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 ${\begin{aligned}f\colon {\mathbb {R}}^{3}&\to {\mathbb {R}}\end{aligned}}$ 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:

$\int \limits _{0}^{1}\int \limits _{0}^{1}\int \limits _{0}^{1}\left(x+y+z\right)\,dx\,dy\,dz=\int \limits _{0}^{1}\int \limits _{0}^{1}\left({\frac 12}+y+z\right)\,dy\,dz=\int \limits _{0}^{1}\left(1+z\right)\,dz={\frac 32}$

External links

Hazewinkel, Michiel, ed. (2001), "Multiple integral", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
Weisstein, Eric W., "Volume integral", MathWorld.

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Volume integral

Example 1

See also

External links