Example: identifying smooth and nonsmooth surfaces

Which of the figures in Figure 8 is smooth?

Figure 8. (a) This surface is smooth. (b) This surface is piecewise smooth.

Show Solution

The surface in Figure 8(a) can be parameterized by

[latex]{\bf{r}}(u,v)=\langle(2+\cos{v})\cos{u},(2+\cos{v})\sin{u},\sin{v}\rangle \ 0\leq{u}\leq2\pi, \ 0\leq{v}\leq2\pi[/latex]

(we can use technology to verify). Notice that vectors

[latex]{\bf{r}}_u=\langle-(2+\cos{v})\sin{u},(2+\cos{v})\cos{u},0\rangle\text{ and }{\bf{r}}_v=\langle-\sin{v}\cos{u},-\sin{v}\sin{u},\cos{v}\rangle[/latex]

exist for any choice of [latex]u[/latex] and [latex]v[/latex] in the parameter domain, and

[latex]\begin{aligned} {\bf{r}}_v&=\begin{vmatrix} {\bf{i}}&{\bf{j}}&{\bf{k}} \\ -(2+\cos{v})\sin{u}&(2+\cos{v})\cos{u}&0 \\ -\sin{v}\cos{u}&-\sin{v}\sin{u}&\cos{v} \end{vmatrix} \\ &=[(2+\cos{v})\cos{u}\cos{v}]{\bf{i}}+[(2+\cos{v})\sin{u}\cos{v}]{\bf{j}}+[(2+\cos{v})\sin{v}\sin^2u+(2+\cos{v})\sin{v}\cos^2u]{\bf{k}} \\ &=[(2+\cos{v})\cos{u}\cos{v}]{\bf{i}}+[(2+\cos{v})\sin{u}\cos{v}]{\bf{j}}+[(2+\cos{v})\sin{v}]{\bf{k}} \end{aligned}[/latex]

.

The [latex]\bf{k}[/latex] component of this vector is zero only if [latex]v=0[/latex] or [latex]v=\pi[/latex]. If [latex]v=0[/latex] or [latex]v=\pi[/latex], then the only choices for [latex]u[/latex] that make the [latex]\bf{j}[/latex] component zero are [latex]u=0[/latex] or [latex]u=\pi[/latex]. But, these choices of [latex]u[/latex] do not make the [latex]\bf{i}[/latex] component zero. Therefore, [latex]{\bf{r}}_u\times{\bf{r}}_v[/latex] is not zero for any choice of [latex]u[/latex] and [latex]v[/latex] in the parameter domain, and the parameterization is smooth. Notice that the corresponding surface has no sharp corners.

In the pyramid in Figure 8(b), the sharpness of the corners ensures that directional derivatives do not exist at those locations. Therefore, the pyramid has no smooth parameterization. However, the pyramid consists of five smooth faces, and thus this surface is piecewise smooth.

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

Note that we can form a grid with lines that are parallel to the [latex]u[/latex]-axis and the [latex]v[/latex]-axis in the [latex]uv[/latex]-plane. These grid lines correspond to a set of grid curves on surface [latex]S[/latex] that is parameterized by [latex]{\bf{r}}(u,v)[/latex]. Without loss of generality, we assume that [latex]P_{ij}[/latex] is located at the corner of two grid curves, as in Figure 9. If we think of [latex]{\bf{r}}[/latex] as a mapping from the [latex]uv[/latex]-plane to [latex]\mathbb{R}^3[/latex], the grid curves are the image of the grid lines under [latex]{\bf{r}}[/latex]. To be precise, consider the grid lines that go through point [latex](u_i,v_j)[/latex]. One line is given by [latex]x=u_i[/latex], [latex]y=v[/latex]; the other is given by [latex]x=u[/latex], [latex]y=v_j[/latex]. In the first grid line, the horizontal component is held constant, yielding a vertical line through [latex](u_i,v_j)[/latex]. In the second grid line, the vertical component is held constant, yielding a horizontal line through [latex](u_i,v_j)[/latex]. The corresponding grid curves are [latex]{\bf{r}}(u_i,v)[/latex] and [latex]{\bf{r}}(u,v_j)[/latex], and these curves intersect at point [latex]P_{ij}[/latex].

Figure 9. Grid lines on a parameter domain correspond to grid curves on a surface.

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

[latex]{\bf{t}}_v(P_{ij})={\bf{r}}_v(u_i,v_j)=\langle{x}_v(u_i,v_j),y_v(u_i,v_j),z_v(u_i,v_j)\rangle[/latex].

For grid curve [latex]{\bf{r}}(u,v_j)[/latex], the tangent vector at [latex]P_{ij}[/latex] is

[latex]{\bf{t}}_u(P_{ij})={\bf{r}}_y(u_i,v_j)=\langle{x}_v(u_i,v_j),y_u(u_i,v_j),z_u(u_i,v_j)\rangle[/latex].

If vector [latex]{\bf{N}}={\bf{t}}_u(P_{ij})\times{\bf{t}}_u(P_{ij}[/latex] exists and is not zero, then the tangent plane at [latex]P_{ij}[/latex] exists (Figure 10). If piece [latex]S_{ij}[/latex] is small enough, then the tangent plane at point [latex]P_{ij}[/latex] is a good approximation of piece [latex]S_{ij}[/latex].

Figure 10. If the cross product of vectors [latex]{\bf{t}}_u[/latex] and [latex]{\bf{t}}_v[/latex] exists, then there is a tangent plane.

The tangent plane at [latex]P_{ij}[/latex] contains vectors [latex]{\bf{t}}_u(P_{ij})[/latex] and [latex]{\bf{t}}_v(P_{ij})[/latex], and therefore the parallelogram spanned by [latex]{\bf{t}}_u(P_{ij})[/latex] and [latex]{\bf{t}}_v(P_{ij})[/latex] is in the tangent plane. Since the original rectangle in the [latex]uv[/latex]-plane corresponding to [latex]S_{ij}[/latex] has width [latex]\Delta{u}[/latex] and length [latex]\Delta{v}[/latex], the parallelogram that we use to approximate [latex]S_{ij}[/latex] is the parallelogram spanned by [latex]\Delta{u}{\bf{t}}_u(P_{ij})[/latex] and [latex]\Delta{v}{\bf{t}}_v(P_{ij})[/latex]. In other words, we scale the tangent vectors by the constants [latex]\Delta{u}[/latex] and [latex]\Delta{v}[/latex] 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 [latex]S_{ij}[/latex] is

[latex]\Delta{S_{ij}}\approx||(\Delta{u}{\bf{t}}_u(P_{ij}))\times(\Delta{v}{\bf{t}}_v(P_{ij}))=||{\bf{t}}_u(P_{ij})\times{\bf{t}}_v(P_{ij})||\Delta{u}\Delta{v}[/latex].

Varying point [latex]P_{ij}[/latex] over all pieces [latex]S_{ij}[/latex] and the previous approximation leads to the following definition of surface area of a parametric surface (Figure 11).

Figure 11. The parallelogram spanned by [latex]{\bf{t}}_u[/latex] and [latex]{\bf{t}}_v[/latex] approximates the piece of surface [latex]S_{ij}[/latex].

Example: calculating surface area

Calculate the lateral surface area (the area of the “side,” not including the base) of the right circular cone with height [latex]h[/latex] and radius [latex]r[/latex].

Show Solution

Solution

Before calculating the surface area of this cone using The Surface Area Equation, we need a parameterization. We assume this cone is in [latex]\mathbb{R}^3[/latex] with its vertex at the origin (Figure 12). To obtain a parameterization, let [latex]\alpha[/latex] be the angle that is swept out by starting at the positive [latex]z[/latex]-axis and ending at the cone, and let [latex]k=\tan\alpha[/latex]. For a height value [latex]v[/latex] with [latex]0\leq{v}\leq{h}[/latex], the radius of the circle formed by intersecting the cone with plane [latex]z=v[/latex] is [latex]kv[/latex]. Therefore, a parameterization of this cone is

[latex]{\bf{s}}(u,v)=\langle{k}v\cos{u},kv\sin{u},v\rangle, \ 0\leq{u}\leq2\pi, \ 0\leq{v}\leq{h}[/latex].

The idea behind this parameterization is that for a fixed [latex]v[/latex] value, the circle swept out by letting [latex]u[/latex] vary is the circle at height [latex]v[/latex] and radius [latex]kv[/latex]. As [latex]v[/latex] increases, the parameterization sweeps out a “stack” of circles, resulting in the desired cone.

Figure 12. The right circular cone with radius [latex]r = kh[/latex] and height [latex]h[/latex] has parameterization [latex]{\bf{s}}(u,v)=\langle{k}v\cos{u},kv\sin{u},v\rangle[/latex], [latex]0\leq{u}\leq2\pi, 0\leq{v}\leq{h}[/latex].

With a parameterization in hand, we can calculate the surface area of the cone using The Surface Area Equation. The tangent vectors are [latex]{\bf{t}}_u=\langle-kv\sin{u},kv\cos{u},0\rangle[/latex] and [latex]{\bf{t}}_v=\langle{k}\cos{u},k\sin{u},1\rangle[/latex]. Therefore,

[latex]\begin{aligned} {\bf{t}}_u\times{\bf{t}}_v&=\begin{vmatrix} {\bf{i}}&{\bf{j}}&{\bf{k}} \\ -kv\sin{u}&kv\cos{u}&0 \\ {k}\cos{u}&k\sin{u}&1 \end{vmatrix} \\ &=\langle{k}v\cos{u},kv\sin{u},-k^2v\sin^2y-k^2v\cos^2u\rangle \\ &=\langle{k}v\cos{u},kv\sin{u},-k^2v\rangle \end{aligned}[/latex].

The magnitude of this vector is

[latex]\begin{aligned} ||\langle{k}v\cos{u},kv\sin{u},-k^2v\rangle||&=\sqrt{k^2v^2\cos^2u+k^2v^2\sin^2u+k^4v^2} \\ &=\sqrt{k^2v^2+k^4v^2} \\ &=kv\sqrt{1+k^2} \end{aligned}[/latex].

By The Surface Area Equation, the surface area of the cone is

[latex]\begin{aligned} \displaystyle\iint_D||{\bf{t}}_u\times{\bf{t}}_v||dA&=\displaystyle\int_0^h\displaystyle\int_0^{2\pi}kv\sqrt{1+k^2}dudv \\ &=2\pi{k}\sqrt{1+k^2}\displaystyle\int_0^hvdv \\ &=2\pi{k}\sqrt{1+k^2}\left[\frac{v^2}2\right]_0^h \\ &=\pi{k}h^2\sqrt{1+k^2} \end{aligned}[/latex].

Since [latex]k=\tan\alpha=r/h[/latex],

[latex]\begin{aligned} \pi{k}h^2\sqrt{1+k^2}&=\pi\frac{r}hh^2\sqrt{1+\frac{r^2}{h^2}} \\ &=\pi{r}h\sqrt{1+\frac{r^2}{h^2}} \\ &=\pi{r}\sqrt{h^2+h^2\left(\frac{r^2}{h^2}\right)} \\ &=\pi{r}\sqrt{h^2+r^2} \end{aligned}[/latex].

Therefore, the lateral surface area of the cone is [latex]\pi{r}\sqrt{h^2+r^2}[/latex].

Analysis

The surface area of a right circular cone with radius [latex]r[/latex] and height [latex]h[/latex] is usually given as [latex]\pi{r}^2+\pi{r}\sqrt{h^2+r^2}[/latex]. The reason for this is that the circular base is included as part of the cone, and therefore the area of the base [latex]\pi{r}^2[/latex] is added to the lateral surface area [latex]\pi{r}\sqrt{h^2+r^2}[/latex] that we found.

Watch the following video to see the worked solution to the above Try It

In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. Therefore, we can calculate the surface area of a surface of revolution by using the same techniques. Let [latex]y=f(x)\geq0[/latex] be a positive single-variable function on the domain [latex]a\leq{x}\leq{b}[/latex] and let [latex]S[/latex] be the surface obtained by rotating [latex]f[/latex] about the [latex]x[/latex]-axis (Figure 13). Let [latex]\theta[/latex] be the angle of rotation. Then, [latex]S[/latex] can be parameterized with parameters [latex]x[/latex] and [latex]\theta[/latex] by

[latex]{\bf{r}}(x,\theta)=\langle{x},f(x)\cos\theta,f(x)\sin\theta\rangle, \ a\leq{x}\leq{b}, \ 0\leq{x}\leq2\pi[/latex].