In Cartesian coordinates where, except for the differentials in $\mathrm{dA}$, the element of surface area is given by $\sqrt{1\+z_x^2\+z_y^2}\=\sqrt{1\+4 x^2\+4 y^2}$.

The disk above which the surface is defined is bounded by the circle whose Cartesian representation is

${\left(x-2\right)}^2\+{\left(y-3\right)}^2\=1$ 

from which is obtained $y_{\pm }\=3 \pm \sqrt{1-{\left(x-2\right)}^2}$.

The requisite surface integral is then

${\int }_1^3{\int }_{y_{-}}^{y_{\+}}x y \left(x^2\+y^2\right) \sqrt{1\+4 x^2\+4 y^2} \mathrm{dy} \mathrm{dx}$ ≐ 2147.26

Alternatively, a solution is obtained by the parametrization

$x\left(r\,\mathrm{\θ}\right)\=2\+r \cos \left(\mathrm{\θ}\right)\,y\left(r\,\mathrm{\θ}\right)\=3\+r \sin \left(\mathrm{\θ}\right)\,z\left(r\,\mathrm{\θ}\right)\=x^2\left(r\,\mathrm{\θ}\right)\+y^2\left(r\,\mathrm{\θ}\right)$ 

Define the surface by the position vector $\mathbf{R}\=x\left(r\,\mathrm{\θ}\right) \mathbf{i}\+y\left(r\,\mathrm{\θ}\right) \mathbf{j}\+z\left(r\,\mathrm{\θ}\right) \mathbf{k}$ so that, except for the differentials in $\mathrm{dA}$, the element of surface area becomes

$∥{\mathbf{R}}_r\times {\mathbf{R}}_{\mathrm{\theta }}∥$= $r\sqrt{4r^2\+16r\cos \left(\mathrm{\θ}\right)\+24r\sin \left(\mathrm{\θ}\right)\+53}$ 

because $$

${\mathbf{R}}_r\times {\mathbf{R}}_{\mathrm{\θ}}\=\left[\begin{array}{c}\cos \(\mathrm{\theta }\)\\ \sin \(\mathrm{\theta }\)\\ 2r\+4\cos \left(t\right)\+6\sin \left(t\right)\end{array}\right]\times \left[\begin{array}{c}-r\sin \left(\mathrm{\θ}\right)\\ r\cos \left(\mathrm{\θ}\right)\\ 2r\left(3\cos \left(\mathrm{\θ}\right)-2\sin \left(\mathrm{\θ}\right)\right)\end{array}\right]\=\left[\begin{array}{c}-2r\left(2\+r\cos \left(\mathrm{\θ}\right)\right)\\ -2r\left(3\+r\sin \left(\mathrm{\θ}\right)\right)\\ r\end{array}\right]$

The requisite surface integral is then

with numeric value approximately 2147.26.$$

Table 9.6.5(a) provides a solution via task template.

Table 9.6.5(b) contains a solution from first principles where Cartesian coordinates are used. Note that Maple's solution in Table 9.6.5(a) uses polar coordinates. Table 9.6.5(b) provides this same solution constructed from first principles. Anticipating that the surface integral has no closed-form solution, the integral is converted to its inert form so that numeric integration can be applied.

Table 9.6.5(c) gives a solution in Cartesian coordinates where, except for the differentials in $\mathrm{dA}$, the element of surface area is given by $\sqrt{1\+z_x^2\+z_y^2}\=\sqrt{1\+4 x^2\+4 y^2}$.

A regression in Maple 2020 forces the use of the evalf command rather than the Approximate option in the Context Panel. Surprisingly, the iterated integral in Table 9.6.5(b) triggers the Approximate option in the Context Panel!

Table 9.6.5(d) contains a solution based on the SurfaceInt command in the Student VectorCalculus package.