Next, we need to compute \({\vec r_u} \times {\vec r_v}\). Here is that work.

\[{\vec r_u} = \left\langle {2,v,0} \right\rangle \hspace{0.5in}{\vec r_v} = \left\langle {0,u, - 2} \right\rangle \] \[{\vec r_u} \times {\vec r_v} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\2&v&0\\0&u&{ - 2}\end{array}} \right| = - 2v\vec i + 4\vec j + 2u\vec k\]

Now, we what we really need is,

\[\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\| = \sqrt {{{\left( { - 2v} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( {2u} \right)}^2}} = \sqrt {4{u^2} + 4{v^2} + 16} = 2\sqrt {{u^2} + {v^2} + 4} \] Show Step 3

The integral for the surface area is then,

\[A = \iint\limits_{D}{{2\sqrt {{u^2} + {v^2} + 4} \,dA}}\]

Where \(D\) is the disk \({u^2} + {v^2} \le 4\).

Show Step 4

Because \(D\) is a disk the best bet for this integral is to use the following “version” of polar coordinates.

\[u = r\cos \theta \hspace{0.25in} v = r\sin \theta \hspace{0.25in} {u^2} + {v^2} = {r^2} \hspace{0.25in} dA = r\,dr\,d\theta \]

The polar coordinate limits for this \(D\) is,

\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\end{array}\]

So, the integral to converting to polar coordinates gives,

\[A = \iint\limits_{D}{{2\sqrt {{u^2} + {v^2} + 4} \,dA}} = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{2r\sqrt {{r^2} + 4} \,dr}}\,d\theta }}\] Show Step 5

Now we just need to evaluate the integral to get the surface area.

\[\begin{align*}A & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{2r\sqrt {{r^2} + 4} \,dr}}\,d\theta }} = \int_{0}^{{2\pi }}{{\left. {\frac{2}{3}{{\left( {{r^2} + 4} \right)}^{\frac{3}{2}}}} \right|_0^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{2}{3}\left( {{8^{\frac{3}{2}}} - {4^{\frac{3}{2}}}} \right)\,d\theta }} = \left. {\frac{2}{3}\left( {{8^{\frac{3}{2}}} - 8} \right)\theta } \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{3}\pi\left( {\sqrt 8 - 1} \right) = 61.2712}}\end{align*}\]