Volume integral

In mathematicsin particular, in multivariable calculusa 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.

In coordinates

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(\rho,\varphi,z)\,\rho\,d\rho\,d\varphi\,dz,
Volume element in spherical coordinates

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

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

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

See also

External links

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