Volume integral

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

$${\displaystyle \iiint _{D}f(r,\theta ,\varphi )r^{2}\sin \theta \,dr\,d\theta \,d\varphi .}$$

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

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