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

Table 6.3.1(a)   Solution by SurfaceArea command implemented in a task template

Tools≻Tasks≻Browse:
Calculus - Vector≻Integration≻Surface Integration≻Surface Defined over a 2-D Region

Table 6.3.1(b)   Solution by the SurfaceInt command implemented in a task template 

Initialize

$F\=x y$$\stackrel{\text{assign}}{\to }$

Obtain $\mathrm{d\σ}\prime \=\sqrt{1\+F_x^2\+F_y^2}$ 

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

Write and evaluate the relevant iterated double integral

Evaluate the inner integral exactly, then form the outer integral and evaluate it numerically

${\int }_0^{x \left(1-x\right)}\sqrt{1\+x^2\+y^2}\ⅆy$

$-\frac{\ln \left(x^2+1\right)x^2}{4}-\frac{\ln \left(x^2+1\right)}{4}+\frac{x\sqrt{x^4-2x^3+2x^2+1}}{2}-\frac{x^2\sqrt{x^4-2x^3+2x^2+1}}{2}+\frac{\ln \left(-x^2+\sqrt{x^4-2x^3+2x^2+1}+x\right)x^2}{2}+\frac{\ln \left(-x^2+\sqrt{x^4-2x^3+2x^2+1}+x\right)}{2}$

$\stackrel{\text{integrate w.r.t. x}}{\to }$

${\int }_0^1\left(-\frac{\ln \left(x^2+1\right)x^2}{4}-\frac{\ln \left(x^2+1\right)}{4}+\frac{x\sqrt{x^4-2x^3+2x^2+1}}{2}-\frac{x^2\sqrt{x^4-2x^3+2x^2+1}}{2}+\frac{\ln \left(-x^2+\sqrt{x^4-2x^3+2x^2+1}+x\right)x^2}{2}+\frac{\ln \left(-x^2+\sqrt{x^4-2x^3+2x^2+1}+x\right)}{2}\right)\ⅆx$

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

$0.1903708066$

Table 6.3.1(c)   Solution from first principles

Initialize

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

$\mathrm{with}\left(\mathrm{IntegrationTools}\right)\:$$$

$F≔x y\:$

$$\begin{gathered} q_1≔\mathrm{SurfaceArea}\left(F\,x\=0..1\, y\=0..x \left(1-x\right)\,\mathrm{output}\=\mathrm{integral}\right)\; \\ q_2≔\mathrm{GetIntegrand}\left(q_1\right)\; \\ q_3≔\mathrm{Int}\left(\mathrm{value}\left(q_2\right)\,x\=0..1\right)\; \\ \mathrm{evalf}\left(q_3\right) \end{gathered}$$

$0.1903708066$

Alternate approach: Include a floating-point number in the argument to the SurfaceArea command

$\mathrm{SurfaceArea}\left(F\,x\=0..1.0\, y\=0..x \left(1-x\right)\right)$ = $0.1903708066$$$

Alternate approach: Apply the SurfaceInt command from the Student VectorCalculus package

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

${\int }_0^1{\int }_0^{x\left(1-x\right)}\sqrt{x^2+y^2+1}\ⅆy\ⅆx$

The top-level Int command returns the inert iterated integral

$Q≔\mathrm{Int}\left(\sqrt{1\+y^2\+x^2}\,\left[y\=0..x \left(1-x\right)\,x\=0..1\right]\right)$

$Q≔{\int }_0^1{\int }_0^{x\left(1-x\right)}\sqrt{x^2+y^2+1}\ⅆy\ⅆx$

Apply the evalf command to obtain a numeric evaluation of the iterated integral

$\mathrm{evalf}\left(Q\right)$ = $0.1903708065$$$