Examples

Common parametric surfaces

Common parametric surfaces

• (a) A vertical cylinder may be given parametrically by: $$\mathbf{r}(\theta ,z)=(R\cos (\theta ),R\sin (\theta ),z).$$
• (b) A sphere of radius $R$ may be parametrized by: $$\mathbf{r}(\theta ,\phi )=(R\cos (\theta )\sin (\phi ),R\sin (\theta )\sin (\phi ),R\cos (\phi )).$$
• (c) The graph of a function of two variables, $z=f(x,y)$, may be parametrized by: $$\mathbf{r}(u,v)=(u,v,f(u,v)).$$

Surface area

Surface area

Problem: Find the approximate surface area of the surface given parametrically by:

$$\mathbf{r}(u,v)=\left(2u\sin (\frac{v}{2}),2u\cos (\frac{v}{2}),3v\right),$$

where $u\in [0,1]$ and $v\in [0,2\pi ]$.
Solution:
First, the tangent vectors are ${\mathbf{r}}_u=\left(2\sin (\frac{v}{2}),2\cos (\frac{v}{2}),0\right)$ and ${\mathbf{r}}_v=(u\cos (\frac{v}{2}),-u\sin (\frac{v}{2}),3)$. Their cross product is $\mathbf{n}=(6\cos (\frac{v}{2}),-6\sin (\frac{v}{2}),-2u)$. The norm is $|\mathbf{n}|=\sqrt{36+4u^2}$. Compute the surface area by integrating $dA$:

$${\int }_{v=0}^{2\pi }{\int }_{u=0}^{1}dA={\int }_{v=0}^{2\pi }{\int }_{u=0}^1\sqrt{36+4u^2}dudv\approx \mathrm{38.39.}$$

Upward flux through hemisphere

Upward flux through hemisphere

Problem: Let $\mathbf{F}=(z,x,1)$ and let $\mathcal{S}$ be the upper hemisphere of the unit sphere, oriented up and out. Compute the total flux of $\mathcal{S}$ across $\mathcal{S}$.

Solution: First we need a parametrization of $\mathcal{S}$ with the correct orientation. Try $\mathbf{r}(\theta ,\phi )=(\cos (\theta )\sin (\phi ),\sin (\theta )\sin (\phi ),\cos (\phi ))$, and check the orientation after computing $\mathbf{n}$. For the upper hemisphere, the domain in the $(\theta ,\phi )$-space will be $\theta \in [0,2\pi ]$ and $\phi \in [0,\pi /2]$. Compute:

$$\begin{array}{rl}{\mathbf{r}}_{\theta }& =(-\sin (\theta )\sin (\phi ),\cos (\theta )\sin (\phi ),0)\\ {\mathbf{r}}_{\phi }& =(\cos (\theta )\cos (\phi ),\sin (\theta )\cos (\phi ),-\sin (\phi ))\\ {\mathbf{r}}_{\theta }\times {\mathbf{r}}_{\phi }& =(-{\sin }^2(\phi )\cos (\theta ),-{\sin }^2(\phi )\sin (\theta ),-\sin (\phi )\cos (\phi ))\\ & =-\sin (\phi )\mathbf{r}.\end{array}$$

Notice that ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_{\phi }$ is $-\sin (\phi )\mathbf{r}$, which points inward, but the problem states that $\mathcal{S}$ should be oriented with normal pointing outward, so we reverse the order, taking $\phi$ first and $\theta$ second.

Now plug $\mathbf{r}$ into the vector field to find its values on the surface:

$$\mathbf{F}(\mathbf{r}(u,v))=(\cos (\phi ),\cos (\theta )\sin (\phi ),1).$$

Then:

$$\begin{array}{rl}& \mathbf{F}(\mathbf{r}(u,v))\cdot \mathbf{n}(u,v)\\ & =(\cos (\phi ),\cos (\theta )\sin (\phi ),1)\cdot ({\sin }^2(\phi )\cos (\theta ),{\sin }^2(\phi )\sin (\theta ),\sin (\phi )\cos (\phi ))\\ & =\cos (\theta ){\sin }^2(\phi )\cos (\phi )+\cos (\theta )\sin (\theta ){\sin }^3(\phi )+\sin (\phi )\cos (\phi ).\end{array}$$

Finally we must evaluate the integral over the domain, using the fact that ${\int }_0^{2\pi }\cos (\theta )d\theta =0$ and ${\int }_0^{\pi /2}\cos (\theta )\sin (\theta )d\theta =0$:

$$\begin{array}{rl}{\iint }_{\mathcal{S}}\mathbf{F}& ={\int }_{\theta =0}^{2\pi }{\int }_{\phi =0}^{\pi /2}\cos (\theta ){\sin }^2(\phi )\cos (\phi )+\cos (\theta )\sin (\theta ){\sin }^3(\phi )+\sin (\phi )\cos (\phi )d\phi d\theta \\ & =2\pi {\int }_0^{\pi /2}\sin (\phi )\cos (\phi )d\phi \\ & =\pi .\end{array}$$