${\int }_0^{\sqrt{3}}{\int }_0^{3-x^2}{\int }_0^{\left(x^2\+z^2\right)\/3}1 \mathrm{dy} \mathrm{dz} \mathrm{dx}$ = $\frac{62}{35}\sqrt{3}$

Initialize

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_0^{\sqrt{3}}{\∫}_0^{-x^2\+3}{\∫}_0^{\frac{1}{3}{x}^2\+\frac{1}{3}{z}^2}1\ⅆy\ⅆz\ⅆx$$\=$$\frac{62}{35}\sqrt{3}$$$

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

Table 8.1.6(a)   Task template integrating in Cartesian coordinates

${\int }_0^{\sqrt{3}}{\int }_0^{3-x^2}{\int }_0^{\left(x^2\+z^2\right)\/3}1 \mathit{\ⅆ}y \mathit{\ⅆ}z \mathit{\ⅆ}x$ = $\frac{62}{35}\sqrt{3}$$$

Table 8.1.6(b)   Integration via first principles

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{MultiInt}\left(1\,y\=0..\left(x^2\+z^2\right)\/3\,z\=0..3-x^2\,x\=0..\sqrt{3}\right)$ = $\frac{62}{35}\sqrt{3}$$$

Table 8.1.6(c)   MultiInt command iterating in Cartesian coordinates in the order $\mathrm{dy} \mathrm{dz} \mathrm{dx}$ 

$$\begin{gathered} \mathrm{Int}\left(1\,\left[y\=0..\left(x^2\+z^2\right)\/3\,z\=0..3-x^2\,x\=0..\sqrt{3}\right]\right)\= \\ \mathrm{int}\left(1\,\left[y\=0..\left(x^2\+z^2\right)\/3\,z\=0..3-x^2\,x\=0..\sqrt{3}\right]\right) \end{gathered}$$

${\∫}_0^{\sqrt{3}}{\∫}_0^{-x^2\+3}{\∫}_0^{\frac{1}{3}{x}^2\+\frac{1}{3}{z}^2}1\ⅆy\ⅆz\ⅆx\=\frac{62}{35}\sqrt{3}$

Table 8.1.6(d)   Top-level Int and int commands