As noted above the region \(D\) is going to be in the \(xz\)-plane and the surface is given in the form \(y = f\left( {x,z} \right)\). The formula for surface area we gave in the notes is only for a region \(D\) that is in the \(xy\)-plane with the surface given by \(z = f\left( {x,y} \right)\) .

However, it shouldn’t be too difficult to see that all we need to do is modify the formula in the following manner to get one for this setup.

\[S = \iint\limits_{D}{{\sqrt {{{\left[ {{f_x}} \right]}^2} + {{\left[ {{f_z}} \right]}^2} + 1} \,dA}}\hspace{0.25in}\,y = f\left( {x,z} \right)\hspace{0.25in}D\,\,{\mbox{is in the }}xz{\mbox{ - plane}}\]

So, the integral for the surface area is,

\[S = \iint\limits_{D}{{\sqrt {{{\left[ {4x} \right]}^2} + {{\left[ {4z} \right]}^2} + 1} \,dA}} = \iint\limits_{D}{{\sqrt {16{x^2} + 16{z^2} + 1} \,dA}}\] Show Step 3

Now, as we noted in Step 1 the region \(D\) is in the \(xz\)-plane and because we are after the portion of the elliptical paraboloid that is inside the cylinder given by \({x^2} + {z^2} = 4\) we can see that the region \(D\) must therefore be the disk \({x^2} + {z^2} \le 4\).

Also as noted in Step 1 we’ll be needing polar coordinates for this integral so here are the limits for the integral in terms of polar coordinates.

\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\end{array}\] Show Step 4

Now, let’s convert the integral into polar coordinates. However, they won’t be the “standard” polar coordinates. Because \(D\) is in the \(xz\)-plane we’ll need to use the following “modified” polar coordinates.

\[\begin{align*}x & = r\cos \theta \\ z & = r\sin \theta \\ {x^2} + {z^2} & = {r^2}\end{align*}\]

Converting the integral to polar coordinates then gives,

\[S = \iint\limits_{D}{{\sqrt {16{x^2} + 16{z^2} + 1} \,dA}} = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{r\sqrt {16{r^2} + 1} \,dr}}\,d\theta }}\]

Don’t forget to pick up the extra \(r\) from converting the \(dA\) into polar coordinates. It is the same \(dA\) as we use for the “standard” polar coordinates.

Show Step 5

Okay, all we need to do then is evaluate the integral.

\[\begin{align*}S & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{r\sqrt {16{r^2} + 1} \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\frac{1}{{48}}{{\left( {16{r^2} + 1} \right)}^{\frac{3}{2}}}} \right|_0^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{1}{{48}}\left( {{{65}^{\frac{3}{2}}} - 1} \right)\,d\theta }}\\ & = \left. {\frac{1}{{48}}\left( {{{65}^{\frac{3}{2}}} - 1} \right)\theta } \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{\pi }{{24}}\left( {{{65}^{\frac{3}{2}}} - 1} \right) = 68.4667}}\end{align*}\]