# Volume integral

In mathematics (particularly in multivariable calculus), a volume integral refers to 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. The volume integral of a function f over a domain D can be denoted abstractly as ${\displaystyle \textstyle \iiint _{D}f\,dV}$.[1][2][3]

## In coordinates

In Cartesian coordinates, a volume integral can be expressed as a triple integral within a region ${\displaystyle D\subset \mathbb {R} ^{3}}$  of a function ${\displaystyle f(x,y,z),}$  and is usually written as:[4]

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

A volume integral in cylindrical coordinates is

${\displaystyle \iiint _{D}f(\rho ,\varphi ,z)\rho \,d\rho \,d\varphi \,dz,}$

and a volume integral in spherical coordinates has the form

${\displaystyle \iiint _{D}f(r,\theta ,\varphi )r^{2}\sin \theta \,dr\,d\theta \,d\varphi .}$

using the ISO convention for angles with ${\displaystyle \varphi }$  as the azimuth, and ${\displaystyle \theta }$  measured from the polar axis (see also Spherical coordinate system § Conventions for more).

## Example

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

${\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}1\,dx\,dy\,dz=\int _{0}^{1}\int _{0}^{1}(1-0)\,dy\,dz=\int _{0}^{1}(1-0)dz=1-0=1}$

So the volume of the unit cube is 1 as expected. This is rather trivial result, and a volume integral is in actuality 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 the density function:

${\displaystyle {\begin{cases}f:\mathbb {R} ^{3}\to \mathbb {R} \\(x,y,z)\longmapsto x+y+z\end{cases}}}$

the total mass of the cube is:

${\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}(x+y+z)\,dx\,dy\,dz=\int _{0}^{1}\int _{0}^{1}\left({\frac {1}{2}}+y+z\right)\,dy\,dz=\int _{0}^{1}(1+z)\,dz={\frac {3}{2}}}$