${\int }_{-2}^2{\int }_{-\frac{\sqrt{4-x^2}}{2}}^{\frac{\sqrt{4-x^2}}{2}}{\int }_{x^2\+4 y^2}^{6-\frac{x^2}{2}-2 y^2}1 \mathit{\ⅆ}z \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $6\mathrm{\π}$$$

Initialize

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_{-2}^2{\∫}_{-\frac{1}{2}\sqrt{-x^2\+4}}^{\frac{1}{2}\sqrt{-x^2\+4}}{\∫}_{x^2\+4y^2}^{6-\frac{1}{2}{x}^2-2y^2}1\ⅆz\ⅆy\ⅆx$$\=$$6\mathrm{\π}$

Tools≻Tasks≻Browse:

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

Table 7.3.19(a)   Solution by visualization task template

${\int }_{-2}^2{\int }_{-\frac{\sqrt{4-x^2}}{2}}^{\frac{\sqrt{4-x^2}}{2}}{\int }_{x^2\+4 y^2}^{6-\frac{x^2}{2}-2 y^2}1 \mathit{\ⅆ}z \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $6\mathrm{\π}$$$

Table 7.3.19(b)   Solution via iterated triple-integral template in the Calculus palette

Tools≻Tasks≻Browse:

Calculus - Vector≻Integration≻Multiple Integration≻3-D≻Over a General 3-D Region

Table 7.3.19(c)   Solution by a VectorCalculus task template

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Multiple Integration≻Cartesian 3-D

Table 7.3.19(d)   Task template with the MultiInt command, Student MultivariateCalculus package

Top-level: Int and int commands

$$\begin{gathered} \mathrm{Int}\left(1\,\left[z\=x^2\+4 y^2..6-x^2\/2-2 y^2\,y\=-\sqrt{4-x^2}\/2..\sqrt{4-x^2}\/2\,x\=-2..2\right]\right)\= \\ \mathrm{int}\left(1\,\left[z\=x^2\+4 y^2..6-x^2\/2-2 y^2\,y\=-\sqrt{4-x^2}\/2..\sqrt{4-x^2}\/2\,x\=-2..2\right]\right) \end{gathered}$$

${\∫}_{-2}^2{\∫}_{-\frac{1}{2}\sqrt{-x^2\+4}}^{\frac{1}{2}\sqrt{-x^2\+4}}{\∫}_{x^2\+4y^2}^{6-\frac{1}{2}{x}^2-2y^2}1\ⅆz\ⅆy\ⅆx\=6\mathrm{\π}$

The MultiInt command in the Student MultivariateCalculus package

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

$\mathrm{MultiInt}\left(1\,z\=x^2\+4 y^2..6-x^2\/2-2 y^2\,y\=-\sqrt{4-x^2}\/2..\sqrt{4-x^2}\/2\,x\=-2..2\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_{-2}^2{\∫}_{-\frac{1}{2}\sqrt{-x^2\+4}}^{\frac{1}{2}\sqrt{-x^2\+4}}{\∫}_{x^2\+4y^2}^{6-\frac{1}{2}{x}^2-2y^2}1\ⅆz\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(1\,z\=x^2\+4 y^2..6-x^2\/2-2 y^2\,y\=-\sqrt{4-x^2}\/2..\sqrt{4-x^2}\/2\,x\=-2..2\right)$ = $6\mathrm{\π}$$$

$\mathrm{MultiInt}\left(1\,z\=x^2\+4 y^2..6-x^2\/2-2 y^2\,y\=-\sqrt{4-x^2}\/2..\sqrt{4-x^2}\/2\,x\=-2..2\,\mathrm{output}\=\mathrm{steps}\right)$

$6\mathrm{\π}$

$\mathrm{MultiInt}\left(1\,\left[x\,y\,z\right]\=\mathrm{Region}\left(-2..2\,-\frac{\sqrt{4-x^2}}{2}..\frac{\sqrt{4-x^2}}{2}\,x^2\+4 y^2..6-\frac{x^2}{2}-2 y^2\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_{-2}^2{\∫}_{-\frac{1}{2}\sqrt{-x^2\+4}}^{\frac{1}{2}\sqrt{-x^2\+4}}{\∫}_{x^2\+4y^2}^{6-\frac{1}{2}{x}^2-2y^2}1\ⅆz\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(1\,\left[x\,y\,z\right]\=\mathrm{Region}\left(-2..2\,-\frac{\sqrt{4-x^2}}{2}..\frac{\sqrt{4-x^2}}{2}\,x^2\+4 y^2..6-\frac{x^2}{2}-2 y^2\right)\right)$ = $6\mathrm{\π}$$$