${\int }_{-2}^2{\int }_{-2\sqrt{4-z^2}}^{2\sqrt{4-z^2}}{\int }_0^{5-y}1 \mathrm{dx} \mathrm{dy} \mathrm{dz}$ = $40 \mathrm{\π}$ 

Initialize

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$Z\=2\sqrt{4-z^2}$$\stackrel{\text{assign}}{\to }$

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_{-2}^2{\∫}_{-2\sqrt{-z^2\+4}}^{2\sqrt{-z^2\+4}}{\∫}_0^{5-y}1\ⅆx\ⅆy\ⅆz$$\=$$40\mathrm{\π}$$$

Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 3-D

Table 8.1.7(a)   Task template integrating in Cartesian coordinates

${\int }_{-2}^2{\int }_{-2\sqrt{4-z^2}}^{2\sqrt{4-z^2}}{\int }_0^{5-y}1 \mathit{\ⅆ}x \mathit{\ⅆ}y \mathit{\ⅆ}z$ = $40\mathrm{\π}$$$

${\int }_{-4}^4{\int }_{-\sqrt{16-y^2}\/2}^{\sqrt{16-y^2}\/2}{\int }_0^{5-y}1 \mathit{\ⅆ}x \mathit{\ⅆ}z \mathit{\ⅆ}y$ = $40\mathrm{\π}$$$

Table 8.1.7(b)   Integration via first principles

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{MultiInt}\left(1\,x\=0..5-y\,y\=-2\sqrt{4-z^2}..2\sqrt{4-z^2}\,z\=-2..2\right)$

$40\mathrm{\π}$

Table 8.1.7(c)   MultiInt command iterating in Cartesian coordinates in the order $\mathrm{dy} \mathrm{dz} \mathrm{dx}$ 

$$\begin{gathered} \mathrm{Int}\left(1\,\left[x\=0..5-y\,y\=-2\sqrt{4-z^2}..2\sqrt{4-z^2}\,z\=-2..2\right]\right)\= \\ \mathrm{int}\left(1\,\left[x\=0..5-y\,y\=-2\sqrt{4-z^2}..2\sqrt{4-z^2}\,z\=-2..2\right]\right) \end{gathered}$$

${\∫}_{-2}^2{\∫}_{-2\sqrt{-z^2\+4}}^{2\sqrt{-z^2\+4}}{\∫}_0^{5-y}1\ⅆx\ⅆy\ⅆz\=40\mathrm{\π}$

Table 8.1.7(d)   Top-level Int and int commands