${\int }_{-2}^2{\int }_{-\sqrt{4-y^2}}^{{\sqrt{4-y^2}}_{}}{\int }_0^{4-x^2-y^2}1 \mathit{\ⅆ}z \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $8\mathrm{\π}$

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Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_{-2}^2{\∫}_{-\sqrt{-y^2\+4}}^{\sqrt{-y^2\+4}}{\∫}_0^{-x^2-y^2\+4}1\ⅆz\ⅆx\ⅆy$$\=$$8\mathrm{\π}$

Tools≻Tasks≻Browse:

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

Table 7.3.15(a)   Solution by visualization task template

${\int }_{-2}^2{\int }_{-\sqrt{4-y^2}}^{{\sqrt{4-y^2}}_{}}{\int }_0^{4-x^2-y^2}1 \mathit{\ⅆ}z \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $8\mathrm{\π}$$$

Table 7.3.15(b)   Solution via iterated triple-integral template in the Calculus palette

Top-level: Int and int commands

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

${\∫}_{-2}^2{\∫}_{-\sqrt{-y^2\+4}}^{\sqrt{-y^2\+4}}{\∫}_0^{-x^2-y^2\+4}1\ⅆz\ⅆx\ⅆy\=8\mathrm{\π}$

The MultiInt command in the Student MultivariateCalculus package

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$

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

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

$\mathrm{MultiInt}\left(1\,z\=0..4-x^2-y^2\, x\=-\sqrt{4-y^2}..\sqrt{4-y^2}\,y\=-2..2\right)$ = $8\mathrm{\π}$$$

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

$8\mathrm{\π}$

The MultiInt command with a pre-defined domain option

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

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

$\mathrm{MultiInt}\left(1\,\left[y\,x\,z\right]\=\mathrm{Region}\left(-2..2\,-\sqrt{4-y^2}..\sqrt{4-y^2}\,0..4-x^2-y^2\right)\right)$ = $8\mathrm{\π}$$$