${\int }_{x\=1}^{x\=2}{\int }_{y\=-1}^{y\=3}{\int }_{z\=2}^{z\=4}f\left(x\,y\,z\right) \mathrm{dz} \mathrm{dy} \mathrm{dx}\=400$ $$

${\int }_{y\=-1}^{y\=3}{\int }_{x\=1}^{x\=2}{\int }_{z\=2}^{z\=4}f\left(x\,y\,z\right) \mathrm{dz} \mathrm{dx} \mathrm{dy}\=400$ $$

${\int }_{z\=2}^{z\=4}{\int }_{x\=1}^{x\=2}{\int }_{y\=-1}^{y\=3}f\left(x\,y\,z\right) \mathrm{dy} \mathrm{dx} \mathrm{dz}\=400$ $$

${\int }_{x\=1}^{x\=2}{\int }_{z\=2}^{z\=4}{\int }_{y\=-1}^{y\=3}f\left(x\,y\,z\right) \mathrm{dy} \mathrm{dz} \mathrm{dx}\=400$ $$

${\int }_{z\=2}^{z\=4}{\int }_{y\=-1}^{y\=3}{\int }_{x\=1}^{x\=2}f\left(x\,y\,z\right) \mathrm{dx} \mathrm{dy} \mathrm{dz}\=400$ $$

${\int }_{y\=-1}^{y\=3}{\int }_{z\=2}^{z\=4}{\int }_{x\=1}^{x\=2}f\left(x\,y\,z\right) \mathrm{dx} \mathrm{dz} \mathrm{dy}\=400$ $$

Table 7.2.3(a)   In Cartesian coordinates, the six iterations of a given triple integral

${\int }_1^2{\int }_{-1}^3{\int }_2^4\left(1\+2 x^2\+3 y^2\+4 z^2\right) \mathrm{dz} \mathrm{dy} \mathrm{dx}$

$\={\int }_1^2{\int }_{-1}^3{\left[\left(2 x^2\+3 y^2\right)z\+\frac{4 z^3}{3}\+z\right]}_{z\=2}^{z\=4} \mathrm{dy} \mathrm{dx}$

$\={\int }_1^2{\int }_{-1}^3\left(\left(2 x^2\+3 y^2\right)\left(4-2\right)\+\frac{4}{3}\left(4^3-2^3\right)\+\left(4-2\right)\right) \mathrm{dy} \mathrm{dx}$

$\={\int }_1^2{\int }_{-1}^3\left(\frac{230}{3}\+4 x^2\+6 y^2\right) \mathrm{dy} \mathrm{dx}$

$\={\int }_1^2{\left[4 x^{2}y\+\frac{6 y^3}{3}\+\frac{230}{3}y\right]}_{y\=-1}^{y\=3} \mathrm{dx}$

$\={\int }_1^2\left(4 x^2\left(3-\left(-1\right)\right)\+2\left(3^3-{\left(-1\right)}^3\right)\+\frac{230}{3}\left(3-\left(-1\right)\right)\right) \mathrm{dx}$

$\={\int }_1^2\left(\frac{1088}{3}\+16 x^2\right) \mathrm{dx}$

$\={\left[\frac{1088}{3}x\+16\frac{x^3}{3}\right]}_{x\=1}^{x\=2}$

$\=\frac{1088}{3}\left(2-1\right)\+\frac{16}{3}\left(2^3-1^3\right)$

$\=\frac{1200}{3}\=400$

Table 7.2.3(b)   Stepwise evaluation of the $\mathrm{dz} \mathrm{dy} \mathrm{dx}$ iteration

Define the integrand $f$ 

$f\=1\+2 x^2\+3 y^2\+4 z^2$$\stackrel{\text{assign}}{\to }$

Write and evaluate all six iterations of the triple integral

${\int }_1^2{\int }_{-1}^3{\int }_2^{4}f \mathit{\ⅆ}z \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $400$$$

${\int }_{-1}^3{\int }_1^2{\int }_2^{4}f \mathit{\ⅆ}z \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $400$$$

${\int }_2^4{\int }_1^2{\int }_{-1}^{3}f \mathit{\ⅆ}y \mathit{\ⅆ}x \mathit{\ⅆ}z$ = $400$$$

${\int }_1^2{\int }_2^4{\int }_{-1}^{3}f \mathit{\ⅆ}y \mathit{\ⅆ}z \mathit{\ⅆ}x$ = $400$$$

${\int }_2^4{\int }_{-1}^3{\int }_1^{2}f \mathit{\ⅆ}x \mathit{\ⅆ}y \mathit{\ⅆ}z$ = $400$$$

${\int }_{-1}^3{\int }_2^4{\int }_1^{2}f \mathit{\ⅆ}x \mathit{\ⅆ}z \mathit{\ⅆ}y$ = $400$$$

Tools≻Tasks≻Browse:

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

Table 7.2.3(c)   Solution by task template

Tools≻Tasks≻Browse:

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

Table 7.2.3(d)   Task template for visualizing regions of integration

Initialize

$f≔1\+2 x^2\+3 y^2\+4 z^2\:$

Write and evaluate all six iterations of the triple integral

$\mathrm{Int}\left(f\,\left[z\=2..4\,y\=-1..3\,x\=1..2\right]\right)\=\mathrm{int}\left(f\,\left[z\=2..4\,y\=-1..3\,x\=1..2\right]\right)$

${\∫}_1^2{\∫}_{-1}^3{\∫}_2^4\left(2x^2\+3y^2\+4z^2\+1\right)\ⅆz\ⅆy\ⅆx\=400$

$\mathrm{Int}\left(f\,\left[z\=2..4\,x\=1..2\,y\=-1..3\right]\right)\=\mathrm{int}\left(f\,\left[z\=2..4\,x\=1..2\,y\=-1..3\right]\right)$

${\∫}_{-1}^3{\∫}_1^2{\∫}_2^4\left(2x^2\+3y^2\+4z^2\+1\right)\ⅆz\ⅆx\ⅆy\=400$

$\mathrm{Int}\left(f\,\left[y\=-1..3\,x\=1..2\,z\=2..4\right]\right)\=\mathrm{int}\left(f\,\left[y\=-1..3\,x\=1..2\,z\=2..4\right]\right)$

${\∫}_2^4{\∫}_1^2{\∫}_{-1}^3\left(2x^2\+3y^2\+4z^2\+1\right)\ⅆy\ⅆx\ⅆz\=400$

$\mathrm{Int}\left(f\,\left[y\=-1..3\,z\=2..4\,x\=1..2\right]\right)\=\mathrm{int}\left(f\,\left[y\=-1..3\,z\=2..4\,x\=1..2\right]\right)$

${\∫}_1^2{\∫}_2^4{\∫}_{-1}^3\left(2x^2\+3y^2\+4z^2\+1\right)\ⅆy\ⅆz\ⅆx\=400$

$\mathrm{Int}\left(f\,\left[x\=1..2\,y\=-1..3\,z\=2..4\right]\right)\=\mathrm{int}\left(f\,\left[x\=1..2\,y\=-1..3\,z\=2..4\right]\right)$

${\∫}_2^4{\∫}_{-1}^3{\∫}_1^2\left(2x^2\+3y^2\+4z^2\+1\right)\ⅆx\ⅆy\ⅆz\=400$

$\mathrm{Int}\left(f\,\left[x\=1..2\,z\=2..4\,y\=-1..3\right]\right)\=\mathrm{int}\left(f\,\left[x\=1..2\,z\=2..4\,y\=-1..3\right]\right)$

${\∫}_{-1}^3{\∫}_2^4{\∫}_1^2\left(2x^2\+3y^2\+4z^2\+1\right)\ⅆx\ⅆz\ⅆy\=400$

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{MultiInt}\left(f\,z\=2..4\,y\=-1..3\,x\=1..2\,\mathrm{output}\=\mathrm{steps}\right)$

$400$

Table 7.2.3(e)   Stepwise evaluation of an iterated triple-integral

$\mathrm{MultiInt}\left(f\,\left[x\,y\,z\right]\=\mathrm{Parallelepiped}\left(1..2\,-1..3\,2..4\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_2^4{\∫}_{-1}^3{\∫}_1^2\left(2x^2\+3y^2\+4z^2\+1\right)\ⅆx\ⅆy\ⅆz$

$\mathrm{MultiInt}\left(f\,\left[x\,y\,z\right]\=\mathrm{Parallelepiped}\left(1..2\,-1..3\,2..4\right)\right)$ = $400$$$

Table 7.2.3(f)   The Parallelepiped option in the MultiInt command, Student MultivariateCalculus