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Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\int }_0^{\frac{10}{3}}{\int }_{-\frac{\sqrt{-6y^2+20y}}{2}}^{\frac{\sqrt{-6y^2+20y}}{2}}{\int }_{2x^2+3y^2}^{10y}1\ⅆz\ⅆx\ⅆy$$\=$$\frac{625\sqrt{6}\mathrm{\pi }}{108}$$$

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

Table 8.1.24(a)   Task template integrating in Cartesian coordinates

${\int }_0^{\frac{10}{3}}{\int }_{-\sqrt{5 y-2 y^2}}^{\sqrt{5 y-\frac{3}{2} y^2}}{\int }_{2 x^2\+3 y^2}^{10 y}1 \mathit{\ⅆ}z \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{8125\sqrt{2}\mathrm{\pi }}{2048}+\frac{625\sqrt{2}\sqrt{3}\mathrm{\pi }}{216}+\frac{1076875\mathrm{I}\sqrt{2}}{124416}-\frac{8125\sqrt{2}\arccos \left(\frac{5}{3}\right)}{2048}$$$

Table 8.1.24(b)   Integration via first principles

$\mathrm{MultiInt}\left(1\,z\=2 x^2\+3 y^2..10 y\,x\=-\sqrt{5 y-3 y^2\/2}..\sqrt{5 y-3 y^2\/2}\,y\=0..10\/3\,\mathrm{output}\=\mathrm{integral}\right)$

${\int }_0^{\frac{10}{3}}{\int }_{-\frac{\sqrt{-6y^2+20y}}{2}}^{\frac{\sqrt{-6y^2+20y}}{2}}{\int }_{2x^2+3y^2}^{10y}1\ⅆz\ⅆx\ⅆy$

$\mathrm{MultiInt}\left(1\,z\=2 x^2\+3 y^2..10 y\,x\=-\sqrt{5 y-3 y^2\/2}..\sqrt{5 y-3 y^2\/2}\,y\=0..10\/3\right)$ = $\frac{625\sqrt{6}\mathrm{\pi }}{108}$$$

Table 8.1.4(c)   MultiInt command iterating in Cartesian coordinates in the order $\mathrm{dz} \mathrm{dx} \mathrm{dy}$ 

$$\begin{gathered} \mathrm{Int}\left(1\,\left[z\=2 x^2\+3 y^2..10 y\,x\=-\sqrt{5 y-3 y^2\/2}..\sqrt{5 y-3 y^2\/2}\,y\=0..10\/3\right]\right)\= \\ \mathrm{int}\left(1\,\left[z\=2 x^2\+3 y^2..10 y\,x\=-\sqrt{5 y-3 y^2\/2}..\sqrt{5 y-3 y^2\/2}\,y\=0..10\/3\right]\right) \end{gathered}$$

${\int }_0^{\frac{10}{3}}{\int }_{-\frac{\sqrt{-6y^2+20y}}{2}}^{\frac{\sqrt{-6y^2+20y}}{2}}{\int }_{2x^2+3y^2}^{10y}1\ⅆz\ⅆx\ⅆy=\frac{625\sqrt{6}\mathrm{\pi }}{108}$

Table 8.1.24(d)   Top-level Int and int commands