Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Surface Area

Table 6.3.5(a)   Solution by SurfaceArea command implemented in a task template

Tools≻Tasks≻Browse:

Calculus - Vector≻Integration≻Surface Integration≻Surface Defined over a Triangle

Table 6.3.5(b)   Task-template implementation of the SurfaceInt command

Loading Student:-Precalculus

$\left[0\,1\right]\,\left[1\,0\right]$$$$\stackrel{\text{equation of line}}{\to }$$y=1-x$

Table 6.3.5(c)   Obtaining the equation of the hypotenuse for the triangle defining region $R$

$F\=5-3 x^2-2 y^2$$\stackrel{\text{assign}}{\to }$

$\sqrt{1\+{\left(\frac{\partial }{\partial x} F\right)}^2\+{\left(\frac{\partial }{\partial y} F\right)}^2}$ = $\sqrt{36x^2+16y^2+1}$$$

Table 6.3.5(d)   Calculation of $\mathrm{\lambda }\=\sqrt{1\+F_x^2\+F_y^2}$

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

${\int }_0^1{\int }_0^{1-x}\sqrt{1\+36 x^2\+16 y^2}\mathit{\ⅆ}y \mathit{\ⅆ}x$

${\int }_0^1{\int }_0^{1-x}\sqrt{36x^2+16y^2+1}\ⅆy\ⅆx$

$\stackrel{\text{at 10 digits}}{\to }$

$1.469876015$

Table 6.2.5(e)   Iterated double-integral for finding the surface area of $F$ over $R$  

Initialize

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

$F≔5-3 x^2-2 y^2\:$

Obtain the equation of the hypotenuse in region $R$ 

$\mathrm{GetRepresentation}\left(\mathrm{Line}\left(\left[0\,1\right]\,\left[1\,0\right]\right)\,\mathrm{form}\=\mathrm{equation}\right)$

$x+y=1$

Obtain $\mathrm{\lambda }\=\sqrt{1\+F_x^2\+F_y^2}$

$\mathrm{\λ}≔\mathrm{sqrt}\left(1\+\mathrm{diff}{\left(F\,x\right)}^2\+\mathrm{diff}{\left(F\,y\right)}^2\right)$

$\mathrm{\lambda }≔\sqrt{36x^2+16y^2+1}$

Top-level, using the Int and evalfcommands

$$\begin{gathered} Q≔\mathrm{Int}\left(\mathrm{\λ}\,\left[y\=0..1-x\,x\=0..1\right]\right)\; \\ \mathrm{evalf}\left(Q\right) \end{gathered}$$

$1.469876015$

Use the MultiInt command from the Student MultivariateCalculus package

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

${\int }_0^1{\int }_0^{1-x}\sqrt{36x^2+16y^2+1}\ⅆy\ⅆx$

Use the MultiInt command with a pre-defined domain option

$\mathrm{MultiInt}\left(\mathrm{\λ}\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨1\,0⟩\,⟨0\,1⟩\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\int }_0^1{\int }_0^{1-x}\sqrt{36x^2+16y^2+1}\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(\mathrm{\lambda }\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨1.0\,0⟩\,⟨0\,1⟩\right)\right)$ = $1.469876015$$$

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

${\int }_0^1{\int }_0^{1-x}\sqrt{36x^2+16y^2+1}\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(\mathrm{\lambda }\,\left[x\,y\right]\=\mathrm{Region}\left(0..1.0\,0.. 1-x\right)\right)$ = $1.469876015$$$

Use the SurfaceInt command in the Student VectorCalculus package

$\mathrm{Student}:-\mathrm{VectorCalculus}:-\mathrm{SurfaceInt}\left(1\,\left[x\,y\,z\right]\=\mathrm{Surface}\left(⟨x\,y\,F⟩\,\left[x\,y\right]\=\mathrm{Triangle}\left(⟨0\,0⟩\,⟨1\,0⟩\,⟨0\,1⟩\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\int }_0^1{\int }_0^{1-x}\sqrt{36x^2+16y^2+1}\ⅆy\ⅆx$

$\mathrm{Student}:-\mathrm{VectorCalculus}:-\mathrm{SurfaceInt}\left(1\,\left[x\,y\,z\right]\=\mathrm{Surface}\left(⟨x\,y\,F⟩\,\left[x\,y\right]\=\mathrm{Region}\left(0..1\,0..1-x\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\int }_0^1{\int }_0^{1-x}\sqrt{36x^2+16y^2+1}\ⅆy\ⅆx$