${\int }_0^1{\int }_0^{1-x^2}{\int }_0^{x^2\+z^2}1 \mathit{\ⅆ}y \mathit{\ⅆ}z \mathit{\ⅆ}x$ = $\frac{2}{7}$$$

Initialize

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_0^1{\∫}_0^{-x^2\+1}{\∫}_0^{x^2\+z^2}1\ⅆy\ⅆz\ⅆx$$\=$$\frac{2}{7}$

Tools≻Tasks≻Browse:

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

Table 7.3.20(a)   Solution by visualization task template

${\int }_0^1{\int }_0^{1-x^2}{\int }_0^{x^2\+z^2}1 \mathit{\ⅆ}y \mathit{\ⅆ}z \mathit{\ⅆ}x$ = $\frac{2}{7}$$$

Table 7.3.20(b)   Solution via iterated triple-integral template in the Calculus palette

Top-level: Int and int commands

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

${\∫}_0^1{\∫}_0^{-x^2\+1}{\∫}_0^{x^2\+z^2}1\ⅆy\ⅆz\ⅆx\=\frac{2}{7}$

The MultiInt command in the Student MultivariateCalculus package

Install the Student MultivariateCalculus package.

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

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

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

$\mathrm{MultiInt}\left(1\,y\=0..x^2\+z^2\,z\=0..1-x^2\,x\=0..1\right)$ = $\frac{2}{7}$$$

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

$\frac{2}{7}$

The MultiInt command with a pre-defined domain option

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

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

$\mathrm{MultiInt}\left(1\,\left[x\,z\,y\right]\=\mathrm{Region}\left(0..1\,0..1-x^2\,0.. x^2\+z^2\right)\right)$ = $\frac{2}{7}$$$