# Volume integral: Wikis

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# Encyclopedia

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

Volume integral is a triple integral of the constant function 1, which gives the volume of the region D, that is, the integral

$\operatorname{Vol}(D)=\iiint\limits_D dx\,dy\,dz.$

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 standard convention for angles) has the form

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

## Example

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

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

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:

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

Topics in Calculus
Fundamental theorem
Limits of functions
Continuity
Mean value theorem
Multivariable calculus
Matrix calculus
Partial derivative
Multiple integral
Line integral
Surface integral
Volume integral
Jacobian

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

Volume integral is a triple integral of the constant function 1, which gives the volume of the region D, that is, the integral

$\operatorname\left\{Vol\right\}\left(D\right)=\iiint\limits_D dx\,dy\,dz.$

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

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

A volume integral in cylindrical coordinates is

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

and a volume integral in spherical coordinates (using the standard convention for angles) has the form

$\iiint\limits_D f\left(r,\theta,\phi\right)\,r^2 \sin\theta \,dr \,d\theta\, d\phi .$

## Example

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

$\iiint \limits_0^1 1 \,dx\, dy \,dz = \iint \limits_0^1 \left(1 - 0\right) \,dy \,dz = \int \limits_0^1 \left(1 - 0\right) dz = 1 - 0 = 1$

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

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