6.6 Surface integrals  (Page 4/27)

Surface area of a parametric surface

Our goal is to define a surface integral, and as a first step we have examined how to parameterize a surface. The second step is to define the surface area of a parametric surface. The notation needed to develop this definition is used throughout the rest of this chapter.

Let S be a surface with parameterization $\text{r}\left(u,v\right)=⟨x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)⟩$ over some parameter domain D . We assume here and throughout that the surface parameterization $\text{r}\left(u,v\right)=⟨x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)⟩$ is continuously differentiable—meaning, each component function has continuous partial derivatives. Assume for the sake of simplicity that D is a rectangle (although the following material can be extended to handle nonrectangular parameter domains). Divide rectangle D into subrectangles $D_{ij}$ with horizontal width $\text{Δ}u$ and vertical length $\text{Δ}v.$ Suppose that i ranges from 1 to m and j ranges from 1 to n so that D is subdivided into mn rectangles. This division of D into subrectangles gives a corresponding division of surface S into pieces $S_{ij}.$ Choose point $P_{ij}$ in each piece $S_{ij}.$ Point $P_{ij}$ corresponds to point $(u_i,v_j)$ in the parameter domain.

Note that we can form a grid with lines that are parallel to the u -axis and the v -axis in the uv -plane. These grid lines correspond to a set of grid curves    on surface S that is parameterized by $\text{r}\left(u,v\right).$ Without loss of generality, we assume that $P_{ij}$ is located at the corner of two grid curves, as in [link] . If we think of r as a mapping from the uv -plane to ${ℝ}^3,$ the grid curves are the image of the grid lines under r . To be precise, consider the grid lines that go through point $(u_i,v_j).$ One line is given by $x=u_i,y=v;$ the other is given by $x=u,y=v_j.$ In the first grid line, the horizontal component is held constant, yielding a vertical line through $(u_i,v_j).$ In the second grid line, the vertical component is held constant, yielding a horizontal line through $(u_i,v_j).$ The corresponding grid curves are $\text{r}(u_i,v)$ and $\text{r}(u,v_j),$ and these curves intersect at point $P_{ij}.$

Now consider the vectors that are tangent to these grid curves. For grid curve $\text{r}(u_i,v),$ the tangent vector at $P_{ij}$ is

For grid curve $\text{r}(u,v_j),$ the tangent vector at $P_{ij}$ is

If vector $\text{N}={\text{t}}_u\left(P_{ij}\right)\times {\text{t}}_v\left(P_{ij}\right)$ exists and is not zero, then the tangent plane at $P_{ij}$ exists ( [link] ). If piece $S_{ij}$ is small enough, then the tangent plane at point $P_{ij}$ is a good approximation of piece $S_{ij}.$

The tangent plane at $P_{ij}$ contains vectors ${\text{t}}_u\left(P_{ij}\right)$ and ${\text{t}}_v\left(P_{ij}\right),$ and therefore the parallelogram spanned by ${\text{t}}_u\left(P_{ij}\right)$ and ${\text{t}}_v\left(P_{ij}\right)$ is in the tangent plane. Since the original rectangle in the uv -plane corresponding to $S_{ij}$ has width $\text{Δ}u$ and length $\text{Δ}v,$ the parallelogram that we use to approximate $S_{ij}$ is the parallelogram spanned by $\text{Δ}u{\text{t}}_u\left(P_{ij}\right)$ and $\text{Δ}v{\text{t}}_v\left(P_{ij}\right).$ In other words, we scale the tangent vectors by the constants $\text{Δ}u$ and $\text{Δ}v$ to match the scale of the original division of rectangles in the parameter domain. Therefore, the area of the parallelogram used to approximate the area of $S_{ij}$ is