Multivariable calculus and vector analysis

Parametrized surfaces

By now, you are too familiar with parameterized curves. If we have a function $\mathbf{c}:\left[a,b\right]\to {\mathbf{R}}^3$, then $\mathbf{c}\left(t\right)$ parametrizes a curve in three dimensions as $t$ ranges from $a$ to $b$. We have frequently used $\mathbf{c}\left(t\right)=\left(cost,sint,t\right)$ as a parametrization of a helix or slinky. For example, here is a CVT of one loop of the helix, with $0\le t\le 2\pi$.

We can parametrize surfaces in the same way. To parametrize surfaces, we simply need a function of two variables (rather than a function of one variable that we needed for parametrized curves). To illustrate the properties of parametrized surfaces, we will use the example function

$$\begin{array}{lll}\hfill \mathbf{\Phi }\left(u,v\right)=\left(ucosv,usinv,v\right).& & \hfill \end{array}$$

If we fix $u=1$, then $\mathbf{\Phi }\left(1,v\right)$ parametrizes the helix as before. If we change $u$ to 0.5, then $\mathbf{\Phi }\left(0.5,v\right)$ also parametrizes a helix, but this time with radius 0.5 rather than radius 1.

If we let $u$ range from 0 to 1, and let $v$ range from 0 to $2\pi$, then $\mathbf{\Phi }\left(u,v\right)$ traces out a whole family of helices, with radii between 0 and 1. The result is a surface called a helicoid, as shown below.

Notice that if you drag the blue slider to change $v$, then the red point traces out a helix. You can drag the green slider to change $u$ and thus the radius of the helix. If you keep $v$ constant and change only $u$, the red point traces out a straight line.

If we let $D$ be the region where $0\le u\le 1$ and $0\le v\le 2\pi$, then the function $\mathbf{\Phi }$ maps the region $D$ onto the helicoid shown above. We sometimes write this as $\mathbf{\Phi }:D\to {\mathbf{R}}^3$. (Note the region $D$ is actually a subset of ${\mathbf{R}}^2$, so we could also say that $\mathbf{\Phi }:{\mathbf{R}}^2\to {\mathbf{R}}^3$.)

The region $D$ is a rectangle in $uv$-space. Any point $\left(u,v\right)$ in $D$ is mapped to the point $\mathbf{\Phi }\left(u,v\right)$ on the helicoid. We can demonstrate this map more clearly by replacing the above blue slider for $v$ and green slider for $u$ with a green point that you can drag in the rectangle $D$ to specify both $u$ and $v$. In the below demo, the region $D$ is shown floating above the helicoid. Notice that if you drag the green point along the bottom of the rectangle, you change $u$ while leaving $v=0$. Similarly, if you drag the green point along the right side of the rectangle, you change $v$ while leaving $u=1$. (Note: you cannot drag the red point on the surface.)

In general, the region $D$ does not have to be a rectangle. But in many examples, we let $D$ be a rectangle to make the calculations simpler.

You can see some examples here.

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

4a. Vector field basics

4b. Parametrized curves

4c. Parametrized surfaces

• Parametrized surfaces
• Parametrized surface examples
• Surface area of parametrized surfaces
• Surface area calculation
• Normal vector of parametrized surfaces
• Orienting surfaces
• A surface that is not orientable
• Surface area example

4d. Path integrals and line integrals

4e. Surface integrals

5. The fundamental theorems of vector calculus

6. Review material