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Calculus - Vector≻Integration≻Surface Integration≻Surface Defined over a 2-D Region

Table 6.3.2(a)   Solution by task template

Initialize

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

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

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

Write and evaluate the relevant iterated integral

${\int }_{-3\/2}^1{\int }_{y^2}^{\left(3-y\right)\/2}\sqrt{1\+16 x^2\+36 y^2}\mathit{\ⅆ}x \mathit{\ⅆ}y$

${\int }_{-\frac{3}{2}}^1{\int }_{y^2}^{\frac{3}{2}-\frac{y}{2}}\sqrt{16x^2+36y^2+1}\ⅆx\ⅆy$

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

$13.91521368$

Table 6.3.2(b)   Solution from first principles

Initialize

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

$F≔2 x^2\+3 y^2\:$

Apply the SurfaceArea command from the Student MultivariateCalculus package

$\mathrm{SurfaceArea}\left(F\,y\=-3\/2..1\,x\=y^2..\left(3-y\right)\/2\,\mathrm{output}\=\mathrm{integral}\right)$

${\int }_{-\frac{3}{2}}^1{\int }_{y^2}^{\frac{3}{2}-\frac{y}{2}}\sqrt{16x^2+36y^2+1}\ⅆx\ⅆy$

$q≔\mathrm{SurfaceArea}\left(F\,y\=-3\/2..1\,x\=y^2..\left(3-y\right)\/2\right)$

$q≔{\int }_{-\frac{3}{2}}^1\left(-\frac{y^2\sqrt{16y^4+36y^2+1}}{2}-\frac{9\ln \left(4y^2+\sqrt{16y^4+36y^2+1}\right)y^2}{2}-\frac{\ln \left(4y^2+\sqrt{16y^4+36y^2+1}\right)}{8}+\frac{3\sqrt{40y^2-24y+37}}{4}-\frac{\sqrt{40y^2-24y+37}y}{4}+\frac{9\ln \left(6-2y+\sqrt{40y^2-24y+37}\right)y^2}{2}+\frac{\ln \left(6-2y+\sqrt{40y^2-24y+37}\right)}{8}\right)\ⅆy$

$\mathrm{evalf}\left(q\right)$ = $13.91521369$

Alternate approach: Include a floating-point number as one of the arguments

$\mathrm{SurfaceArea}\left(F\,y\=-3\/2..1.0\,x\=y^2..\left(3-y\right)\/2\right)$ = $13.91521369$

Apply the SurfaceInt command from the Student VectorCalculus package

$\mathrm{Student}:-\mathrm{VectorCalculus}:-\mathrm{SurfaceInt}\left(1\,\left[y\,x\,z\right]\=\mathrm{Surface}\left(⟨x\,y\,F⟩\,y\=-3\/2..1\, x\=y^2..\left(3-y\right)\/2\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\int }_{-\frac{3}{2}}^1{\int }_{y^2}^{\frac{3}{2}-\frac{y}{2}}\sqrt{16x^2+36y^2+1}\ⅆx\ⅆy$

The top-level Int command returns the inert iterated integral

$Q≔\mathrm{Int}\left(\sqrt{1\+16 x^2\+36 y^2}\,\left[x\=y^2..\left(3-y\right)\/2\,y\=-3\/2..1\right]\right)$

$Q≔{\int }_{-\frac{3}{2}}^1{\int }_{y^2}^{\frac{3}{2}-\frac{y}{2}}\sqrt{16x^2+36y^2+1}\ⅆx\ⅆy$

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

$\mathrm{evalf}\left(Q\right)$ = $13.91521368$