${\int }_{x\=0}^{x\=1}{\int }_{y\=0}^{y\=1}{\int }_{z\=0}^{z\=1}f\left(x\,y\,z\right) \mathrm{dz} \mathrm{dy} \mathrm{dx}\=\frac{1}{8}$ $$

${\int }_{y\=0}^{y\=1}{\int }_{x\=0}^{x\=1}{\int }_{z\=0}^{z\=1}f\left(x\,y\,z\right) \mathrm{dz} \mathrm{dx} \mathrm{dy}\=\frac{1}{8}$ $$

${\int }_{z\=0}^{z\=1}{\int }_{x\=0}^{x\=1}{\int }_{y\=0}^{y\=1}f\left(x\,y\,z\right) \mathrm{dy} \mathrm{dx} \mathrm{dz}\=\frac{1}{8}$ $$

${\int }_{x\=0}^{x\=1}{\int }_{z\=0}^{z\=1}{\int }_{y\=0}^{y\=1}f\left(x\,y\,z\right) \mathrm{dy} \mathrm{dz} \mathrm{dx}\=\frac{1}{8}$ $$

${\int }_{z\=0}^{z\=1}{\int }_{y\=0}^{y\=1}{\int }_{x\=0}^{x\=1}f\left(x\,y\,z\right) \mathrm{dx} \mathrm{dy} \mathrm{dz}\=\frac{1}{8}$ $$

${\int }_{y\=0}^{y\=1}{\int }_{z\=0}^{z\=1}{\int }_{x\=0}^{x\=1}f\left(x\,y\,z\right) \mathrm{dx} \mathrm{dz} \mathrm{dy}\=\frac{1}{8}$ $$

Table 7.2.1(a)   In Cartesian coordinates, the six iterations of a given triple integral

${\int }_0^1{\int }_0^1{\int }_0^1\left(x y z\right) \mathrm{dz} \mathrm{dy} \mathrm{dx}$

$\={\int }_0^1{\int }_0^1{\left[\frac{z^2}{2} x y\right]}_{z\=0}^{z\=1} \mathrm{dy} \mathrm{dx}$

$\={\int }_0^1{\int }_0^1\left(\frac{1^1-0^2}{2}\right) x y \mathrm{dy} \mathrm{dx}$

$\=\frac{1}{2}{\int }_0^1{\int }_0^{1}x y \mathrm{dy} \mathrm{dx}$

$\=\frac{1}{2}{\int }_0^1{\left[\frac{y^2}{2} x\right]}_{y\=0}^{y\=1} \mathrm{dx}$

$\=\frac{1}{2}{\int }_0^1\left(\frac{1^2-0^2}{2}\right) x \mathrm{dx}$

$\=\frac{1}{4}{\int }_0^{1}x \mathrm{dx}$

$\=\frac{1}{4}{\left[\frac{x^2}{2}\right]}_{x\=0}^{x\=1}$

$\=\frac{1}{4}\left(\frac{1^2-0^2}{2}\right)$

$\=1\/8$

Table 7.2.1(b)   Stepwise evaluation of the $\mathrm{dz} \mathrm{dy} \mathrm{dx}$ iteration

Define the integrand $f$ 

$f\=x y z$$\stackrel{\text{assign}}{\to }$

Write and evaluate all six iterations of the triple integral

${\int }_0^1{\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}z \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{1}{8}$$$

${\int }_0^1{\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}z \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{1}{8}$$$

${\int }_0^1{\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}y \mathit{\ⅆ}x \mathit{\ⅆ}z$ = $\frac{1}{8}$$$

${\int }_0^1{\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}y \mathit{\ⅆ}z \mathit{\ⅆ}x$ = $\frac{1}{8}$$$

${\int }_0^1{\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}x \mathit{\ⅆ}y \mathit{\ⅆ}z$ = $\frac{1}{8}$$$

${\int }_0^1{\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}x \mathit{\ⅆ}z \mathit{\ⅆ}y$ = $\frac{1}{8}$$$

Tools≻Tasks≻Browse:

Calculus - Vector≻Integration≻Multiple Integration≻3-D≻Over a Cube

Table 7.2.1(c)   Task template based on the int command in the Student VectorCalculus package

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 3-D

Table 7.2.1(d)   Task template for visualizing regions of integration

Initialize

$f≔x y z\:$

Write and evaluate all six iterations of the triple integral

$\mathrm{Int}\left(f\,\left[z\=0..1\,y\=0..1\,x\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[z\=0..1\,y\=0..1\,x\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^1{\∫}_0^{1}xyz\ⅆz\ⅆy\ⅆx\=\frac{1}{8}$

$\mathrm{Int}\left(f\,\left[z\=0..1\,x\=0..1\,y\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[z\=0..1\,x\=0..1\,y\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^1{\∫}_0^{1}xyz\ⅆz\ⅆx\ⅆy\=\frac{1}{8}$

$\mathrm{Int}\left(f\,\left[y\=0..1\,x\=0..1\,z\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[y\=0..1\,x\=0..1\,z\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^1{\∫}_0^{1}xyz\ⅆy\ⅆx\ⅆz\=\frac{1}{8}$

$\mathrm{Int}\left(f\,\left[y\=0..1\,z\=0..1\,x\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[y\=0..1\,z\=0..1\,x\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^1{\∫}_0^{1}xyz\ⅆy\ⅆz\ⅆx\=\frac{1}{8}$

$\mathrm{Int}\left(f\,\left[x\=0..1\,y\=0..1\,z\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[x\=0..1\,y\=0..1\,z\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^1{\∫}_0^{1}xyz\ⅆx\ⅆy\ⅆz\=\frac{1}{8}$

$\mathrm{Int}\left(f\,\left[x\=0..1\,z\=0..1\,y\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[x\=0..1\,z\=0..1\,y\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^1{\∫}_0^{1}xyz\ⅆx\ⅆz\ⅆy\=\frac{1}{8}$

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{MultiInt}\left(f\,z\=0..1\,y\=0..1\,x\=0..1\,\mathrm{output}\=\mathrm{steps}\right)$

$\frac{1}{8}$

Table 7.2.1(e)   Stepwise evaluation of an iterated triple-integral

$\mathrm{MultiInt}\left(x y z\,\left[x\,y\,z\right]\=\mathrm{Parallelepiped}\left(0..1\,0..1\,0..1\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^1{\∫}_0^1{\∫}_0^{1}xyz\ⅆx\ⅆy\ⅆz$

$\mathrm{MultiInt}\left(x y z\,\left[x\,y\,z\right]\=\mathrm{Parallelepiped}\left(0..1\,0..1\,0..1\right)\right)$

$\frac{1}{8}$

Table 7.2.1(f)   The Parallelepiped option in the MultiInt command, Student MultivariateCalculus