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.

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