Initialize

Loading Student:-MultivariateCalculus

$F\=2 x^2\+3 y^2$$\stackrel{\text{assign}}{\to }$

$X_L\=y^2$$\stackrel{\text{assign}}{\to }$

$X_R\=\left(3-y\right)\/2$$\stackrel{\text{assign}}{\to }$

Obtain the intersections of the curves bounding $R$ 

$x\=y^2\,y\=3-2 x$$\stackrel{\text{solve}}{\to }$$\left\{x\=1\,y\=1\right\}\,\left\{x\=\frac{9}{4}\,y\=-\frac{3}{2}\right\}$

Access the MultiInt command via the Context Panel

$F$ = $2x^2\+3y^2$$\stackrel{\text{MultiInt}}{\to }$${\∫}_{-\frac{3}{2}}^1{\∫}_{y^2}^{\frac{3}{2}-\frac{1}{2}y}\left(2x^2\+3y^2\right)\ⅆx\ⅆy$$\=$$\frac{16875}{1792}$$$

Iterate in the order $\mathrm{dx} \mathrm{dy}$ via the template in the Calculus palette

${\int }_{-3\/2}^1{\int }_{y^2}^{\left(3-y\right)\/2}F \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{16875}{1792}$$$

Iterate in the order $\mathrm{dy} \mathrm{dx}$ via the template in the Calculus palette

${\int }_0^1{\int }_{-\sqrt{x}}^{{\sqrt{x}}_{}}F \ⅆy \ⅆx\+{\int }_1^{9\/4}{\int }_{-\sqrt{x}}^{3-2 x}F \ⅆy \ⅆx$ = $\frac{16875}{1792}$$$$$

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 2-D

Table 6.2.2(a)   Iteration in the order $\mathrm{dx} \mathrm{dy}$ via visualization task-template

Initialize

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

$F≔2 x^2\+3 y^2\:$

Top-level, using the Int and int commands

$\mathrm{Int}\left(F\,\left[x\=y^2..\left(3-y\right)\/2\,y\=-3\/2..1\right]\right)\=\mathrm{int}\left(F\,\left[x\=y^2..\left(3-y\right)\/2\,y\=-3\/2..1\right]\right)$

${\∫}_{-\frac{3}{2}}^1{\∫}_{y^2}^{\frac{3}{2}-\frac{1}{2}y}\left(2x^2\+3y^2\right)\ⅆx\ⅆy\=\frac{16875}{1792}$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(F\,x\=y^2..\left(3-y\right)\/2\,y\=-3\/2..1\right)$ = $\frac{16875}{1792}$$$

$\mathrm{MultiInt}\left(F\,x\=y^2..\left(3-y\right)\/2\,y\=-3\/2..1\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_{-\frac{3}{2}}^1{\∫}_{y^2}^{\frac{3}{2}-\frac{1}{2}y}\left(2x^2\+3y^2\right)\ⅆx\ⅆy$

$\mathrm{MultiInt}\left(F\,x\=y^2..\left(3-y\right)\/2\,y\=-3\/2..1\,\mathrm{output}\=\mathrm{steps}\right)$

$\frac{16875}{1792}$

Use the MultiInt command with a pre-defined domain option

$\mathrm{MultiInt}\left(F\,\left[y\,x\right]\=\mathrm{Region}\left(-3\/2..1\,y^2..\left(3-y\right)\/2\right)\right)$ = $\frac{16875}{1792}$$$

$\mathrm{MultiInt}\left(F\,\left[y\,x\right]\=\mathrm{Region}\left(-3\/2..1\,y^2..\left(3-y\right)\/2\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_{-\frac{3}{2}}^1{\∫}_{y^2}^{\frac{3}{2}-\frac{1}{2}y}\left(2x^2\+3y^2\right)\ⅆx\ⅆy$