${\int }_0^5{\int }_0^{\left(5-x\right)\/3}{\int }_0^{5-x-3 y}1 \mathrm{dz} \mathrm{dy} \mathrm{dx}$ = $\frac{125}{18}$

Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Multiple Integration≻Cartesian 3-D

Table 8.1.1(a)   Task template implementing the MultiInt command iterating in the order $\mathrm{dz} \mathrm{dy} \mathrm{dx}$

Initialize

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_0^5{\∫}_0^{\frac{5}{3}-\frac{1}{3}x}{\∫}_0^{5-x-3y}1\ⅆz\ⅆy\ⅆx$$\=$$\frac{125}{18}$

${\int }_0^5{\int }_0^{\left(5-x\right)\/3}{\int }_0^{5-x-3 y}1 \mathit{\ⅆ}z \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{125}{18}$$$

Table 8.1.1(b)   Integration via first principles

$$\begin{gathered} V≔⟨0\,0\,0⟩\,⟨5\,0\,0⟩\,⟨0\,5\/3\,0⟩\,⟨0\,0\,5⟩\: \\ \mathrm{MultiInt}\left(1\,\left[x\,y\,z\right]\=\mathrm{Tetrahedron}\left(V\right)\,\mathrm{output}\=\mathrm{integral}\right) \end{gathered}$$

$5\left({\∫}_0^1{\∫}_0^{5-5t}{\∫}_0^{\frac{\left(-\frac{5}{3}\+\frac{5}{3}t\right)\left(x-5\+5t\right)}{5-5t}}1\ⅆy\ⅆx\ⅆt\right)$

$\mathrm{MultiInt}\left(1\,\left[x\,y\,z\right]\=\mathrm{Tetrahedron}\left(V\right)\right)$ = $\frac{125}{18}$$$

Table 8.1.1(c)   Integration over a tetrahedron

$\mathrm{MultiInt}\left(1\,z\=0..5-x-3 y\,y\=0..\left(5-x\right)\/3\,x\=0..5\right)$ = $\frac{125}{18}$$$$$

Table 8.1.1(d)   MultiInt command iterating in the order $\mathrm{dz} \mathrm{dy} \mathrm{dx}$

$$\begin{gathered} \mathrm{Int}\left(1\,\left[z\=0..5-x-3 y\,y\=0..\left(5-x\right)\/3\,x\=0..5\right]\right)\= \\ \mathrm{int}\left(1\,\left[z\=0..5-x-3 y\,y\=0..\left(5-x\right)\/3\,x\=0..5\right]\right) \end{gathered}$$

${\∫}_0^5{\∫}_0^{\frac{5}{3}-\frac{1}{3}x}{\∫}_0^{5-x-3y}1\ⅆz\ⅆy\ⅆx\=\frac{125}{18}$

Table 8.1.1(e)   Top-level Int and int commands