${\int }_0^a{\int }_0^{b \left(1-x\/a\right)}1 \mathrm{dy} \mathrm{dx}$ = $\frac{a b}{2}$ 

${\int }_0^b{\int }_0^{a\left(1-y\/b\right)}1 \mathrm{dx} \mathrm{dy}$ = $\frac{a b}{2}$ 

Loading Student:-Precalculus

$\left[0\,b\right]\,\left[a\,0\right]$$\stackrel{\text{equation of line}}{\to }$$y\=-\frac{bx}{a}\+b$$\stackrel{\text{isolate for x}}{\to }$$x\=\frac{a\left(b-y\right)}{b}$

Table 6.1.5(a)   Obtaining the equation of the hypotenuse for the triangle defining region $R$

Initialize

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_0^a{\∫}_0^{b\left(1-\frac{x}{a}\right)}1\ⅆy\ⅆx$$\=$$\frac{1}{2}ba$

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

${\int }_0^a{\int }_0^{b \left(1-x\/a\right)}1 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{1}{2}ba$

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

${\int }_0^b{\int }_0^{a \left(1-y\/b\right)}1 \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{1}{2}ba$

Table 6.1.5(b)   Iterated double-integrals for finding the area of region $R$ 

Initialize

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

Obtain the equation of the hypotenuse in region $R$ 

Apply the isolate command to the equation returned by the Line and GetRepresentation commands.

$\mathrm{isolate}\left(\mathrm{GetRepresentation}\left(\mathrm{Line}\left(\left[0\,b\right]\,\left[a\,0\right]\right)\,\mathrm{form}\=\mathrm{equation}\right)\,x\right)$

$x\=\frac{ab-ay}{b}$

Top-level, using the Int and int commands

$\mathrm{Int}\left(1\,\left[y\=0..b \left(1-x\/a\right)\,x\=0..a\right]\right)\=\mathrm{int}\left(1\,\left[y\=0..b \left(1-x\/a\right)\,x\=0..a\right]\right)$

${\∫}_0^a{\∫}_0^{b\left(1-\frac{x}{a}\right)}1\ⅆy\ⅆx\=\frac{1}{2}ba$

$\mathrm{Int}\left(1\,\left[x\=0..a \left(1-y\/b\right)\,y\=0..b\right]\right)\=\mathrm{int}\left(1\,\left[x\=0..a \left(1-y\/b\right)\,y\=0..b\right]\right)$

${\∫}_0^b{\∫}_0^{a\left(1-\frac{y}{b}\right)}1\ⅆx\ⅆy\=\frac{1}{2}ba$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(1\,y\=0..b \left(1-x\/a\right)\,x\=0..a\right)$ = $\frac{1}{2}ba$

$\mathrm{MultiInt}\left(1\,y\=0..b \left(1-x\/a\right)\,x\=0..a\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^a{\∫}_0^{b\left(1-\frac{x}{a}\right)}1\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(1\,y\=0..b \left(1-x\/a\right)\,x\=0..a\,\mathrm{output}\=\mathrm{steps}\right)$

$\frac{1}{2}ba$

Use the MultiInt command with a pre-defined domain option

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(0..a\,0.. b \left(1-x\/a\right)\right)\right)$ = $\frac{1}{2}ba$

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(0..a\,0.. b \left(1-x\/a\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^a{\∫}_0^{b\left(1-\frac{x}{a}\right)}1\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(1\,\left[y\,x\right]\=\mathrm{Region}\left(0..b\,0..a \left(1-y\/b\right)\right)\right)$ = $\frac{1}{2}ba$

$\mathrm{MultiInt}\left(1\,\left[y\,x\right]\=\mathrm{Region}\left(0..b\,0..a \left(1-y\/b\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^b{\∫}_0^{a\left(1-\frac{y}{b}\right)}1\ⅆx\ⅆy$