Let’s get the integral set up now. In this case the surface is in the form,

$$y = g\left( {x,z} \right) = 3{x^2} + 3{z^2}$$

so we’ll use the following formula for the surface integral.

$$\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {x,g\left( {x,z} \right),z} \right)\sqrt {{{\left( {\frac{{\partial g}}{{\partial x}}} \right)}^2} + 1 + {{\left( {\frac{{\partial g}}{{\partial z}}} \right)}^2}} \,dA}}$$

The integral is then,

$$\begin{align*}\iint\limits_{S}{{40y\,dS}} & = \iint\limits_{D}{{40\left( {3{x^2} + 3{z^2}} \right)\sqrt {{{\left( {6x} \right)}^2} + 1 + {{\left( {6z} \right)}^2}} \,dA}}\\ & = \iint\limits_{D}{{120\left( {{x^2} + {z^2}} \right)\sqrt {36\left( {{x^2} + {z^2}} \right) + 1} \,dA}}\end{align*}$$

Don’t forget to plug the equation of the surface into $y$ in the integrand and recall that $D$ is the disk we found in Step 1.

Show Step 3

Now, for this problem it should be pretty clear that we’ll want to use polar coordinates to do the integral. We’ll use the following set of polar coordinates.

$$x = r\cos \theta \hspace{0.25in}\hspace{0.25in}z = r\sin \theta \hspace{0.25in}\hspace{0.25in}{x^2} + {z^2} = {r^2}$$

Also, because $D$ is the disk ${x^2} + {z^2} \le 2$ the limits for the integral will be,

$$\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le \sqrt 2 \end{array}$$

Converting the integral to polar coordinates gives,

$$\begin{align*}\iint\limits_{S}{{40y\,dS}} & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 2 }}{{120{r^2}\sqrt {36{r^2} + 1} \,\left( r \right)dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 2 }}{{120{r^3}\sqrt {36{r^2} + 1} \,dr}}\,d\theta }}\end{align*}$$

Don’t forget to pick up the extra $r$ when converting the $dA$ into polar coordinates.

Show Step 4

Now all that we need to do is evaluate the double integral and this one can be a little tricky unless you’ve seen this kind of integral done before.

We’ll use the following substitution to do the integral.

$$u = 36{r^2} + 1\hspace{0.25in} \to \hspace{0.25in}du = 72r\,dr\hspace{0.25in} \to \hspace{0.25in}\frac{1}{{72}}du = r\,dr$$

The problem is that this doesn’t seem to work at first glance because the differential will only get rid of one of the three $r$’s in front of the root. However, we can also solve the substitution for ${r^2}$ to get,

$${r^2} = \frac{1}{{36}}\left( {u - 1} \right)$$

and we can now convert the remaining two $r$’s into $u$’s.

So, using the substitution the integral becomes,

$$\begin{align*}\iint\limits_{S}{{40y\,dS}}& = \int_{0}^{{2\pi }}{{\int_{1}^{{73}}{{120\left( {\frac{1}{{72}}} \right)\left( {\frac{1}{{36}}} \right)\left( {u - 1} \right){u^{\frac{1}{2}}}\,du}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{1}^{{73}}{{\frac{5}{{108}}\left( {{u^{\frac{3}{2}}} - {u^{\frac{1}{2}}}} \right)\,du}}\,d\theta }}\end{align*}$$

Note that we also converted the $r$ limits in the original integral into $u$ limits simply by plugging the “old” $r$ limits into the substitution to get “new” $u$ limits.

We can now easily finish evaluating the integral.

$$\begin{align*}\iint\limits_{S}{{40y\,dS}} & = \int_{0}^{{2\pi }}{{\left. {\frac{5}{{108}}\left( {\frac{2}{5}{u^{\frac{5}{2}}} - \frac{2}{3}{u^{\frac{3}{2}}}} \right)} \right|_1^{73}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{5}{{108}}\left[ {\frac{2}{5}\left( {{{73}^{\frac{5}{2}}}} \right) - \frac{2}{3}\left( {{{73}^{\frac{3}{2}}}} \right) - \left( { - \frac{4}{{15}}} \right)} \right]\,d\theta }}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{5\pi }}{{54}}\left[ {\frac{2}{5}\left( {{{73}^{\frac{5}{2}}}} \right) - \frac{2}{3}\left( {{{73}^{\frac{3}{2}}}} \right) + \frac{4}{{15}}} \right]\, = 5176.8958}}\end{align*}$$

Kind of messy integral with a messy answer but that will happen on occasion so we shouldn’t get too excited about it when that does happen.