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 on page 1018. 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 \[ \begin{eqnarray*} \iint\limits_{\kern-13ptR}( 3+\cos v)\, du\,dv&=&\int_{0}^{2\pi }\int_{0}^{2\pi }( 3+\cos v)\, dv\,du=\int_{0}^{2\pi }\big[ 3v+\sin v \big] _{0}^{2\pi }du\\[4pt] &=&\int_{0}^{2\pi }6\pi du=12\pi ^{2} \end{eqnarray*} \]