Multivariable calculus and vector analysis

Surface area of parametrized surfaces

We can calculate the surface area of a parametrized surface much in the same way that we calculated the path length of a parametrized path. You may recall how we chopped up the interval $\left[a,b\right]$ into small segments, and approximated the path as a bunch of straight line segments in order to estimate its length.

The procedure for calculating surface area is similar. In this case, we chop up the region $D$ into a bunch of small rectangles, as illustrated below with our example helicoid surface.

The small rectangles form a grid on $D$. The rectangles are mapped by $\mathbf{\Phi }$ onto the helicoid, forming a grid on the surface. In fact, the grid on the surface was shown in all the above examples because that is how the computer typically draws surfaces.

Notice how the function $\mathbf{\Phi }$ maps each small rectangle in $D$ onto the surface. You can see below how each rectangle in $D$ (outlined in green) maps onto a small part of the surface, called the image of the rectangle (outlined in red).

In reality, the image of each rectangle is some “curvy rectangle” on the surface. (In the above demos, the image of each rectangle appears to have straight edges, but that is due to how the computer draws surfaces. What is shown by the computer is only an approximation of the actual helicoid.)

To estimate the surface area, we approximate each curvy rectangle as a parallelogram. You can read how the sides of parallelogram are proportional to $\frac{\partial \mathbf{\Phi }}{\partial u}$ and $\frac{\partial \mathbf{\Phi }}{\partial v}$. We need to calculate the area of the parallelogram spanned by these two vectors. We take the magnitude of their cross product.

If we would form a Riemann sum defining this approximation of surface area and take the limit as the size of the rectangles go to zero, we would see that the total surface area is the double integral

$$\begin{array}{lll}\hfill A={\iint }_D∥\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)∥dudv.& & \hfill \end{array}$$

Note the similarity to the expression we derived for the length of a parametrized path,

$$\begin{array}{lll}\hfill L={\int \; <!--nolimits\; -->}_a^b\left|\right|\mathbf{c}{}^{\text{'}}\left(t\right)\left|\right|dt.& & \hfill \end{array}$$

For surface area, instead of integrating over an interval $\left[a,b\right]$, we integrate over the region $D$. In the formula for path length, the expression $\left|\right|\mathbf{c}{}^{\text{'}}\left(t\right)\left|\right|$ estimates how much $\mathbf{c}$ is stretching the interval $\left[a,b\right]$ as it maps $\left[a,b\right]$ onto the path (or, in other words, how fast a particle with position $\mathbf{c}\left(t\right)$ is moving). The ugly expression

$$\begin{array}{lll}\hfill ∥\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)∥& & \hfill \end{array}$$

is analogous to $\left|\right|\mathbf{c}{}^{\text{'}}\left(t\right)\left|\right|$. The expression indicates how much $\mathbf{\Phi }$ is stretching $D$ as it maps $D$ onto the surface.

You can read an example 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