Finding a Surface Integral

Find \(\iint\limits_{\kern-3ptS}ye^{x^{2}+z^{2}}dS,\) where \(S\) is the part of the surface of the cylinder \(x^{2}+z^{2}=9\) that lies between \(y=0\) and \(y=2.\)

Solution The surface \(S\) is shown in Figure 52. Since the surface is a piece of a cylinder, we use a variation of cylindrical coordinates to parametrize the surface. Here, we use \(x=r\cos \theta ,\) \(y=y,\) and \(z=r\sin \theta .\) (Do you see why?) Then \(r^{2}=x^{2}+z^{2}=9,\) so \(r=3.\) We parametrize the surface \(S\) using \[ \begin{equation*} \mathbf{r}( \theta ,y) =3\cos \theta\ \mathbf{i}+y\mathbf{j}+3\sin \theta\ \mathbf{k} \end{equation*} \]

where the parameter domain \({R}\) is \(0\leq \theta \leq 2\pi \) and \(0\leq y\leq 2.\)

The partial derivatives of \(\mathbf{r}\) are \[ \mathbf{r}_{\theta }( \theta ,y) =-3\sin \theta\ \mathbf{i}+3\cos \theta\ \mathbf{k}\qquad\hbox{and}\qquad \mathbf{r}_{y}( \theta ,y) =\mathbf{j} \]

and their cross product is \[ \begin{equation*} \mathbf{r}_{\theta }\times \mathbf{r}_{y}= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3\sin \theta & 0 & 3\cos \theta \\ 0 & 1 & 0 \end{array}\right| =-3\cos \theta\ \mathbf{i}-3\sin \theta\ \mathbf{k} \end{equation*} \]

Then \(\left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{y}\right\Vert =\sqrt{9\cos ^{2}\theta +9\sin ^{2}\theta }=3.\) So, we have \[ \begin{eqnarray*} \iint\limits_{\kern-3ptS}ye^{x^{2}+z^{2}}dS &=&\iint\limits_{\kern-3ptR}ye^{9}\left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{y}\right\Vert d\theta dy=\int_{0}^{2}\int_{0}^{2\pi }( ye^{9}) 3\,d\theta\, dy \notag \\[4pt] &=&3e^{9}\int_{0}^{2}y\big[ \theta \big] _{0}^{2\pi }dy=3e^{9}\int_{0}^{2}2\pi y\,dy=6\pi e^{9}\left[ \dfrac{y^{2}}{2}\right] _{0}^{2}=12\pi e^{9} \end{eqnarray*} \]