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Loading Student:-MultivariateCalculus

$f\=1\+2 x^2\+3 y^2$$\stackrel{\text{assign}}{\to }$

$Y\=6\+x-x^2$$\stackrel{\text{assign}}{\to }$

$y\=Y$

$y\=-x^2\+x\+6$

$\stackrel{\text{solutions for x}}{\to }$

$\frac{1}{2}\+\frac{1}{2}\sqrt{25-4y}\,\frac{1}{2}-\frac{1}{2}\sqrt{25-4y}$

$\stackrel{\text{assign to a name}}{\to }$

$X$

$\mathrm{XL}\=X_2$$\stackrel{\text{assign}}{\to }$

$\mathrm{XR}\=X_1$$\stackrel{\text{assign}}{\to }$

Access the MultiInt command via the Context Panel

$f$ = $2x^2\+3y^2\+1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_{-2}^3{\∫}_0^{-x^2\+x\+6}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx$$\=$$\frac{53875}{84}$$$

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻

Table 5.3.2(a)   Visualizing $R$ and the resulting volume for iteration in the order $\mathrm{dy} \mathrm{dx}$ 

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻

Table 5.3.2(b)   Visualizing $R$ and the resulting volume for iteration in the order $\mathrm{dx} \mathrm{dy}$ 

Iterate $\mathrm{dy} \mathrm{dx}$ and $\mathrm{dx} \mathrm{dy}$ 

${\int }_{-2}^3{\int }_0^{Y}f \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{53875}{84}$$$

${\int }_0^{25\/4}{\int }_{\mathrm{XL}}^{\mathrm{XR}}f \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{53875}{84}$$$

Display the iterated integrals

${\int }_{-2}^3{\int }_0^{Y}f \ⅆy \ⅆx$

${\∫}_{-2}^3{\∫}_0^{-x^2\+x\+6}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx$

$\=$

$\frac{53875}{84}$

${\int }_0^{25\/4}{\int }_{\mathrm{XL}}^{\mathrm{XR}}f \ⅆx \ⅆy$

${\∫}_0^{\frac{25}{4}}{\∫}_{\frac{1}{2}-\frac{1}{2}\sqrt{25-4y}}^{\frac{1}{2}\+\frac{1}{2}\sqrt{25-4y}}\left(2x^2\+3y^2\+1\right)\ⅆx\ⅆy$

$\=$

$\frac{53875}{84}$

Initialize

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

$f≔1\+2 x^2\+3 y^2\:$

$Y≔6\+x-x^2\:$

Iterate in the order $\mathrm{dy} \mathrm{dx}$

$\mathrm{Int}\left(f\,\left[y\=0..Y\,x\=-2..3\right]\right)\=\mathrm{int}\left(f\,\left[y\=0.. Y\,x\=-2..3\right]\right)$

${\∫}_{-2}^3{\∫}_0^{-x^2\+x\+6}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx\=\frac{53875}{84}$

Iterate in the order $\mathrm{dx} \mathrm{dy}$ 

$X≔\mathrm{solve}\left(y\=Y\,x\right)$

$\frac{1}{2}\+\frac{1}{2}\sqrt{25-4y}\,\frac{1}{2}-\frac{1}{2}\sqrt{25-4y}$

$\mathrm{Int}\left(f\,\left[x\=X_2..X_1\,y\=0..\frac{25}{4}\right]\right)\=\mathrm{int}\left(f\,\left[x\=X_2..X_1\,y\=0..\frac{25}{4}\right]\right)$

${\∫}_0^{\frac{25}{4}}{\∫}_{\frac{1}{2}-\frac{1}{2}\sqrt{25-4y}}^{\frac{1}{2}\+\frac{1}{2}\sqrt{25-4y}}\left(2x^2\+3y^2\+1\right)\ⅆx\ⅆy\=\frac{53875}{84}$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(f\,y\=0..Y\,x\=-2..3\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_{-2}^3{\∫}_0^{-x^2\+x\+6}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(f\,x\=X_2..X_1\,y\=0..\frac{25}{4}\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^{\frac{25}{4}}{\∫}_{\frac{1}{2}-\frac{1}{2}\sqrt{25-4y}}^{\frac{1}{2}\+\frac{1}{2}\sqrt{25-4y}}\left(2x^2\+3y^2\+1\right)\ⅆx\ⅆy$

$\mathrm{MultiInt}\left(f\,y\=0..Y\,x\=-2..3\right)$

$\frac{53875}{84}$

$\mathrm{MultiInt}\left(f\,x\=X_2..X_1\,y\=0..\frac{25}{4}\right)$

$\frac{53875}{84}$

Obtain stepwise evaluations via the MultiInt command

$\mathrm{MultiInt}\left(f\,y\=0..Y\,x\=-2..3\,\mathrm{output}\=\mathrm{steps}\right)$

$\begin{array}{ccc}\multicolumn{3}{c}{{\int }_{\mathrm{-2}}^3{\int }_0^{-x^2+x+6}\left(2x^2+3y^2+1\right)\ⅆy\ⅆx}\\ & \text{=}& {\int }_{\mathrm{-2}}^3\left(\genfrac{}{}{0ex}{}{\left(2x^{2}y+y^3+y\right)}{\phantom{y=0..-x^2+x+6}}|\genfrac{}{}{0ex}{}{\phantom{\left(2x^{2}y+y^3+y\right)}}{y=0..-x^2+x+6}\right)\ⅆx\hfill \\ & \text{=}& {\int }_{\mathrm{-2}}^3\left(2x^2\left(-x^2+x+6\right)+{\left(-x^2+x+6\right)}^3-x^2+x+6\right)\ⅆx\hfill \\ & \text{=}& \genfrac{}{}{0ex}{}{\left(\frac{13}{5}{x}^5-\frac{33}{4}{x}^4-\frac{79}{3}{x}^3-\frac{1}{7}{x}^7+\frac{1}{2}{x}^6+\frac{109}{2}{x}^2+222x\right)}{\phantom{x=\mathrm{-2}..3}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{13}{5}{x}^5-\frac{33}{4}{x}^4-\frac{79}{3}{x}^3-\frac{1}{7}{x}^7+\frac{1}{2}{x}^6+\frac{109}{2}{x}^2+222x\right)}}{x=\mathrm{-2}..3}\hfill \end{array}$

$\mathrm{MultiInt}\left(f\,x\=X_2..X_1\,y\=0..\frac{25}{4}\,\mathrm{output}\=\mathrm{steps}\right)$

$\begin{array}{ccc}\multicolumn{3}{c}{{\int }_0^{\frac{25}{4}}{\int }_{\frac{1}{2}-\frac{\sqrt{25-4y}}{2}}^{\frac{1}{2}+\frac{\sqrt{25-4y}}{2}}\left(2x^2+3y^2+1\right)\ⅆx\ⅆy}\\ & \text{=}& {\int }_0^{\frac{25}{4}}\left(\genfrac{}{}{0ex}{}{\left(\frac{2}{3}{x}^3+3y^{2}x+x\right)}{\phantom{x=\frac{1}{2}-\frac{\sqrt{25-4y}}{2}..\frac{1}{2}+\frac{\sqrt{25-4y}}{2}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{2}{3}{x}^3+3y^{2}x+x\right)}}{x=\frac{1}{2}-\frac{\sqrt{25-4y}}{2}..\frac{1}{2}+\frac{\sqrt{25-4y}}{2}}\right)\ⅆy\hfill \\ & \text{=}& {\int }_0^{\frac{25}{4}}\left(\frac{2{\left(\frac{1}{2}+\frac{\sqrt{25-4y}}{2}\right)}^3}{3}-\frac{2{\left(\frac{1}{2}-\frac{\sqrt{25-4y}}{2}\right)}^3}{3}+3y^2\sqrt{25-4y}+\sqrt{25-4y}\right)\ⅆy\hfill \\ & \text{=}& \genfrac{}{}{0ex}{}{-\frac{{\left(25-4y\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(90y^2+422y+2155\right)}{420}}{\phantom{y=0..\frac{25}{4}}}|\genfrac{}{}{0ex}{}{\phantom{-\frac{{\left(25-4y\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(90y^2+422y+2155\right)}{420}}}{y=0..\frac{25}{4}}\hfill \end{array}$