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.

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, i.e. with φ as the azimuth) has the form

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

Example

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:

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

See also

• Divergence theorem
• Surface integral

External links

• Weisstein, Eric W., "Volume integral" from MathWorld.