Initialize

$F\=3 x^2\+2 y^2\+1$$\stackrel{\text{assign}}{\to }$

$f\=\arctan \left(x\+1\right)-1\/2$$\stackrel{\text{assign}}{\to }$

$g\=\sin \left(x\right)$$\stackrel{\text{assign}}{\to }$

Obtain the intersections of the curves bounding $R$ 

$f\=g$

$\arctan \left(x\+1\right)-\frac{1}{2}\=\sin \left(x\right)$

$\stackrel{\text{solve}}{\to }$

$0.5059163080$

$\stackrel{\text{assign to a name}}{\to }$

$X$

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

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

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

${\int }_0^X{\int }_g^f\sqrt{1\+36 x^2\+16 y^2} \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_X^{\mathrm{\pi }\/2}{\int }_f^g\sqrt{1\+36 x^2\+16 y^2} \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $1.898369230$ 

Table 6.3.4(a)   Solution from first principles

Initialize

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

$F≔3 x^2\+2 y^2\+1\:$

$$\begin{gathered} f≔\arctan \left(x\+1\right)-1\/2\: \\ g≔\sin \left(x\right)\: \end{gathered}$$

Solve the equation $f\left(x\right)\=g\left(x\right)$ for $x$ 

$X≔\mathrm{fsolve}\left(f\=g\,x\right)$

$0.5059163080$

Apply the SurfaceArea command from the Student MultivariateCalculus package

$$\begin{gathered} \mathrm{SurfaceArea}\left(F\,x\=0..X\,y\=g..f\,\mathrm{output}\=\mathrm{integral}\right)\+ \\ \mathrm{SurfaceArea}\left(F\,x\=X..\mathrm{\π}\/2\,y\=f..g\,\mathrm{output}\=\mathrm{integral}\right) \end{gathered}$$

${\∫}_0^{0.5059163080}{\∫}_{\sin \left(x\right)}^{\arctan \left(x\+1\right)-\frac{1}{2}}\sqrt{36x^2\+16y^2\+1}\ⅆy\ⅆx\+{\∫}_{0.5059163080}^{\frac{1}{2}\mathrm{\π}}{\∫}_{\arctan \left(x\+1\right)-\frac{1}{2}}^{\sin \left(x\right)}\sqrt{36x^2\+16y^2\+1}\ⅆy\ⅆx$

$\mathrm{SurfaceArea}\left(F\,x\=0..X\,y\=g..f\right)\+\mathrm{SurfaceArea}\left(F\,x\=X..\mathrm{\π}\/2\,y\=f..g\right)$ = $1.898369230$

Apply the SurfaceInt command from the Student VectorCalculus package

$$\begin{gathered} \mathrm{Student}:-\mathrm{VectorCalculus}:-\mathrm{SurfaceInt}\left(1\,\left[x\,y\,z\right]\=\mathrm{Surface}\left(⟨x\,y\,F⟩\,x\=0..X\,y\=g..f\right)\,\mathrm{output}\=\mathrm{integral}\right)\+ \\ \mathrm{Student}:-\mathrm{VectorCalculus}:-\mathrm{SurfaceInt}\left(1\,\left[x\,y\,z\right]\=\mathrm{Surface}\left(⟨x\,y\,F⟩\,x\=X..\mathrm{\π}\/2\,y\=f..g\right)\,\mathrm{output}\=\mathrm{integral}\right) \end{gathered}$$

${\∫}_0^{0.5059163080}{\∫}_{\sin \left(x\right)}^{\arctan \left(x\+1\right)-\frac{1}{2}}\sqrt{36x^2\+16y^2\+1}\ⅆy\ⅆx\+{\∫}_{0.5059163080}^{\frac{1}{2}\mathrm{\π}}{\∫}_{\arctan \left(x\+1\right)-\frac{1}{2}}^{\sin \left(x\right)}\sqrt{36x^2\+16y^2\+1}\ⅆy\ⅆx$

$\mathrm{Int}\left(\sqrt{1\+36 x^2\+16 y^2}\,\left[y\=g..f\,x\=0..X\right]\right)\+\mathrm{Int}\left(\sqrt{1\+36 x^2\+16 y^2}\,\left[y\=f..g\,x\=X..\mathrm{\π}\/2\right]\right)\=\mathrm{int}\left(\sqrt{1\+36 x^2\+16 y^2}\,\left[y\=g..f\,x\=0..X\right]\right)\+\mathrm{int}\left(\sqrt{1\+36 x^2\+16 y^2}\,\left[y\=f..g\,x\=X..\mathrm{\pi }\/2\right]\right)$

${\∫}_0^{0.5059163080}{\∫}_{\sin \left(x\right)}^{\arctan \left(x\+1\right)-\frac{1}{2}}\sqrt{36x^2\+16y^2\+1}\ⅆy\ⅆx\+{\∫}_{0.5059163080}^{\frac{1}{2}\mathrm{\π}}{\∫}_{\arctan \left(x\+1\right)-\frac{1}{2}}^{\sin \left(x\right)}\sqrt{36x^2\+16y^2\+1}\ⅆy\ⅆx\=1.898369230$