In Cartesian coordinates, the element of surface area is $\mathrm{d\σ}\=\sqrt{1\+z_x^2\+z_y^2} \mathrm{dA}\=\sqrt{1\+4 x^2\+4 y^2} \mathrm{dA}$, where $\mathrm{dA}$ is either $\mathrm{dy} \mathrm{dx}$ or $\mathrm{dx} \mathrm{dy}$. The equation of the ellipse in Cartesian coordinates is

${\left(x-2\right)}^2\+4{\left(y-3\right)}^2\=1$ 

Solving this equation for $y\=y\left(x\right)$ gives $y_{\pm }\=3 \pm \sqrt{1-{\left(x-2\right)}^2}\/2$. Hence the requisite surface integral is

${\int }_1^3{\int }_{y_{-}}^{y_{\+}}x y \left(x^2\+y^2\right) \sqrt{1\+4 x^2\+4 y^2} \mathrm{dy} \mathrm{dx}$ ≐ 1001.38

Maple can provide an exact evaluation of the inner integral, but then the outer integral must be evaluated numerically. Alternatively, simply evaluate the double integral numerically.

Another approach to integrating over the ellipse makes use of the change of coordinates

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

This is not a change to polar coordinates, but rather, a translation of the ellipse to the origin. In these coordinates, the equation of the ellipse itself becomes $r^2\left({\cos }^2\left(\mathrm{\θ}\right)\+4 {\sin }^2\left(\mathrm{\θ}\right)\right)\=1$, so that

$r\=1\/\sqrt{{\cos }^2\left(\mathrm{\θ}\right)\+4 {\sin }^2\left(\mathrm{\θ}\right)}$ 

Note that when Maple implements this same change of coordinates, it writes the ellipse as shown on the left below. The calculations relating that form to the form shown above are given on the right.

To obtain the element of surface area, define the surface by the position vector

$\mathbf{R}\=\left[\begin{array}{c}2\+r \cos \(\mathrm{\θ}\)\\ 3\+r \sin \(\mathrm{\θ}\)\\ \(2\+r \cos \(\mathrm{\θ}\)\)\(3\+r \sin \(\mathrm{\θ}\)\)\end{array}\right]$ 

and obtain $∥{\mathbf{R}}_r\times {\mathbf{R}}_{\mathrm{\θ}}∥\=r \sqrt{4r^2\+16r\cos \left(\mathrm{\theta }\right)\+24r\sin \left(\mathrm{\theta }\right)\+53}$ as that part of $\mathrm{d\σ}$ free of the differentials in $\mathrm{dA}$. The requisite surface integral is then

${\int }_0^{2 \mathrm{\π}}{\int }_0^{1\/\sqrt{{\cos }^2\left(\mathrm{\θ}\right)\+4 {\sin }^2\left(\mathrm{\θ}\right)}}\left(2\+r \cos \left(\mathrm{\θ}\right)\right)\left(3\+r \sin \left(\mathrm{\θ}\right)\right)\left({\left(2\+r \cos \left(\mathrm{\θ}\right)\right)}^2\+{\left(3\+r \sin \left(\mathrm{\θ}\right)\right)}^2\right) r \sqrt{4r^2\+16r\cos \left(\mathrm{\theta }\right)\+24r\sin \left(\mathrm{\theta }\right)\+53} \mathrm{dr} d\mathrm{\θ}$

which, when evaluated numerically, has approximate value 1001.38.

Table 9.6.7(a) provides a solution via task template. Anticipating that the surface integral cannot be evaluated exactly, a numeric evaluation appears in the table.

Table 9.6.7(b) contains a solution from first principles where Cartesian coordinates are used. Note that Maple's solution in Table 9.6.7(a) uses what appear to be polar coordinates but in reality are just the change of coordinates $x\=2\+r \cos \left(\mathrm{\θ}\right)\,y\=3\+r \sin \left(\mathrm{\θ}\right)$. Table 9.6.7(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 9.6.7(c) contains a solution in Cartesian coordinates.

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 9.6.7(b) triggers the Approximate option in the Context Panel!

Table 9.6.7(d) provides a solution based on the SurfaceInt command in the Student VectorCalculus package.

Table 9.6.7(e) provides a solution from first principles. This solution makes use of the change of coordinates defined by $x\=2\+r \cos \left(\mathrm{\θ}\right)\,y\=3\+r \sin \left(\mathrm{\θ}\right)$.