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In coordinates

Often the volume integral is represented in terms of a differential volume element d V = d x d y d z {\displaystyle dV=dx\,dy\,dz} . ∭ D f ( x , y , z ) d V . {\displaystyle \iiint _{D}f(x,y,z)\,dV.} It can also mean a triple integral within a region D ⊂ R 3 {\displaystyle D\subset \mathbb {R} ^{3}} of a function f ( x , y , z ) , {\displaystyle f(x,y,z),} and is usually written as: ∭ D f ( x , y , z ) d x d y d z . {\displaystyle \iiint _{D}f(x,y,z)\,dx\,dy\,dz.} A volume integral in cylindrical coordinates is ∭ D f ( ρ , φ , z ) ρ d ρ d φ d z , {\displaystyle \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 φ {\displaystyle \varphi } as the azimuth and θ {\displaystyle \theta } measured from the polar axis (see more on conventions)) has the form ∭ D f ( r , θ , φ ) r 2 sin ⁡ θ d r d θ d φ . {\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 the Jacobian 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. ∭ D f ( x , y , z ) d x d y d z = ∭ D f ( u , v , w ) | ∂ ( x , y , z ) ∂ ( u , v , w ) | d u d v d w {\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 ) = | ∂ x ∂ u ∂ x ∂ v ∂ x ∂ w ∂ y ∂ u ∂ y ∂ v ∂ y ∂ w ∂ z ∂ u ∂ z ∂ v ∂ z ∂ w | {\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

Integrating the equation f ( x , y , z ) = 1 {\displaystyle f(x,y,z)=1} over a unit cube yields the following result: ∫ 0 1 ∫ 0 1 ∫ 0 1 1 d x d y d z = ∫ 0 1 ∫ 0 1 ( 1 − 0 ) d y d z = ∫ 0 1 ( 1 − 0 ) d z = 1 − 0 = 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 : R 3 → R f : ( 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: ∫ 0 1 ∫ 0 1 ∫ 0 1 ( x + y + z ) d x d y d z = ∫ 0 1 ∫ 0 1 ( 1 2 + y + z ) d y d z = ∫ 0 1 ( 1 + z ) d z = 3 2 {\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

• Mathematics portal

• Divergence theorem
• Surface integral
• Volume element
• Line element
• Line integral

External links

• "Multiple integral", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Weisstein, Eric W. "Volume integral". MathWorld.