${\int }_0^2{\int }_0^4{\int }_0^{\sqrt{4-x^2}}1 \mathrm{dz} \mathrm{dy} \mathrm{dx}$ = $4 \mathrm{\π}$

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Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_0^2{\∫}_0^4{\∫}_0^{\sqrt{-x^2\+4}}1\ⅆz\ⅆy\ⅆx$$\=$$4\mathrm{\π}$$$

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

Table 8.1.27(a)   Task template integrating in Cartesian coordinates

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

Table 8.1.27(b)   Integration via first principles

$\mathrm{MultiInt}\left(1\,z\=0..\sqrt{4-x^2}\,y\=0..4\,x\=0..2\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^2{\∫}_0^4{\∫}_0^{\sqrt{-x^2\+4}}1\ⅆz\ⅆy\ⅆx$

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

Table 8.1.27(c)   MultiInt command iterating in Cartesian coordinates in the order $\mathrm{dy} \mathrm{dz} \mathrm{dx}$ 

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

${\∫}_0^2{\∫}_0^4{\∫}_0^{\sqrt{-x^2\+4}}1\ⅆz\ⅆy\ⅆx\=4\mathrm{\π}$

Table 8.1.27(d)   Top-level Int and int commands