# Volume integral

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In mathematics—in particular, in multivariable calculus—a volume integral refers to an integral over a 3-dimensional domain.

It can also mean a triple integral within a region D in R3 of a function $f(x,y,z),$ and is usually written as:

$\iiint\limits_D f(x,y,z)\,dx\,dy\,dz.$

A volume integral in cylindrical coordinates is

$\iiint\limits_D f(r,\theta,z)\,r\,dr\,d\theta\,dz,$

and a volume integral in spherical coordinates (using the convention for angles with $\theta$ as the azimuth and $\phi$ measured from the polar axis (see more on conventions)) has the form

$\iiint\limits_D f(\rho,\theta,\phi)\,\rho^2 \sin\phi \,d\rho \,d\theta\, d\phi .$

## Example 1

Integrating the function $f(x,y,z) = 1$ over a unit cube yields the following result:

$\int\limits_0^1\int\limits_0^1\int\limits_0^1 1 \,dx\, dy \,dz = \int\limits_0^1\int\limits_0^1 (1 - 0) \,dy \,dz = \int\limits_0^1 (1 - 0) dz = 1 - 0 = 1$

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 function \begin{align} f\colon \mathbb{R}^3 &\to \mathbb{R} \end{align} describing the density of the cube at a given point $(x,y,z)$ by $f = x+y+z$ then performing the volume integral will give the total mass of the cube:

$\int\limits_0^1\int\limits_0^1\int\limits_0^1 \left(x + y + z\right) \, dx \,dy \,dz = \int\limits_0^1\int\limits_0^1 \left(\frac 12 + y + z\right) \, dy \,dz = \int \limits_0^1 \left(1 + z\right) \, dz = \frac 32$

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