the polar axis (see more on conventions)) has the form
∭ D f ( r , θ , φ ) r 2 sin ⁡ θ d r d θ d φ . {\displaystyle \iiint _{D}f(r,\theta ,\varphi )r^{2}\sin \theta \,dr\,d\theta \,d\varphi .}

Example

Integrating the equation f ( x , y , z ) = 1 {\displaystyle f(x,y,z)=1} over a unit cube yields the following result:

∫ 0 1 ∫ 0 1 ∫ 0 1 1 d x d y d z = ∫ 0 1 ∫ 0 1 ( 1 − 0 ) d y d z = ∫ 0 1 ( 1 − 0 ) d z = 1 − 0 = 1 {\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}

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:

{ f : R 3 → R f : ( x , y , z ) ↦ x + y + z {\displaystyle {\begin{cases}f:\mathbb {R} ^{3}\to \mathbb {R} \\f:(x,y,z)\mapsto x+y+z\end{cases}}}
the total mass of the cube is:
∫ 0 1 ∫ 0 1 ∫ 0 1 ( x + y + z ) d x d y d z = ∫ 0 1 ∫ 0 1 ( 1 2 + y + z ) d y d z = ∫ 0 1 ( 1 + z ) d z = 3 2 {\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}}}

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