imported>LegendoftheGoldenAges85: /* In coordinates */ incorrect mapping

2025-05-12T21:03:07Z

In coordinates: incorrect mapping

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{{Short description|Integral over a 3-D domain}}
{{Calculus |Multivariable}}

In [[mathematics]] (particularly [[multivariable calculus]]), a '''volume integral''' (∭) is an [[integral]] over a [[Three-dimensional space|3-dimensional]] domain; that is, it is a special case of [[multiple integral]]s. Volume integrals are especially important in [[physics]] for many applications, for example, to calculate [[flux]] densities, or to calculate mass from a corresponding density function.

==In coordinates==

Often the volume integral is represented in terms of a differential volume element <math> dV=dx\, dy\, dz </math>.
<math display="block">\iiint_D f(x,y,z)\,dV.</math>
It can also mean a [[multiple integral|triple integral]] within a region <math>D \subset \R^3</math> of a [[function (mathematics)|function]] <math>f(x,y,z),</math> and is usually written as:
<math display="block">\iiint_D f(x,y,z)\,dx\,dy\,dz.</math>
A volume integral in [[cylindrical coordinates]] is
<math display="block">\iiint_D f(\rho,\varphi,z) \rho \,d\rho \,d\varphi \,dz,</math>
and a volume integral in [[spherical coordinates]] (using the ISO convention for angles with <math>\varphi</math> as the azimuth and <math>\theta</math> measured from the polar axis (see more on [[Spherical coordinate system#Conventions|conventions]])) has the form
<math display="block">\iiint_D f(r,\theta,\varphi) r^2 \sin\theta \,dr \,d\theta\, d\varphi .</math>
The triple integral can be transformed from Cartesian coordinates to any arbitrary coordinate system using the [[Jacobian matrix and determinant]]. Suppose we have a transformation of coordinates from <math> (x,y,z)\mapsto(u,v,w) </math>. We can represent the integral as the following.
<math display="block">\iiint_D f(x,y,z)\,dx\,dy\,dz=\iiint_D f(u,v,w)\left|\frac{\partial (x,y,z)}{\partial (u,v,w)}\right|\,du\,dv\,dw</math>
Where we define the Jacobian determinant to be.
<math display="block">
\mathbf{J}=\frac{\partial (x,y,z)}{\partial (u,v,w)}=
\begin{vmatrix}
\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}& \frac{\partial x}{\partial w}\\
\frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}& \frac{\partial y}{\partial w}\\
\frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}& \frac{\partial z}{\partial w}\\
\end{vmatrix}
</math>

== Example ==

Integrating the equation <math> f(x,y,z) = 1 </math> over a unit cube yields the following result:
<math display="block">\int_0^1 \int_0^1 \int_0^1 1 \,dx \,dy \,dz = \int_0^1 \int_0^1 (1 - 0) \,dy \,dz = \int_0^1 \left(1 - 0\right) dz = 1 - 0 = 1</math>

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 density function on the unit cube then the volume integral will give the total mass of the cube. For example for density function:
<math display="block"> \begin{cases}
f: \R^3 \to \R \\
f: (x,y,z) \mapsto x+y+z
\end{cases}</math>
the total mass of the cube is:
<math display="block">\int_0^1 \int_0^1 \int_0^1 (x+y+z) \,dx \,dy \,dz = \int_0^1 \int_0^1 \left(\frac 1 2 + y + z\right) dy \,dz = \int_0^1 (1 + z) \, dz = \frac 3 2</math>

==See also==
{{Portal|Mathematics}}
*[[Divergence theorem]]
*[[Surface integral]]
*[[Volume element]]
*[[Line element]]
*[[Line integral]]

==External links==
* {{springer|title=Multiple integral|id=p/m065370}}
* {{MathWorld|VolumeIntegral|Volume integral}}

{{Calculus topics}}

[[Category:Multivariable calculus]]

Volume integral - Revision history

imported>LegendoftheGoldenAges85: /* In coordinates */ incorrect mapping