Find the surface area of the torus parametrized by $$\begin{equation*} \mathbf{r}(u,v) =( 3+\cos v) \cos u\,\mathbf{i} +( 3+\cos v) \sin u\,\mathbf{j}+\sin v\mathbf{k} \end{equation*}$$

where $0\leq u\leq 2\pi$ and $0\leq v\leq 2\pi .$

Solution This is the torus graphed in Figure 42. To use formula (1) to find the area of a parametrized surface, we need to find the partial derivatives of $\mathbf{r=r}(u,v) .$ $$\begin{eqnarray*} \mathbf{r}_{u}(u,v) &=&( 3+\cos v) \left( -\sin u\right) \,\mathbf{i}+( 3+\cos v) \cos u\,\mathbf{j} \\[4pt] \mathbf{r}_{v}(u,v) &=& (-\sin v) \cos u \mathbf{i} -\sin v\sin u\,\mathbf{j}+\cos v\mathbf{k} \end{eqnarray*}$$

Next we find the cross product: $$\begin{eqnarray*} \mathbf{r}_{u}\times \mathbf{r}_{v} &=& \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] -( 3+\cos v) \sin u & ( 3+\cos v) \cos u & 0 \\[3pt] -\sin v\cos u & -\sin v\sin u & \cos v \end{array}\right|\\[5pt] &=&( 3+\cos v) \left( \cos u\cos v\mathbf{i}+\sin u\cos v\mathbf{j }+\sin v\mathbf{k}\right) \end{eqnarray*}$$

Then $$\begin{eqnarray*} \hspace{-2pc}\left\Vert \mathbf{r}_{u}\times \mathbf{r}_{v}\right\Vert &=&\sqrt{( 3+\cos v) ^{2}( \cos ^{2}u\cos ^{2}v+\sin ^{2}u\cos ^{2}v+\sin ^{2}v) } \\ &=&( 3+\cos v) \sqrt{\cos ^{2}v( \cos ^{2}u+\sin ^{2}u) +\sin ^{2}v}=( 3+\cos v) \sqrt{\cos ^{2}v+\sin ^{2}v}\\ &=&3+\cos v \end{eqnarray*}$$

Now using formula (1), the surface area of $S$ is $$\begin{eqnarray*} \iint\limits_{\kern-3ptR}( 3+\cos v)\, du\,dv&=&\int_{0}^{2\pi }\int_{0}^{2\pi }( 3+\cos v)\, dv\,du=\int_{0}^{2\pi }\big[ 3v+\sin v \big] _{0}^{2\pi }du\\[4pt] &=&\int_{0}^{2\pi }6\pi du=12\pi ^{2} \end{eqnarray*}$$