APEX Calculus

Subsection 14.5.1 Parametrizing surfaces

We are accustomed to describing surfaces as functions of two variables, usually written as \(z=f(x,y)\text{.}\) For our coming needs, this method of describing surfaces will prove to be insufficient. Instead, we will parametrize our surfaces, describing them as the set of terminal points of some vector-valued function \(\vec r(u,v) =\langle f(u,v),g(u,v),h(u,v)\rangle\text{.}\) The bulk of this section is spent practicing the skill of describing a surface \(\surfaceS\)using a vector-valued function. Once this skill is developed, we'll show how to find the surface area \(S\) of a parametrically-defined surface \(\surfaceS\text{,}\) a skill needed in the remaining sections of this chapter.

Given a point \((u_0,v_0)\) in the domain of a vector-valued function \(\vec r\text{,}\) the vectors \(\vec r_u(u_0,v_0)\) and \(\vec r_v(u_0,v_0)\) are tangent to the surface \(\surfaceS\) at \(\vec r(u_0,v_0)\) (a proof of this is developed later in this section). The definition of smoothness dictates that \(\vec r_u\times \vec r_v \neq \vec 0\text{;}\) this ensures that neither \(\vec r_u\) nor \(\vec r_v\) are \(\vec 0\text{,}\) nor are they ever parallel. Therefore smoothness guarantees that \(\vec r_u\) and \(\vec r_v\) determine a plane that is tangent to \(\surfaceS\text{.}\)

A surface \(\surfaceS\) is said to be orientable if a field of normal vectors can be defined on \(\surfaceS\) that vary continuously along \(\surfaceS\text{.}\) This definition may be hard to understand; it may help to know that orientable surfaces are often called “two sided.” A sphere is an orientable surface, and one can easily envision an “inside” and “outside” of the sphere. A paraboloid is orientable, where again one can generally envision “inside” and “outside” sides (or “top” and “bottom” sides) to this surface. Just about every surface that one can imagine is orientable, and we'll assume all surfaces we deal with in this text are orientable.

It is enlightening to examine a classic non-orientable surface: the Möbius band, shown in Figure 14.5.3. Vectors normal to the surface are given, starting at the point indicated in the figure. These normal vectors “vary continuously” as they move along the surface. Letting each vector indicate the “top” side of the band, we can easily see near any vector which side is the “top”.

However, if as we progress along the band, we recognize that we are labeling “both sides” of the band as the top; in fact, there are not two “sides” to this band, but one. The Möbius band is a non-orientable surface.

We now practice parametrizing surfaces.

Examples 14.5.4 and Example 14.5.6 demonstrate an important principle when parametrizing surfaces given in the form \(z=f(x,y)\) over a region \(R\text{:}\) if one can determine \(x\) and \(y\) in terms of \(u\) and \(v\text{,}\) then \(z\) follows directly as \(z=f(x,y)\text{.}\)

In the following two examples, we parametrize the same surface over triangular regions. Each will use \(v\) as a “scaling factor” as done in Example 14.5.6.

Example 14.5.8. Parametrizing a surface over a triangle.

Parametrize the surface \(z=x^2+2y^2\) over the triangular region \(R\) enclosed by the coordinate axes and the line \(y=2-2x/3\text{,}\) as shown in Figure 14.5.9.(a).

Solution

We may begin by letting \(x=u\text{,}\) \(0\leq u\leq 3\text{,}\) and \(y = 2-2u/3\text{.}\) This gives only the line on the “upper” side of the triangle. To get all of the region \(R\text{,}\) we can once again scale \(y\) by a variable factor, \(v\text{.}\)

Still letting \(x = u\text{,}\) \(0\leq u\leq 3\text{,}\) we let \(y = v(2-2u/3)\text{,}\) \(0\leq v\leq 1\text{.}\) When \(v=0\text{,}\) all \(y\)-values are 0, and we get the portion of the \(x\)-axis between \(x=0\) and \(x=3\text{.}\) When \(v=1\text{,}\) we get the upper side of the triangle. When \(v=1/2\text{,}\) we get the line \(y=1/2(2-2u/3) = 1-u/3\text{,}\) which is the line “halfway up” the triangle, shown in the figure with a dashed line.

Letting \(z = f(x,y) = x^2+2y^2\text{,}\) we have \(\vec r(u,v) = \langle u, v(2-2u/3), u^2+2\big(v(2-2u/3)\big)^2\rangle\text{,}\) \(0\leq u\leq 3\text{,}\) \(0\leq v\leq 1\text{.}\) This surface is graphed in Figure 14.5.9.(b). Again, when one looks from above, we can see the scaling effects of \(v\text{:}\) the series of lines that run to the point \((3,0)\) each represent a different value of \(v\text{.}\)

Another common way to parametrize the surface is to begin with \(y=u\text{,}\) \(0\leq u\leq 2\text{.}\) Solving the equation of the line \(y=2-2x/3\) for \(x\text{,}\) we have \(x = 3-3y/2\text{,}\) leading to using \(x=v(3-3u/2)\text{,}\) \(0\leq v\leq 1\text{.}\) With \(z=x^2+2y^2\text{,}\) we have \(\vec r(u,v) = \langle v(3-3u/2),u, \big(v(3-3u/2)\big)^2+2v^2\rangle\text{,}\) \(0\leq u\leq 2\text{,}\) \(0\leq v\leq 1\text{.}\)

Example 14.5.10. Parametrizing a surface over a triangle.

Parametrize the surface \(z=x^2+2y^2\) over the triangular region \(R\) enclosed by the lines \(y=3-2x/3\text{,}\) \(y=1\) and \(x=0\) as shown in Figure 14.5.11.(a).

Solution

While the region \(R\) in this example is very similar to the region \(R\) in the previous example, and our method of parametrizing the surface is fundamentally the same, it will feel as though our answer is much different than before.

We begin with letting \(x=u\text{,}\) \(0\leq u\leq 3\text{.}\) We may be tempted to let \(y = v(3-2u/3)\text{,}\) \(0\leq v\leq 1\text{,}\) but this is incorrect. When \(v = 1\text{,}\) we obtain the upper line of the triangle as desired. However, when \(v=0\text{,}\) the \(y\)-value is 0, which does not lie in the region \(R\text{.}\)

We will describe the general method of proceeding following this example. For now, consider \(y = 1+v(2-2u/3)\text{,}\) \(0\leq v\leq 1\text{.}\) Note that when \(v=1\text{,}\) we have \(y=3-2u/3\text{,}\) the upper line of the boundary of \(R\text{.}\) Also, when \(v=0\text{,}\) we have \(y=1\text{,}\) which is the lower boundary of \(R\text{.}\) With \(z=x^2+2y^2\text{,}\) we determine \(\vec r(u,v) = \langle u, 1+v(2-2u/3), u^2+2\big(1+v(2-2u/3)\big)^2\rangle\text{,}\) \(0\leq u\leq 3\text{,}\) \(0\leq v\leq 1\text{.}\)

The surface is graphed in Figure 14.5.11.(b).

Given a surface of the form \(z=f(x,y)\text{,}\) one can often determine a parametrization of the surface over a region \(R\) in a manner similar to determining bounds of integration over a region \(R\text{.}\) Using the techniques of Section 13.1, suppose a region \(R\) can be described by \(a\leq x\leq b\text{,}\) \(g_1(x) \leq y\leq g_2(x)\text{,}\) i.e., the area of \(R\) can be found using the iterated integral

When parametrizing the surface, we can let \(x=u\text{,}\) \(a\leq u\leq b\text{,}\) and we can let \(y = g_1(u)+v\big(g_2(u)-g_1(u)\big)\text{,}\) \(0\leq v\leq 1\text{.}\) The parametrization of \(x\) is straightforward, but look closely at how \(y\) is determined. When \(v=0\text{,}\) \(y=g_1(u) = g_1(x)\text{.}\) When \(v=1\text{,}\) \(y= g_2(u)=g_2(x)\text{.}\)

As a specific example, consider the triangular region \(R\) from Example 14.5.10, shown in Figure 14.5.11.(a). Using the techniques of Section 13.1, we can find the area of \(R\) as

Following the above discussion, we can set \(x=u\text{,}\) where \(0\leq u\leq 3\text{,}\) and set \(y = 1+ v\big(3-2u/3-1\big) = 1+v(2-2u/3)\text{,}\) \(0\leq v\leq 1\text{,}\) as used in that example.

One can do a similar thing if \(R\) is bounded by \(c\leq y\leq d\text{,}\) \(h_1(y)\leq x\leq h_2(y)\text{,}\) but for the sake of simplicity we leave it to the reader to flesh out those details. The principles outlined above are given in the following Key Idea for reference.

Parametrization is a powerful way to represent surfaces. One of the advantages of the methods of parametrization described in this section is that the domain of \(\vec r(u,v)\) is always a rectangle; that is, the bounds on \(u\) and \(v\) are constants. This will make some of our future computations easier to evaluate.

Just as we could parametrize curves in more than one way, there will always be multiple ways to parametrize a surface. Some ways will be more “natural” than others, but these other ways are not incorrect. Because technology is often readily available, it is often a good idea to check one's work by graphing a parametrization of a surface to check if it indeed represents what it was intended to.

Subsection 14.5.2 Surface Area

It will become important in the following sections to be able to compute the surface area of a surface \(\surfaceS\) given a smooth parametrization \(\vec r(u,v)\text{,}\) \(a\leq u\leq b\text{,}\) \(c\leq v\leq d\text{.}\) Following the principles given in the integration review at the beginning of this chapter, we can say that

where \(dS\) represents a small amount of surface area. That is, to compute total surface area \(S\text{,}\) add up lots of small amounts of surface area \(dS\) across the entire surface \(\surfaceS\text{.}\) The key to finding surface area is knowing how to compute \(dS\text{.}\) We begin by approximating.

In Section 13.5 we used the area of a plane to approximate the surface area of a small portion of a surface. We will do the same here.

Let \(R\) be the region of the \(u\)-\(v\) plane bounded by \(a\leq u\leq b\text{,}\) \(c\leq v\leq d\) as shown in Figure 14.5.21.(a). Partition \(R\) into rectangles of width \(\Delta u = \frac{b-a}n\) and height \(\Delta v = \frac{d-c}n\text{,}\) for some \(n\text{.}\) Let \(p=(u_0,v_0)\) be the lower left corner of some rectangle in the partition, and let \(m\) and \(q\) be neighboring corners as shown.

The point \(p\) maps to a point \(P = \vec r(u_0,v_0)\) on the surface \(\surfaceS\text{,}\) and the rectangle with corners \(p\text{,}\) \(m\) and \(q\) maps to some region (probably not rectangular) on the surface as shown in Figure 14.5.21.(b), where \(M = \vec r(m)\) and \(Q = \vec r(q)\text{.}\) We wish to approximate the surface area of this mapped region.

Let \(\vec u = M-P\) and \(\vec v = Q-P\text{.}\) These two vectors form a parallelogram, illustrated in Figure 14.5.21.(c), whose area approximates the surface area we seek. In this particular illustration, we can see that parallelogram does not particularly match well the region we wish to approximate, but that is acceptable; by increasing the number of partitions of \(R\text{,}\) \(\Delta u\) and \(\Delta v\) shrink and our approximations will become better.

From Section 10.4 we know the area of this parallelogram is \(\snorm{\vec u\times \vec v}\text{.}\) If we repeat this approximation process for each rectangle in the partition of \(R\text{,}\) we can sum the areas of all the parallelograms to get an approximation of the surface area \(S\text{:}\)

where \(\vec u_{i,j} = \vec r(u_i+\Delta u,v_j) - \vec r(u_i,v_j)\) and \(\vec v_{i,j} = \vec r(u_i,v_j+\Delta v)-\vec r(u_i,v_j)\text{.}\)

From our previous calculus experience, we expect that taking a limit as \(n\to \infty\) will result in the exact surface area. However, the current form of the above double sum makes it difficult to realize what the result of that limit is. The following rewriting of the double summation will be helpful:

We now take the limit as \(n\to\infty\text{,}\) forcing \(\Delta u\) and \(\Delta v\) to 0. As \(\Delta u\to 0\text{,}\)

(This limit process also demonstrates that \(\vec r_u(u,v)\) and \(\vec r_v(u,v)\) are tangent to the surface \(\surfaceS\) at \(\vec r(u,v)\text{.}\) We don't need this fact now, but it will be important in the next section.)

Thus, in the limit, the double sum leads to a double integral:

In Section 14.1, we recalled the arc length differential \(ds=\snorm{\vrp(t)}dt\text{.}\) In subsequent sections, we used that differential, but in most applications the “\(\snorm{\vrp(t)}\)” part of the differential canceled out of the integrand (to our benefit, as integrating the square roots of functions is generally difficult). We will find a similar thing happens when we use the surface area differential \(dS\) in the following sections. That is, our main goal is not to be able to compute surface area; rather, surface area is a tool to obtain other quantities that are more important and useful. In our applications, we will use \(dS\text{,}\) but most of the time the “\(\snorm{\vec r_u\times \vec r_v}\)” part will cancel out of the integrand, making the subsequent integration easier to compute.