Volume Integral (∭)

Volume Integral - Definition

The volume integral of a scalar-valued function 𝑓(𝑥,𝑦,𝑧) over a region 𝐷⊂ℝ³ is defined as:

• ${\iiint }_{D}f(x,y,z)dxdydz$

Volume Integral - Examples

Integrating the scalar-valued function 𝑓(𝑥,𝑦,𝑧) = 1 over a unit cube yields the following result:

• ${\int }_0^1{\int }_0^1{\int }_0^{1}1dxdydz={\int }_0^1{\int }_0^1(1-0)dydz={\int }_0^1(1-0)dz=1-0=1$

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:

• $f(x,y,z)=x+y+z$

the total mass of the cube is:

• ${\int }_0^1{\int }_0^1{\int }_0^1(x+y+z)dxdydz={\int }_0^1{\int }_0^1[\frac{1}{2}x+yx+zx{]}_0^{1}dydz$
• ${\int }_0^1{\int }_0^1{\int }_0^1(x+y+z)dxdydz={\int }_0^1{\int }_0^1[(\frac{1}{2}1+y1+z1)-(\frac{1}{2}0+y0+z0){]}_0^{1}dydz$
• ${\int }_0^1{\int }_0^1{\int }_0^1(x+y+z)dxdydz={\int }_0^1{\int }_0^1(\frac{1}{2}+y+z)dydz={\int }_0^1(1+z)dz=1-0=\frac{3}{2}$