Volume integral

In mathematics (particularly multivariable calculus), a volume integral (∭) is 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, or to calculate mass from a corresponding density function.

In coordinates

Often the volume integral is represented in terms of a differential volume element $dV=dx\, dy\, dz$.
$\iiint_D f(x,y,z)\,dV.$
It can also mean a triple integral within a region $D \subset \R^3$ of a function $f(x,y,z),$ and is usually written as:
$\iiint_D f(x,y,z)\,dx\,dy\,dz.$
A volume integral in cylindrical coordinates is
$\iiint_D f(\rho,\varphi,z) \rho \,d\rho \,d\varphi \,dz,$
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
$\iiint_D f(r,\theta,\varphi) r^2 \sin\theta \,dr \,d\theta\, d\varphi .$
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 $(x,y,z)\mapsto(u,v,w)$. We can represent the integral as the following.
$\iiint_D f(x,y,z)\,dx\,dy\,dz=\iiint_D f(u,v,w)\left, \frac\\,du\,dv\,dw$
Where we define the Jacobian determinant to be.
$\mathbf=\frac= \begin \frac& \frac& \frac\\ \frac& \frac& \frac\\ \frac& \frac& \frac\\ \end$

Example

Integrating the equation $f(x,y,z) = 1$ over a unit cube yields the following result:
$\int_0^1 \int_0^1 \int_0^1 1 \,dx \,dy \,dz = \int_0^1 \int_0^1 (1 - 0) \,dy \,dz = \int_0^1 \left(1 - 0\right) dz = 1 - 0 = 1$

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:
$\begin f: \R^3 \to \R \\ f: (x,y,z) \mapsto x+y+z \end$
the total mass of the cube is:
$\int_0^1 \int_0^1 \int_0^1 (x+y+z) \,dx \,dy \,dz = \int_0^1 \int_0^1 \left(\frac 1 2 + y + z\right) dy \,dz = \int_0^1 (1 + z) \, dz = \frac 3 2$

See also

* Divergence theorem
*Surface integral
* Volume element
* Line element
*Line integral

External links

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{{Calculus topics

Multivariable calculus