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

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

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: $${\displaystyle {\begin{cases}f:\mathbb {R} ^{3}\to \mathbb {R} \\f:(x,y,z)\mapsto x+y+z\end{cases}}}$$  the total mass of the cube is: $${\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}(x+y+z)\,dx\,dy\,dz=\int _{0}^{1}\int _{0}^{1}\left({\frac {1}{2}}+y+z\right)dy\,dz=\int _{0}^{1}(1+z)\,dz={\frac {3}{2}}}$$ 

• 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.