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.

Often the volume integral is represented in terms of a differential volume element ${\displaystyle dV=dx,円dy,円dz}${\displaystyle dV=dx,円dy,円dz}. $${\displaystyle \iiint _{D}f(x,y,z),円dV.}$${\displaystyle \iiint _{D}f(x,y,z),円dV.} It can also mean a triple integral within a region ${\displaystyle D\subset \mathbb {R} ^{3}}${\displaystyle D\subset \mathbb {R} ^{3}} of a function ${\displaystyle f(x,y,z),}${\displaystyle f(x,y,z),} and is usually written as: $${\displaystyle \iiint _{D}f(x,y,z),円dx,円dy,円dz.}$${\displaystyle \iiint _{D}f(x,y,z),円dx,円dy,円dz.} A volume integral in cylindrical coordinates is $${\displaystyle \iiint _{D}f(\rho ,\varphi ,z)\rho ,円d\rho ,円d\varphi ,円dz,}$${\displaystyle \iiint _{D}f(\rho ,\varphi ,z)\rho ,円d\rho ,円d\varphi ,円dz,} and a volume integral in spherical coordinates (using the ISO convention for angles with ${\displaystyle \varphi }${\displaystyle \varphi } as the azimuth and ${\displaystyle \theta }${\displaystyle \theta } measured from the polar axis (see more on conventions)) has the form $${\displaystyle \iiint _{D}f(r,\theta ,\varphi )r^{2}\sin \theta ,円dr,円d\theta ,円d\varphi .}$${\displaystyle \iiint _{D}f(r,\theta ,\varphi )r^{2}\sin \theta ,円dr,円d\theta ,円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 ${\displaystyle (x,y,z)\mapsto (u,v,w)}${\displaystyle (x,y,z)\mapsto (u,v,w)}. We can represent the integral as the following. $${\displaystyle \iiint _{D}f(x,y,z),円dx,円dy,円dz=\iiint _{D}f(u,v,w)\left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|,円du,円dv,円dw}$${\displaystyle \iiint _{D}f(x,y,z),円dx,円dy,円dz=\iiint _{D}f(u,v,w)\left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|,円du,円dv,円dw} Where we define the Jacobian determinant to be. $${\displaystyle \mathbf {J} ={\frac {\partial (x,y,z)}{\partial (u,v,w)}}={\begin{vmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}&{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}&{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial u}}&{\frac {\partial z}{\partial v}}&{\frac {\partial z}{\partial w}}\\\end{vmatrix}}}$${\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}}}

Integrating the equation ${\displaystyle f(x,y,z)=1}${\displaystyle f(x,y,z)=1} over a unit cube yields the following result: $${\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}1,円dx,円dy,円dz=\int _{0}^{1}\int _{0}^{1}(1-0),円dy,円dz=\int _{0}^{1}\left(1-0\right)dz=1-0=1}$${\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}1,円dx,円dy,円dz=\int _{0}^{1}\int _{0}^{1}(1-0),円dy,円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: $${\displaystyle {\begin{cases}f:\mathbb {R} ^{3}\to \mathbb {R} \\f:(x,y,z)\mapsto x+y+z\end{cases}}}$${\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: $${\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}(x+y+z),円dx,円dy,円dz=\int _{0}^{1}\int _{0}^{1}\left({\frac {1}{2}}+y+z\right)dy,円dz=\int _{0}^{1}(1+z),円dz={\frac {3}{2}}}$${\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}(x+y+z),円dx,円dy,円dz=\int _{0}^{1}\int _{0}^{1}\left({\frac {1}{2}}+y+z\right)dy,円dz=\int _{0}^{1}(1+z),円dz={\frac {3}{2}}}

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

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

• Multivariable calculus

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