Next, we need to compute ${\vec r_x} \times {\vec r_y}$. Here is that work.

$${\vec r_x} = \left\langle {1,0, - \frac{1}{3}} \right\rangle \hspace{0.5in}{\vec r_y} = \left\langle {0,1, - \frac{1}{2}} \right\rangle$$ $${\vec r_x} \times {\vec r_y} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\1&0&{ - \frac{1}{3}}\\0&1&{ - \frac{1}{2}}\end{array}} \right| = \frac{1}{3}\vec i + \frac{1}{2}\vec j + \vec k$$

Now, we what we really need is,

$$\left\| {{{\vec r}_x} \times {{\vec r}_y}} \right\| = \sqrt {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2} + 1} = \sqrt {\frac{{49}}{{36}}} = \frac{7}{6}$$ Show Step 3

The integral for the surface area is then,

$$A = \iint\limits_{D}{{\frac{7}{6}\,dA}}$$

In this case $D$ is just the restriction on $x$ and $y$ that we noted in Step 1. So, $D$ is just the disk ${x^2} + {y^2} \le 7$.

Show Step 4

Computing the integral in this case is very simple. All we need to do is take advantage of the fact that,

$$\iint\limits_{D}{{dA}} = {\mbox{Area of }}D$$

So, the surface area is simply,

$$A = \iint\limits_{D}{{\frac{7}{6}\,dA}} = \frac{7}{6}\iint\limits_{D}{{dA}} = \frac{7}{6}\left[ {{\mbox{Area of }}D} \right] = \frac{7}{6}\left[ {\pi {{\left( {\sqrt 7 } \right)}^2}} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{49}}{6}\pi }}$$