The chosen order of iteration results in the sum

${\int }_1^5{\int }_{Y_{\mathrm{AB}}}^{Y_{\mathrm{AC}}}1 \mathrm{dy} \mathrm{dx}\+{\int }_5^7{\int }_{Y_{\mathrm{AB}}}^{Y_{\mathrm{BC}}}1 \mathrm{dy} \mathrm{dx}$ = 16

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$\left[1\,3\right]\,\left[7\,4\right]$$\stackrel{\text{equation of line}}{\to }$$y\=\frac{1}{6}x\+\frac{17}{6}$$\stackrel{\text{right hand side}}{\to }$$\frac{1}{6}x\+\frac{17}{6}$$\stackrel{\text{assign to a name}}{\to }$$Y_{\mathrm{AB}}$$$

$\left[7\,4\right]\,\left[5\,9\right]$$\stackrel{\text{equation of line}}{\to }$$y\=-\frac{5}{2}x\+\frac{43}{2}$$\stackrel{\text{right hand side}}{\to }$$-\frac{5}{2}x\+\frac{43}{2}$$\stackrel{\text{assign to a name}}{\to }$$Y_{\mathrm{BC}}$

$\left[1\,3\right]\,\left[5\,9\right]$$\stackrel{\text{equation of line}}{\to }$$y\=\frac{3}{2}x\+\frac{3}{2}$$$$\stackrel{\text{right hand side}}{\to }$$\frac{3}{2}x\+\frac{3}{2}$$\stackrel{\text{assign to a name}}{\to }$$Y_{\mathrm{AC}}$

Table 6.1.6(a)   Obtaining the equations of the edges of the triangle defining region $R$

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

${\int }_1^5{\int }_{Y_{\mathrm{AB}}}^{Y_{\mathrm{AC}}}1 \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_5^7{\int }_{Y_{\mathrm{AB}}}^{Y_{\mathrm{BC}}}1 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $16$$$

Table 6.1.6(b)   Iterated double-integrals for finding the area of region $R$ 

Tools≻Tasks≻Browse:

Calculus - Vector≻Integration≻Multiple Integration≻2-D≻Over a Triangle

Table 6.1.6(c)   Task-template for integrating over a triangle

Initialize

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

Obtain the equations of the edges of the triangle forming region $R$ 

$$\begin{gathered} Y_{\mathrm{AB}}≔\mathrm{rhs}\left(\mathrm{isolate}\left(\mathrm{GetRepresentation}\left(\mathrm{Line}\left(\left[1\,3\right]\,\left[7\,4\right]\right)\,\mathrm{form}\=\mathrm{equation}\right)\,y\right)\right)\: \\ {Y}_{\mathrm{BC}}≔\mathrm{rhs}\left(\mathrm{isolate}\left(\mathrm{GetRepresentation}\left(\mathrm{Line}\left(\left[7\,4\right]\,\left[5\,9\right]\right)\,\mathrm{form}\=\mathrm{equation}\right)\,y\right)\right)\: \\ {Y}_{\mathrm{AC}}≔\mathrm{rhs}\left(\mathrm{isolate}\left(\mathrm{GetRepresentation}\left(\mathrm{Line}\left(\left[1\,3\right]\,\left[5\,9\right]\right)\,\mathrm{form}\=\mathrm{equation}\right)\,y\right)\right)\: \end{gathered}$$

Top-level, using the Int and int commands

$$\begin{gathered} \mathrm{Int}\left(1\,\left[y\=Y_{\mathrm{AB}}..Y_{\mathrm{AC}}\,x\=1..5\right]\right)\+\mathrm{Int}\left(1\,\left[y\=Y_{\mathrm{AB}}..Y_{\mathrm{BC}}\,x\=5..7\right]\right)\= \\ \mathrm{int}\left(1\,\left[y\=Y_{\mathrm{AB}}..Y_{\mathrm{AC}}\,x\=1..5\right]\right)\+\mathrm{int}\left(1\,\left[y\=Y_{\mathrm{AB}}..Y_{\mathrm{BC}}\,x\=5..7\right]\right) \end{gathered}$$

${\∫}_1^5{\∫}_{\frac{1}{6}x\+\frac{17}{6}}^{\frac{3}{2}x\+\frac{3}{2}}1\ⅆy\ⅆx\+{\∫}_5^7{\∫}_{\frac{1}{6}x\+\frac{17}{6}}^{-\frac{5}{2}x\+\frac{43}{2}}1\ⅆy\ⅆx\=16$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(1\,y\=Y_{\mathrm{AB}}..Y_{\mathrm{AC}}\,x\=1..5\right)\+\mathrm{MultiInt}\left(1\,y\=Y_{\mathrm{AB}}..Y_{\mathrm{BC}}\,x\=5..7\right)$ = $16$$$

$$\begin{gathered} \mathrm{MultiInt}\left(1\,y\=Y_{\mathrm{AB}}..Y_{\mathrm{AC}}\,x\=1..5\,\mathrm{output}\=\mathrm{integral}\right)\+ \\ \mathrm{MultiInt}\left(1\,y\=Y_{\mathrm{AB}}..Y_{\mathrm{BC}}\,x\=5..7\,\mathrm{output}\=\mathrm{integral}\right) \end{gathered}$$

${\∫}_1^5{\∫}_{\frac{17}{6}\+\frac{1}{6}x}^{\frac{3}{2}\+\frac{3}{2}x}1\ⅆy\ⅆx\+{\∫}_5^7{\∫}_{\frac{17}{6}\+\frac{1}{6}x}^{\frac{43}{2}-\frac{5}{2}x}1\ⅆy\ⅆx$

Use the MultiInt command with a pre-defined domain option

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨1\,3⟩\,⟨7\,4⟩\,⟨5\,9⟩\right)\right)$ = $16$$$

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨1\,3⟩\,⟨7\,4⟩\,⟨5\,9⟩\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_1^5{\∫}_{\frac{3}{2}\+\frac{3}{2}x}^{\frac{17}{6}\+\frac{1}{6}x}\left(-1\right)\ⅆy\ⅆx\+{\∫}_5^7{\∫}_{\frac{43}{2}-\frac{5}{2}x}^{\frac{17}{6}\+\frac{1}{6}x}\left(-1\right)\ⅆy\ⅆx$