Volume integral
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}}}
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