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Volume integral

From Wikipedia, the free encyclopedia
Integral over a 3-D domain
Part of a series of articles about
Calculus
abf(t)dt=f(b)f(a){\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}

Inmathematics (particularlymultivariable calculus), avolume integral (∭) is anintegral over a3-dimensional domain; that is, it is a special case ofmultiple integrals. Volume integrals are especially important inphysics for many applications, for example, to calculateflux densities, or to calculate mass from a corresponding density function.

In coordinates

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Often the volume integral is represented in terms of a differential volume elementdV=dxdydz{\displaystyle dV=dx\,dy\,dz}.Df(x,y,z)dV.{\displaystyle \iiint _{D}f(x,y,z)\,dV.}It can also mean atriple integral within a regionDR3{\displaystyle D\subset \mathbb {R} ^{3}} of afunctionf(x,y,z),{\displaystyle f(x,y,z),} and is usually written as:Df(x,y,z)dxdydz.{\displaystyle \iiint _{D}f(x,y,z)\,dx\,dy\,dz.}A volume integral incylindrical coordinates isDf(ρ,φ,z)ρdρdφdz,{\displaystyle \iiint _{D}f(\rho ,\varphi ,z)\rho \,d\rho \,d\varphi \,dz,}and a volume integral inspherical coordinates (using the ISO convention for angles withφ{\displaystyle \varphi } as the azimuth andθ{\displaystyle \theta } measured from the polar axis (see more onconventions)) has the formDf(r,θ,φ)r2sinθdrdθdφ.{\displaystyle \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 theJacobian matrix and determinant. Suppose we have a transformation of coordinates from(x,y,z)(u,v,w){\displaystyle (x,y,z)\mapsto (u,v,w)}. We can represent the integral as the following.Df(x,y,z)dxdydz=Df(u,v,w)|(x,y,z)(u,v,w)|dudvdw{\displaystyle \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}Where we define the Jacobian determinant to be.J=(x,y,z)(u,v,w)=|xuxvxwyuyvywzuzvzw|{\displaystyle \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}}}

Example

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Integrating the equationf(x,y,z)=1{\displaystyle f(x,y,z)=1} over a unit cube yields the following result:0101011dxdydz=0101(10)dydz=01(10)dz=10=1{\displaystyle \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:{f:R3Rf:(x,y,z)x+y+z{\displaystyle {\begin{cases}f:\mathbb {R} ^{3}\to \mathbb {R} \\f:(x,y,z)\mapsto x+y+z\end{cases}}} the total mass of the cube is:010101(x+y+z)dxdydz=0101(12+y+z)dydz=01(1+z)dz=32{\displaystyle \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

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External links

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Precalculus
Limits
Differential calculus
Integral calculus
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Multivariable calculus
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