Surface Area of a Parametric Surface

Now that we can parameterize surfaces, we need to define the surface area of a parametric surface. This definition provides the foundation for surface integrals.

Let [latex]S[/latex] be a surface with parameterization [latex]\mathbf{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle[/latex] over some parameter domain [latex]D[/latex]. We assume the surface parameterization is continuously differentiable—meaning each component function has continuous partial derivatives. For simplicity, assume [latex]D[/latex] is a rectangle (this material extends to nonrectangular domains). Divide rectangle [latex]D[/latex] into subrectangles [latex]D_{ij}[/latex] with horizontal width [latex]\Delta u[/latex] and vertical length [latex]\Delta v[/latex].

Suppose [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 [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.

Grid Curves and Tangent Vectors

We can form a grid with lines parallel to the [latex]u[/latex]-axis and [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]\mathbf{r}(u,v)[/latex].

Without loss of generality, assume [latex]P_{ij}[/latex] is located at the corner of two grid curves, as in Figure 9. If we think of [latex]\mathbf{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]\mathbf{r}[/latex].

Consider the grid lines 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]\mathbf{r}(u_i, v)[/latex] and [latex]\mathbf{r}(u, v_j)[/latex], and these curves intersect at point [latex]P_{ij}[/latex].

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].

Approximating Surface Area

The tangent plane at [latex]P_{ij}[/latex] contains vectors [latex]\mathbf{t}_u(P_{ij})[/latex] and [latex]\mathbf{t}_v(P_{ij})[/latex]. Therefore, the parallelogram spanned by these two vectors lies 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], we approximate [latex]S_{ij}[/latex] using 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).

surface area of a parametric surface

Let [latex]{\bf{r}}(u,v)=\langle(x(u,v),y(u,v),z(u,v)\rangle[/latex] with parameter domain [latex]D[/latex] be a smooth parameterization of surface [latex]S[/latex]. Furthermore, assume that [latex]S[/latex] is traced out only once as [latex](u, v)[/latex] varies over [latex]D[/latex]. The surface area of [latex]S[/latex] is:

[latex]\large{\displaystyle\iint_D||{\bf{t}}_u\times{\bf{t}}_v||dA}[/latex],

where [latex]{\bf{t}}_u=\left\langle\frac{\partial{x}}{\partial{u}},\frac{\partial{y}}{\partial{u}},\frac{\partial{z}}{\partial{u}}\right\rangle\text{ and }{\bf{t}}_v=\left\langle\frac{\partial{x}}{\partial{v}},\frac{\partial{y}}{\partial{v}},\frac{\partial{z}}{\partial{v}}\right\rangle[/latex].

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

Before calculating the surface area of this cone using the surface area of a parametric surface 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.

With a parameterization in hand, we can calculate the surface area of the cone using the surface area of a parametric surface 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 of a parametric surface 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.

Show that the surface area of the sphere [latex]x^{2}+y^{2}+z^{2}=r^{2}[/latex] is [latex]4\pi{r}^2[/latex].

Show Solution

The sphere has parameterization

[latex]\langle{r}\cos\theta\sin\phi,r\sin\theta\sin\phi,r\cos\phi\rangle, \ 0\leq\theta\leq2\pi, \ 0\leq\phi\leq\pi[/latex].

The tangent vectors are

[latex]{\bf{t}}_\theta=\langle-r\sin\theta\sin\phi,r\cos\theta\sin\phi,0\rangle\text{ and }{\bf{t}}_\phi=\langle{r}\cos\theta\cos\phi,r\sin\theta\cos\phi,-r\sin\phi\rangle[/latex].

Therefore,

[latex]\begin{aligned} {\bf{t}}_\phi\times{\bf{t}}_\theta&=\langle{r}^2\cos\theta\sin^2\phi,r^2\sin\theta\sin^2\phi,r^2\sin^2\theta\sin\phi\cos\phi+r^2\cos^2\theta\sin\phi\cos\phi\rangle \\ &=\langle{r}^2\cos\theta\sin^2\phi,r^2\sin\theta\sin^2\phi,r^2\sin\phi\cos\phi\rangle \end{aligned}[/latex].

Now,

[latex]\begin{aligned} ||{\bf{t}}_\phi\times{\bf{t}}_\theta||&=\sqrt{r^4\sin^4\phi\cos^2\theta+r^4\sin^4\phi\sin^2\theta+r^4\sin^2\phi\cos^2\phi} \\ &=\sqrt{r^4\sin^4\phi+r^4\sin^2\phi\cos^2\phi} \\ &=r^2\sqrt{\sin^2\phi} \\ &=r^2\sin\phi \end{aligned}[/latex].

Notice that [latex]\phi\geq0[/latex] on the parameter domain because [latex]0\leq\phi\leq\pi[/latex], and this justifies equation [latex]\sqrt{\sin^2\phi}=\sin\phi[/latex]. The surface area of the sphere is

[latex]\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi}r^2\sin\phi{d}\phi{d}\theta=r^2\displaystyle\int_0^{2\pi}2d\theta=4\pi{r}^2[/latex].

We have derived the familiar formula for the surface area of a sphere using surface integrals.