Volume integral: Definition from Answers.com

Fundamental theorem
Limits of functions
Continuity
Mean value theorem

Differential calculus
Derivative Change of variables Implicit differentiation Taylor's theorem Related rates Identities Rules: Power rule, Product rule, Quotient rule, Chain rule

Integral calculus
Integral Lists of integrals Improper integrals Integration by: parts, disks, cylindrical shells, substitution, trigonometric substitution, partial fractions, changing order

Vector calculus
Gradient Divergence Curl Laplacian Gradient theorem Green's theorem Stokes' theorem Divergence theorem

Multivariable calculus
Matrix calculus Partial derivative Multiple integral Line integral Surface integral Volume integral Jacobian

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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 (using the standard convention for angles) has the form

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

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:

$\iiint \limits_0^1 \left(x + y + z\right) \, dx \,dy \,dz = \iint \limits_0^1 \left(\frac 12 + y + z\right) \, dy \,dz = \int \limits_0^1 \left(1 + z\right) \, dz = \frac 32$

External links

MathWorld article on volume integrals

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Volume integral

Example

See also

External links