Volume integral

From Wikipedia, the free 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 has the form

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

[edit] 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 x + y + z \, dx \,dy \,dz = \iint \limits_0^1 \frac 12 + y + z \, dy \,dz = \int \limits_0^1 1 + z \, dz = \frac 32

[edit] See also

[edit] External links

This mathematical analysis-related article is a stub. You can help Wikipedia by expanding it.