Chapter 4

Surfaces

In this chapter we turn to surfaces in general. We discuss the following topics.

Describing surfaces with equations and parametric descriptions. • Some constructions of surfaces: surfaces of revolution and ruled surfaces. • Tangent planes to surfaces. • Surface area of surfaces. • Curvature of surfaces. •

Figure 4.1: In this design for the 1972 Olympic Games in Munich, Germany, Frei Otto illustrates the role of surfaces in modern architecture. (Photo: Wim Huisman)

91 92 Surfaces

4.1 Describing general surfaces

4.1.1 After degree 2 surfaces, we could proceed to degree 3, but instead we turn to surfaces in general. We highlight some important aspects of these surfaces and ways to construct them.

4.1.2 Surfaces given as graphs of functions Suppose f : R2 R is a differentiable function, then the graph of f consisting of all points of the form→(x, y, f(x, y)) is a surface in 3–space with a smooth appearance due to the differentiability of f. Since we are not tied to degree 1 or 2 expressions for f, the resulting surfaces may have all sorts of fascinating shapes. The description as a graph can

1 1 0.5 0.5 0 0 2 2 -0.5 -0.5 -1 -1 0 -6 0 -2 -4 -2 0 -2 0 -2 2 2

Figure 4.2: Two ‘sinusoidal’ graphs: the graphs of sin(xy) and sin(x).

be seen as a special case of a parametric description:

x = u y = v z = f(u, v),

where u and v are the parameters. Just like in parametric descriptions of planes, two parameters are needed. This reflects the 2–dimensionality of a surface. Although we are used to thinking of graphs for which the points are of the form (x, y, f(x, y)), it is sometimes more convenient to think of y as a function of x and z, or of x as a function of y and z. For instance, a portion of the hyperboloid x2 + y2 z2 = 1 is described by the graph of x = 1 + z2 y2. So the function is a function of −y and z. − So points of the graph are of the formp ( 1 + z2 y2, y, z). − p 4.1.3 Surfaces given by an equation In principle, a single equation in x, y and z describes a surface in R3. Depending on the structure of the equation, the surface may look more or less smooth. Examples we have come across before include planes, like x + 2y 3z = 5, and quadrics, like the sphere x2 + y2 + z2 = 3. − 4.1 Describing general surfaces 93

4.1.4 Surfaces given by a parametric description A surface is 2–dimensional. This 2–dimensionality is the reason why surfaces or parts of surfaces can be described using two parameters. So, if x, y and z all three depend on the two parameters u and v, then x, y and z will run through the points of a surface if u and v vary. We denote such a parametric description by a boldface symbol, usually x. Note that x is a ‘vector–valued function’, so that x(u, v) is a 3d vector for every pair u, v. The coordinate functions are ordinary functions of two variables, whose values at (u, v) are denoted by x(u, v), y(u, v) and z(u, v), respectively. Whether the surface looks nice depends on the explicit expressions for x(u, v), y(u, v) and z(u, v). In general, if x(u, v), y(u, v) and z(u, v) depend differentiably on u and v, the surface usually does look nice.

4.1.5 Curves in space and their tangents Just like representing a curve in the plane by a parametric description of the form, x(t) = (x(t), y(t)), curves in 3–space can be described in the following way:

x(t) = (x(t), y(t), z(t)),

where t runs through (part of) the real numbers. For instance, x(t) = (cos t, sin t, 0) (with 0 t 2π) is a circle in the x, y–plane, but x(t) = (cos t, sin t, t) is a kind of spiral in 3–space.≤ ≤ By the way, both of these curves lie on the cylinder with equation x2 + y2 = 1, since cos2 t + sin2 t = 1. Just like in 2–space, a tangent vector to the curve at x(t0) is computed by differentiating the components of the curve, i.e.,

0 0 0 0 x (t0) = (x (t0), y (t0), z (t0))

is the tangent vector (at least if not all three components are 0). The tangent line at x(t0) has parametric description 0 x(t0) + λx (t0). For example, in the case x(t) = (cos t, sin t, t), the tangent vector is x0(t) = ( sin t, cos t, 1). In particular, the tangent vector at (1, 0, 0) corresponding to t = 0, is the −vector (0, 1, 1). The tangent line is then described by (1, 0, 0) + λ(0, 1, 1). The length of a (part of a) curve can be computed through an integral:

t=b x0(t) dt Zt=a | |

computes, under some mild conditions, the length of the part of the curve, where t ranges from a to b. In our example, the length of the curve if t varies from 0 to 4 is 4 ( sin t)2 + cos2 t + 12 dt = 4 √2 dt = 4√2. 0 − 0 R pThe tangent vectors themselvRes can be used to compute the curvature of a curve, but we will not explore that topic in these notes. 94 Surfaces

4.2 Some constructions of surfaces

4.2.1 In this section we present a few ways of constructing surfaces: surfaces of revolution and ruled surfaces.

Surfaces of revolution 4.2.2 Rotating a curve around a coordinate axis A surface of revolution arises by rotating a curve around a line. We have briefly discussed surfaces of revolution in Chapter 1, but only in a special case. Since the descriptions of surfaces of revolution in terms of equations or parametric descriptions can be quite complicated, we discuss this topic using an example. Suppose, we have a curve whose parametric description is as follows:

x(t) = (t, t2, t3).

Now suppose we rotate this curve around the x–axis and ask for a parametric description and an equation of the resulting surface of revolution.

Constructing a parametric description. • To construct a parametric description, take t to be fixed for the moment. The point (t, t2, t3) rotates along the circle in the plane x = t, whose radius is √t4 + t6 (the distance between (t, 0, 0) and (t, t2, t3)) and whose center is (t, 0, 0). A parametric description of this circle is (t, √t4 + t6 cos u, √t4 + t6 sin u). Varying t again, we find the parametric description of the surface of revolution:

x(t, u) = (t, √t4 + t6 cos u, √t4 + t6 sin u),

where 0 u 2π for instance. Here is maybe the most direct method: use a rotation ≤ ≤

Figure 4.3: Rotating the curve (t, t2, t3) around the x–axis.

matrix. The matrix 1 0 0  0 cos u sin u  0 sin u −cos u   4.2 Some constructions of surfaces 95

describes a rotation around the x–axis over an angle of u (radians, say). So we multiply each vector (t, t2, t3) with this matrix:

1 0 0 t t  0 cos u sin u   t2  =  t2 cos u t3 sin u  , 0 sin u −cos u t3 t2 sin u +−t3 cos u       and find the parametric description x(t, u) = (t, t2 cos u t3 sin u, t2 sin u + t3 cos u). − Constructing an equation. • There are several ways to find an equation. One is to start with the first parametric description obtained above and try to eliminate the parameters t and u. Since cos2 u+ sin2 u = 1 (for every u), the parameter u can be eliminated by taking the sum of the squares of the y– and z–coordinates:

y2 + z2 = (√t4 + t6 cos u)2 + (√t4 + t6 sin u)2 = t4 + t6.

Since x = t, we finally find y2 + z2 = x4 + x6. (Try eliminating t and u from the second parametric description yourself.) Here is another approach, without using the parametric description. Every point (t, t2, t3) will move along a circle in the plane x = t. This circle can be viewed as the intersection of the sphere with center (0, 0, 0) and radius x(t) , and the plane x = t. In terms of equations: | | x2 + y2 + z2 = t2 + t4 + t6, x = t. To get the equation of the surface of revolution, we have to eliminate the parameter t so that an equation in terms of x, y and z remains. This is easily done using the second equation: we replace t in the first equation by x, so we get x2 +y2 +z2 = x2 +x4 +x6. This simplifies to an equation of degree 6:

x4 x6 + y2 + z2 = 0. − − In general, if you rotate the curve x(t) = (x(t), y(t), z(t)) around the x–axis, you have to eliminate t from x2 + y2 + z2 = x(t)2 + y(t)2 + z(t)2, x = x(t).

Even for relatively simple curves, this may turn out to be a complicated matter.

4.2.3 The torus or ‘dough–nut’ Take the circle (x 2)2 + z2 = 1, y = 0 in the x, z–plane and rotate this circle around the z–axis. To get a parametric− description of the resulting shape in our hands, we start by parametrizing the circle: x = 2 + cos t, z = sin t, y = 0. 96 Surfaces

Figure 4.4: A torus is obtained by rotating a circle around an axis.

If we rotate this point around the z–axis over the angle u, then we obtain

cos u sin u 0 2 + cos t (2 + cos t) cos u −  sin u cos u 0   0  =  (2 + cos t) sin u  , 0 0 1 sin t sin t       where the first matrix on the left is the rotation matrix around the z–axis over the angle u. In this way we get a parametric description of a so–called torus. An equation for the torus is 2 x2 + y2 2 + z2 = 1. ³p − ´ This equation is obtained by eliminating t and u from the parametric description in the following two steps.

a) To eliminate u we square the expressions for x and y and add them:

x2 + y2 = (2 + cos t)2 cos2 u + (2 + cos t)2 sin2 u = (2 + cos t)2.

So x2 + y2 = 2 + cos t (the alternative x2 + y2 = (2 + cos t) is impossible since − xp2 + y2 0 and (2 + cos t) < 0). p ≥ − p b) To eliminate t, square x2 + y2 2 = cos t and z = sin t. Add the results to get − p 2 x2 + y2 2 + z2 = 1. ³p − ´ The equation becomes less pleasant if you expand the squares and/or try to get rid of the square sign.

4.2.4 Rotating a curve around an arbitrary line Suppose we rotate a curve around an arbitrary line `. Then any point on the rotating 4.2 Some constructions of surfaces 97

curve describes a circle within a plane perpendicular to the line. The center of the circle is the intersection of ` with this plane. The circle is also the intersection of any sphere with center on ` and passing through the point on the curve. For example, suppose we rotate the z–axis around the line ` : λ(1, 0, 1). A point on the z–axis is given by (0, 0, t). The sphere with center (0, 0, 0) (on `!) containing this point has equation x2 + y2 + z2 = t2. The plane perpendicular to ` and containing (0, 0, t) has equation x + z = t. So we have to eliminate t from x2 + y2 + z2 = t2, x + z = t. This is easily done, using the second equation: substitute x + z for t in the first equation and we get: x2 + y2 + z2 = (x + z)2. This can be rewritten as y2 = 2xz. Any idea what type of surface this is?

Ruled surfaces 4.2.5 Constructing surfaces using a moving line Ruled surfaces arise when you move a line in space according to certain rules. Basically, it comes down to the following. Start with a curve C in space. The surface consists of a ‘nicely’ varying family of lines all meeting the curve C and satisfying additional requirements. The curve C is called a directrix of the ruled surface. The lines are called the rulings. We demonstrate this technique of constructing surfaces in a few examples.

4.2.6 Cone over a curve A cone over a curve arises in the following way. Take a point P in space and a curve C parametrized by x(t) in space (not passing through P ). Now take the lines which connect P with a point on C. For instance, let C be the curve given by (0, cos t, sin t) (a circle in the y, z–plane), and let P be the point (1, 2, 3). A parametric description of the line passing through (1, 2, 3) and (0, cos t, sin t) (for fixed t) is (1, 2, 3) + λ(0 1, cos t 2, sin t 3). − − − So, x = 1 λ, y = 2 +− λ(cos t 2), z = 3 + λ(sin t −3) − To find an equation of the surface, we have to eliminate the parameters λ and t. From the first equation we conclude that we can replace λ by 1 x. The second and third equation then become y 2 = (1 x)(cos t 2) and z 3 = (1− x)(sin t 3). To eliminate t, we ‘isolate’ cos t and− sin t in−order to exploit− the relation− cos−2 t + sin2 t−= 1. Therefore, we first rewrite the equations as follows. y 2 + 2(1 x) = (1 x) cos t, z − 3 + 3(1 − x) = (1 − x) sin t. − − − 98 Surfaces

Figure 4.5: A ruled surface is constructed of (parts of) straight lines.

Now simplify the two left–hand sides, then square both equations and add the results:

(y 2x)2 + (z 3x)2 = (1 x)2. − − − Obviously, this equation represents a quadric.

4.2.7 A ruled surface all of whose lines are parallel to a given plane In this example, we construct a surface by moving a line in such a way that it always meets two given lines and is always parallel to a given plane. We take as lines the x–axis (`) and the line m which is the intersection of x = 0 and z = 2 (this is a translate of the y–axis). Now connect a point of ` with a point of m, but in such a way that the resulting line is parallel to the plane x + y + z = 0. You can imagine, that if n is such a connecting line, and if you change the point of intersection with ` a bit, you also have to change the point on m in order to keep the connecting line parallel to the plane. In this way the connecting lines which are also parallel to the plane x + y + z = 0 sweep out a surface. Here are the computational details. Take a point (t, 0, 0) on ` and a point (0, u, 2) on m. The line connecting these points has direction vector b = (0, u, 2) (t, 0, 0) = ( t, u, 2). Now the condition that b be − − parallel to the plane x+y +z = 0 means that b (1, 1, 1) = 0. So t+u+2 = 0. Therefore a parametric description of the connecting line•is −

(t, 0, 0) + λ( t, t 2, 2). − − So x = t λt, y = λ(t 2), z = 2λ. To find an equation of this surface, i.e., a relation between x−, y and z in whic− h λ and t do not appear, we have to eliminate t and λ. We use z = 2λ to replace λ by z/2, i.e., 2x = 2t zt and 2y = z(t 2). If we rewrite this as 2x = t(2 z) and 2y + 2z = tz, substitute (2−y + 2z)/z for t in −2x = t(2 z), and get rid of denominators− we get − 2xz = (2y + 2z)(2 z). − So we end up with the quadric yz + xz + z2 2y 2z = 0. − − 4.3 Surfaces: tangent vectors and tangent planes 99

A note of warning: In substituting (2y + 2z)/z for t the scrutinous reader may notice that we are a bit sloppy, since the z in the denominator may attain the value 0. In practice, however, this way of dealing with the elimination will almost always lead to the correct equation. It is beyond the scope of these notes to provide the relevant details.

4.3 Surfaces: tangent vectors and tangent planes

4.3.1 The tangent plane to a surface: parametric description A tangent line through a point P of a curve is a line that has the same direction as the curve in P . If P is a point on a surface, then the tangent lines of curves on the surface through P span a plane. This plane is called the tangent plane to the surface at P . Since two direction vectors are enough to determine the directions of a plane, we can use the parametric curves through the point arising from a parametric description of a surface. Suppose x(u, v) = (x(u, v), y(u, v), z(u, v)) is the parametric description of a (part of a) surface S, where the three coordinates depend differentiably on u and v. Differentiating x(u, v) with respect to u and v, respectively, produces in general two direction vectors of the tangent plane to S. More precisely, for given parameter values (u0, v0), the vector p = (x(u0, v0), y(u0, v0), z(u0, v0)) is a vector on S. It can serve as support vector for the tangent plane to S at p. The partial derivatives

xu(u0, v0) = (xu(u0, v0), yu(u0, v0), zu(u0, v0)), xv(u0, v0) = (xv(u0, v0), yv(u0, v0), zv(u0, v0))

of x(u, v) are direction vectors of the tangent plane. We denote the tangent plane to S at P (the point corresponding to p) by TP (S). A parametric description is given by

TP (S) : p + λxu(u0, v0) + µxv(u0, v0).

4.3.2 Example: the tangent plane to a plane Of course, the tangent plane to a plane should be the plane itself. Let us take an example to illustrate this. Suppose the plane S is given by the parametric description x(u, v) = (2, 3, 1)+u(1, 0, 1)+v(2, 1, 5). For u = v = 1 the corresponding point on S is P = (5, 2, 7). So p = (5, 2, 7) is the supp− ort vector for the tangent plane at P . To find the direction vectors, we rewrite the parametric description as x(u, v) = (2 + u + 2v, 3 v, 1 + u + 5v). − Differention with respect to u and v, respectively, yields the two vectors xu(u, v) = (1, 0, 1) and xv(u, v) = (2, 1, 5). So, in particular, −

xu(1, 1) = (1, 0, 1) and xv(1, 1) = (2, 1, 5). − You probably notice that these are exactly the direction vectors of S. Now, the tangent plane is given by (5, 2, 7) + λ(1, 0, 1) + µ(2, 1, 5). Since p = (5, 2, 7) is on S, we conclude − that the tangent plane TP (S) coincides with S, just like we expected. 100 Surfaces

4.3.3 Example: the tangent plane to a sphere A parametric description of the sphere S with center (0, 0, 0) and radius 2 is given by

x(u, v) = (2 cos v sin u, 2 sin v sin u, 2 cos u)

(note the factor 2). Now consider the point P on S corresponding to u = π/3 and v = 0, i.e., P = (√3, 0, 1). We expect the tangent plane at P to be a plane perpendicular to p = (√3, 0, 1). To verify this, let us compute the direction vectors. First we compute the two partial derivatives:

xu(u, v) = (2 cos v cos u, 2 sin v cos u, 2 sin u) and xv(u, v) = ( 2 sin v sin u, 2 cos v sin u, 0). − − Next, we substitute u = π/3 and v = 0 and find:

Figure 4.6: A sphere with a tangent plane.

(1, 0, √3) and (0, √3, 0). − It is easily checked that these two vectors are indeed perpendicular to p = (√3, 0, 1). So the tangent plane TP (S) is the plane (√3, 0, 1) + λ( 1, 0, √3) + µ(0, 1, 0). − 4.3.4 The tangent plane to a surface: equation Given the parametric description of the tangent plane to a surface, the tangent plane’s equation can of course be obtained by using the techniques of Chapter 2: find a vector normal to the direction vectors of the tangent plane. If, however, the surface is given by means of an equation, the equation of the tangent plane can be obtained directly: if S is given by the equation f(x, y, z) = 0, and P = (x0, y0, z0) is a point on S, then the vector

(fx(x0, y0, z0), fy(x0, y0, z0), fz(x0, y0, z0)) is perpendicular to the surface at P . (The proof is not hard, but we will not discuss it here.) So, the equation of the tangent plane is

fx(x , y , z )(x x ) + fy(x , y , z )(y y ) + fz(x , y , z )(z z ) = 0. 0 0 0 − 0 0 0 0 − 0 0 0 0 − 0 For the sphere x2 + y2 + z2 = 4, and the point P = (√3, 0, 1), the tangent plane becomes

2√3(x √3) + 2(z 1) = 0, or √3 x + z = 4. − − 4.4 Surface area 101

(This is the plane that we computed above using a parametric description.) For a plane, say 2x 3y + 5z = 5, the tangent plane at P = (1, 1, 0) has equation − − 2(x 1) 3(y 1) + 5(z 0) = 0, − − − − − or 2x 3y + 5z = 5, i.e., the plane we started with. − 4.3.5 Remark. A few subtleties are worth pointing out. The technique to compute the tangent plane from a parametric description of the • surface works if the three functions x(u, v), y(u, v) and z(u, v) are differentiable and the direction vectors produced from them are independent.

Likewise, computing the tangent plane in a point (x , y , z ) only works if the vector • 0 0 0 (fx(x0, y0, z0), fy(x0, y0, z0), fz(x0, y0, z0)) of partial derivatives is non–zero.

4.3.6 Example. A tangent plane of a hyperboloid To compute the tangent plane at (1, 1, 1) of the hyperboloid x2 + y2 z2 = 1, we first compute the three partial derivatives 2x, 2y and 2z. Therefore, the −tangent plane has − equation 2(x 1) + 2(y 1) 2(z 1) = 0 or x + y z = 1. − − − − − Maybe you expect the tangent plane to meet the hyperboloid only in (1, 1, 1), but, surpris- ingly enough, the tangent plane intersects the hyperboloid along the two lines 1) x = 1, y = z, and 2) y = 1, x = z. It is easy to verify that these lines are on the hyperboloid. In the exercises you will go through the necessary computations.

4.4 Surface area

4.4.1 If you decide to use a curved surface in a design, you probably need to know how ‘big’ it is in order to be able to tell how much material you need and what the weight is. This section is devoted to the issue of computing surface areas of curved surfaces. The result is that the surface area of a surface given by a parametric description (where the parameters run through the region R) can be given by a surface integral:

xu xv dudv. Z ZR | × | Below, we explain how this integral is obtained. Basic ingredients are the surface area of a parallellogram and the cross product of vectors.

4.4.2 The area of a rectangle and a parallellogram If a rectangle has length ` and width w, then its area is `w. Next, we consider a parallel- logram R with sides along the two vectors a = (a1, a2) and b = (b1, b2) in the plane. In this case, the area equals the length of one side, say a , multiplied by the distance of b | | 102 Surfaces

b φ a Figure 4.7: The area of a parallellogram spanned by the vectors a and b: Multiply the length a of a by the distance of b to the line spanned by a. This distance is b sin φ. | | | | ·

to this side. If the angle between a and b is φ (with 0 φ π), then this distance is b sin φ. So the area A(R) of R is ≤ ≤ | | a b sin φ. | | · | | · Our next goal is to find a way to express this area directly in terms of the coordinates of a and b. To do this, we first note that the area is related to the inner product a b = a b cos φ of a and b, an expression which involves cos φ rather than sin φ. •Since | | · | | · cos2 φ + sin2 φ = 1, we find the following relation between the inner product and the area:

(a b)2 + A(R)2 = a 2 b 2 cos2 φ + a 2 b 2 sin2 φ • = |a|2 · |b|2(cos· 2 φ + sin| |2 φ·)| =| a· 2 b 2. | | · | | | | · | | Since both a b and a 2 b 2 can be expressed in terms of coordinates, we conclude: • | | · | | A(R)2 = a 2 b 2(1 cos2 φ) = a 2 b 2 (a b)2 | |2 · | 2| 2− 2 | | · | | 2− • = (a1 + a2)(b1 + b2) (a1b1 + a2b2) = (a b a b )2. − 1 2 − 2 1 Therefore, A(R) = a b a b . | 1 2 − 2 1| (An area is positive, hence the absolute value signs.) A similar formula, but much more involved, can be given for the area of a parallellogram spanned by two vectors in 3–space. To describe the result we make a sidetour to discuss the so–called cross product of two vectors in 3–space.

4.4.3 Intermezzo: The cross product of two vectors in 3-space The direction of a plane in 3–space is determined as soon as you know a vector perpen- dicular to the plane. We have used this observation before in describing the equation of a plane. Now this idea of a normal vector has been extended to an explicit construction of a vector perpendicular to two given vectors, the so–called cross product. Moreover, there 4.4 Surface area 103

exists an easy way to compute the coordinates of the cross product from the coordinates of the two given vectors. The cross product of two non–zero vectors a and b in 3–space is a vector, which is usually written as a b, determined by the following properties: × a) It is perpendicular to a and b.

b) The three vectors a, b, a b (in this order) form a positively oriented triple of vectors × (just like the three standard basis vectors e1, e2, e3).

c) The length of the cross product is a b sin φ , where φ is the angle between a | | · | | · | | and b, i.e., it depends on the length of the two vectors and the angle they make.

In case one of the two vectors is the zero–vector 0 we define the cross product to be the zero–vector 0. A computation similar in nature to the one in 2–space leads to the following expression for the cross product in terms of coordinates. Given two vectors a = (a1, a2, a3) and b = (b1, b2, b3) in 3–space, the cross product is given by

a b = (a b a b , a b a b , a b a b ). × 2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 This expression is mostly used in computations.

4.4.4 Example. Let a = (1, 0, 1) and b = (2, 1, 3). The cross product a b is then − × (0 3 1 ( 1), 1 2 1 3, 1 ( 1) 0 2) = (1, 1, 1). · − · − · − · · − − · − − As a check, one can verify that this vector is indeed perpendicular to (1, 0, 1) and (2, 1, 3): − (1, 0, 1) (1, 1, 1) = 1 1 + 0 ( 1) + 1 ( 1) = 0, (2, 1, 3)• (1−, 1−, 1) = ·2 1 + · 1− ( 1) +· 3− ( 1) = 0. − • − − · − · − · − The (coordinates of the) cross product can of course also be used to find equations of planes. For example, the plane spanned by a and b has parametric description λ(1, 0, 1) + µ(2, 1, 3) and equation x y z = 0. The coefficients 1, 1, 1 occurring in the equation are the− coordinates of the −cross− product (1, 1, 1) of the−two−direction vectors a and b. − − 4.4.5 The cross product and the area of a parallellogram spanned by two vectors Just like in 2–space, the area of a parallellogram spanned by two vectors a and b in 3–space equals a b sin φ , i.e., the length of the cross product of a and b. Since we have a way of |computing| · | | · | the| cross product in terms of the coordinates of a and b, we find that the area A of the parallellogram spanned by the vectors a = (a1, a2, a3) and b = (b1, b2, b3) equals (a b a b , a b a b , a b a b ) . | 2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 | 104 Surfaces

4.4.6 Approximating surface areas using parametric descriptions Suppose x(u, v) is a parametric description of a surface S, where the parameters u and v range through some region R in u, v–space. So if u and v range through R, then x(u, v) moves through the points of (part of) the surface S. Using tangent vectors and tangent planes, we will indicate how the surface area of a (piece of a) parametrized surface can be computed. Here is the basic idea. Fix a vector r0 on S. The tangent plane at r0 is a good linear (flat) approximation to the surface. It is spanned by the tangent vectors xu and xv, evaluated at r0. If du and dv span a small rectangle in u, v-space, then xu du and xv dv span a small parallollogram in the tangent plane. The area of this parallellogram· is a lo·cal approximation of a piece of S. Now the area of this parallellogram is given by the absolute value of the cross product of the two tangent vectors multiplied by dudv, i.e.,

xu xv dudv. | × | Using these parallellograms we get a first order approximation of the area. The surface area itself is given by an integral, where u and v run through R,

xu xv dudv. Z ZR | × |

4.4.7 Computing xu xv in the case of a graph of a function | × | If our surface is the graph of f(x, y), then the most natural parametric description is x(u, v) = (u, v, f(u, v)). The two partial derivatives are

xu(u, v) = (1, 0, fu(u, v)) and xv(u, v) = (0, 1, fv(u, v)).

2 2 The cross product is ( fu, fv, 1) whose length is 1 + f + f . So the surface area of − − u v the part of the graph where x and y are in the rectanglep 0 x 3, 0 y 2 is given by ≤ ≤ ≤ ≤ 3 2 2 2 1 + fx + fy dydx. Z0 Z0 q For example, to compute the surface area of the graph of f(x, y) = 2x+3y, where 0 x 3 and 0 y 2, we evaluate the integral ≤ ≤ ≤ ≤ 3 2 √1 + 22 + 32 dy dx = 3 2 √14 = 6√14 . Z0 µZ0 ¶ · · 4.5 Curvature of surfaces

4.5.1 Curvature is an intuitively clear notion. For instance, you will agree that a sphere with radius 1 is more curved than a sphere with radius 10. Yet, in more complicated situa- tions you may loose your grip on the notion of curvature. For instance, in the case of a hyperboloid the surface seems to bend differently in different directions. This section is devoted to the mathematical notion of total curvature, which makes the intuitive notion 4.5 Curvature of surfaces 105

more precise. There are more subtle notions of curvature, but those are beyond the scope of these notes. One aspect of curvature we mention right at the beginning: If we rotate or translate a surface, its curvature does not change. This is sometimes useful in actual computations, where you can move the surface to your favourite position before doing the computations. We will make use of this observation when convenient. For the graph of a (differentiable) function g, the curvature at (p, q) turns out to be

2 gxx(p, q)gyy(p, q) g (p, q) − xy 2 2 2 . (1 + gx(p, q) + gy(p, q) ) You may have come across the expression in the numerator before, since it is used in analysing extreme values of a function of two variables (maximum, minimum, saddle point).

4.5.2 Curvature of curves in the plane We say that a circle of radius r has (everywhere constant) curvature 1/r. So, the larger the radius, the smaller the curvature, in accordance with common intuition. For a general curve, the curvature in a point of the curve is related to the circle which ‘touches best’ to the curve in the given point. Suppose that the curve C passes through the origin O = (0, 0) and that the x–axis is the tangent line to the curve at O. Let us assume that the curve is the graph of a function f(x). So f 0(0) = 0 since we are assuming that the x–axis is the tangent line at O. Any circle whose center is on the y–axis, and which passes through O, has the x–axis as tangent line at O. Now which of these fits ‘best’? One way to deal with this problem is to look at the Taylor expansion of f and compare it with the Taylor expansion of the circle. First, we note that the lower part of the circle with center (0, r) and radius r is described by the function g(x) = r √r2 x2 (for r x r). To find this, we started from the equation x2 + (y r)2 = r−2. The−Taylor expansion− ≤ ≤of f(x) starts x2f 00(0) − x2f 00(0) with f(0) + xf 0(0) + + , and in our case this simplifies to . The Taylor 2 · · · 2 expansion of g(x) around x = 0 looks like 1 x2 + 2 · r · · · So the circle with radius r = 1/f 00(0) fits best in the sense that the graph and the circle agree ‘up to order 2’. This approach is at the basis of the definition of curvature of a curve in the plane: in the situation we just discussed, the (absolute) curvature is said to be 1/ f 00(0) . This definition assumes that f 0(0) = 0 and works only if f 00(0) = 0. | | 6 In a sense, curvature comes down to finding the radius of the circle which fits best. Since there is more room in 3–space, for surfaces, the situation is more involved.

4.5.3 Intermezzo: second order approximations to functions For functions of one variable f(x) you are familiar with its Taylor expansions. For instance, x3 x5 sin x = x + − 3! 5! − · · · 106 Surfaces

is the Taylor expansion of sin x around x = 0. There are similar expansions for functions of two variables. Usually, they can be derived from well–known series for functions of one variable. Here is an example. To get an expansion for sin(x2 + y2) around (0, 0), we use the above expansion and subsitute x2 + y2 for x:

(x2 + y2)3 sin(x2 + y2) = x2 + y2 + − 3! · · · The second order approximation contains exactly all linear and degree two terms of the expansion. So for sin(x2 + y2), the second order approximation is x2 + y2 (around 0). Here are a few examples and remarks:

To compute the second order approximation of 1 x2 y2 around (0, 0), we use • − − the first order expansion of √1 t around t = 0:p1 t/2 + (plus higher order). If we substitute x2 + y2 for t we−already get the desired− second· · ·order approximation: 1 (x2 + y2)/2. − In general, the 2nd order approximation around x = 0, y = 0 of a differentiable • function f(x, y) is 1 f(0, 0) + x f (0, 0) + y f (0, 0) + x2f (0, 0) + 2xyf (0, 0) + y2f (0, 0) . x y 2 xx xy yy ¡ ¢ Around the point x = a, y = b, the second order expansion is

f(a, b)+ (x a)fx(a, b) + (y b)fy(a, b)+ 1 − 2 − 2 ((x a) fxx(a, b) + 2(x a)(y b)fxy(a, b) + (y b) fyy(a, b)) . 2 − − − − The second order expansion of f(x, y) = cos xy around (0, 0) is computed via its • partial derivatives:

fx = y sin(xy), fy = x sin(xy), − 2 − 2 fxx = y cos(xy), fxy = sin(xy) xy cos(xy), fyy = x cos(xy). − − − − So the 2nd order expansion becomes 1.

The second order expansion is also used to investigate (local) extrema of functions. • 4.5.4 Curvature of surfaces: first considerations The idea of a ‘best touching circle’ is easily generalized to a ‘best touching sphere’ in the case of surfaces, but alas, that is too simple. For instance, take a point on the cylinder x2 + y2 = 1. Suppose its z–coordinate is 3. It is intuitively clear that the best touching sphere is a sphere with radius 1 and center at (0, 0, 3). But the sphere does not seem to take into account that along a vertical line on the cylinder, the cylinder is not curved. For a horizontal cross section the sphere seems fine, but not for a vertical cross section. If we use the radius of the sphere as a measure for the curvature of the cylinder, then both 4.5 Curvature of surfaces 107

sphere and cylinder would have the same curvature, which early does not correspond to our intuition. It seems that the 2–dimensionality of the surface makes things more subtle. Instead we turn to ‘best 2nd order approximation’ in the sense of Taylor expansions. Or, more geometrically, best quadric approximation. First consider the sphere x2 +y2 +(z − r)2 = r2. It passes through the origin and its tangent plane is horizontal. The lower half of this sphere can be described as the graph of the function f(x, y) = r r2 x2 y2. − − − The expansion around (0, 0) looks like p

1 x2 y2 f(x, y) + . ∼ 2 µ r r ¶

In this expression we see appearing a curvature in the x–direction (where y = 0, and x is allowed to change) and in the y–direction. We say that the total curvature is the product of the coefficients 1/r of the term x2 and y2 in the Taylor expansion (so we ignore the factor 1/2), i.e., 1 1 1 = . r · r r2 Now let us take a surface passing through the origin, with a horizontal tangent plane at (0, 0, 0). Suppose the surface is given as the graph of the function g(x, y). Then the Taylor expansion up to order 2 looks like 1 1 g(0, 0) + xg (0, 0) + yg (0, 0) + (g x2 + 2g xy + g y2) = (g x2 + 2g xy + g y2) x y 2 xx xy yy 2 xx xy yy

(where we have used g(0, 0) = 0, gx(0, 0) = gy(0, 0) = 0). This quadratic expression is the best 2nd order approximation to the surface. Unfortunately, there is a mixed term in the expression. By rotating the graph we can get rid of this term (we will not do this computation here, it is similar in spirit to the computation in (3.1.4) on page 61). This complicates matters in the sense that the correct definition of the total curvature is not the product of the coefficients of x2 and y2 (multiplied by 2), but is the following more elaborate expression: 2 gxx(0, 0)gyy(0, 0) g (0, 0) − xy (which simplifies to gxx(0, 0)gyy(0, 0) if the mixed term is absent). But this expression is only correct if the tangent plane at O is horizontal. By using suitable rotations, the curvature in a point (p, q) of the graph of g(x, y) can be shown to be

2 gxx(p, q)gyy(p, q) g (p, q) − xy 2 2 2 . (1 + gx(p, q) + gy(p, q) )

For short, we usually write 2 gxxgyy g − xy 2 2 2 , (1 + gx + gy) omitting the coordinates of the point. 108 Surfaces

4.5.5 Example: The curvature of an ellipsoid x2 y2 z2 To study the curvature of the standard ellipsoid + + = 1, at (0, 0, 5), we first 22 32 52 − note that the tangent plane is horizontal at (0, 0, 5). So we move on to describing part of the ellipsoid as the graph of a function. The relev− ant part is the graph of the function

x2 y2 f(x, y) = 5 1 − r − 22 − 32 The second order approximation at (0, 0) of this function is 1 x2 1 y2 1 5x2 5y2 5(1 ) = 5 + + . − − 2 · 22 − 2 · 32 − 2 µ 22 32 ¶ 5 5 25 So the curvature is · = (note again that we ignore the factor 1/2). In general, the 22 32 36 · x2 y2 z2 curvature of the ellipsoid with equation + + = 1 at (0, 0, c) is a2 b2 c2 − c2 K(0, 0, c) = . − a2 b2 · This is in agreement with intuition: if a and b get larger, then the ellipsoid becomes ‘flatter’ near (0, 0, c), whereas the curvature increases, if c gets larger. −

x2 y2 z2 Figure 4.8: From left to right: The curvature of the ellipsoid + + = 1 at the ‘south a2 b2 c2 pole’ increases with increasing c.

In the special case of a sphere of radius r, the coefficients a, b and c are all equal to r. The resulting total curvature of a sphere in (0, 0, r) is − r2 1 K(0, 0, r) = = . − r2 r2 r2 · Since a sphere is fully symmetric, we conclude that the curvature of the sphere is 1/r2 at any point. 4.5 Curvature of surfaces 109

4.5.6 Example: The curvature of a right circular cylinder To compute the curvature of the cylinder x2 + y2 = 1 in a point, we first note that we cannot describe this cylinder as the graph of a function of x and y. But we can describe part of the cylinder as the graph of y = √1 x2 (a function of x and z). Let us compute the curvature at the point P = (0, 1,−0). So− we start computing partial derivatives of f(x, z) = √1 x2 at x = 0 and z =−0. The first partial derivatives are both 0: − − x fx(0, 0) = (0,0) = 0, fz(0, 0) = 0 √1 x2 | − (the partial derivative of f with respect to z is 0 anyway, since √1 x2 does not conatin − − the variable z). Next, we need fxx(0, 0), fzz(0, 0) and fxz(0, 0). Since the expression 2 √1 x for f(x, z) does not involve z, we conclude that fzz(0, 0) and fxz(0, 0) are both 0.− So− 2 2 fxx(0, 0) fzz(0, 0) fxz(0, 0) = fxx(0, 0) 0 0 = 0. − · − Hence, the curvature is 0.

4.5.7 On the meaning of curvature Here are some remarks on curvature. We use the graph of a function f(x, y) to support our observations. We assume f(0, 0) = 0, fx(0, 0) = 0 and fy(0, 0) = 0 and consider the graph near (0, 0, 0) in 3–space.

a) Suppose the curvature is positive at (0, 0, 0). This means that fxx(0, 0)fyy(0, 0) 2 − fxy(0, 0) > 0. Now this condition means that f(x, y) has a (local) minimum or maximum at (0, 0), so the graph tends to be above the x, y–plane or below it near (0, 0, 0). In general, positive curvature means that the surface tends to on one side of the tangent plane.

b) Negative curvature corresponds to the fact that f(x, y) has neither a minimum nor a maximum at (0, 0). The function has a saddle point at (0, 0).

c) If the curvature is 0, there is not much one can say in general. If the curvature is 0 for all points near a given point, the surface tends to be flat in at least one direction. For instance, the cylinder x2 + y2 = 1 has curvature 0 in every point. It is not curved in vertical directions.

4.5.8 Other approaches to curvature: via unit normal vectors1 The approach via Taylor expansions is not the only way to analyse the notion of curvature. Two other approaches use unit normal vectors to the surface. A unit normal vector to a surface S in P is a vector N of length 1 which is perpendicular to the tangent plane TP (S) at P . If P runs through the surface, the vector N becomes a vector valued function. For example, for the sphere x2 + y2 + z2 = 4, the unit normal vector to the sphere at P = (x, y, z) is N(P ) = (x/2, y/2, z/2). In terms of vectors: N(p) = p/2. To study the

1This part can be omitted. 110 Surfaces

curvature of a surface, one analyses the variation of N. The idea is that the more curved a surface is, the more N will vary. This is worked out in two ways (which we will not elaborate any further). Measuring the derivative • To analyse the variation of N, the derivative of N is studied. This derivative turns out to be a 2 by 2 matrix for each point of the surface. The so–called determinant of the matrix, a suitable combination of the coefficients of the matrix, is the curvature. Comparing surface areas • The vector N(P ) lies on the unit sphere x2 + y2 + z2 = 1 for all P . If P runs through a small portion of S, then the unit normal vector will run through some portion of the unit sphere. For example, if the surface S is a plane, the unit normal vector is a constant vector of length one. No matter what portion of the plane P runs through, the unit normal vector is just a fixed single vector on the unit sphere. The area of this single point is evidently 0 times the area of the portion of the plane. This ‘explains’ that the curvature of a plane is 0. For the sphere S with equation x2 + y2 + z2 = 4 the situation is different. If P runs through the part of the sphere with x 0, y 0, the unit normal vector runs through the part of x2 + y2 + z2 = 1 satisfying≥ x ≥ 0 and y 0. In this case the area of the portion on the unit sphere is 1/4 times≥the area of≥the part on S. This ‘explains’ that the curvature of a sphere of radius 2 is 1/4. In general, if a point runs through a part of a surface containing point P with area R and the unit normal vector sweeps out a piece of area T , then the quotient T/R is an approximation of the curvature at P . By taking smaller and smaller pieces around P , i.e., taking the limit where R 0, we get the curvature at P . → 4.5.9 Other approaches to curvature: via curves Yet another approach to curvature uses curves on the surface. The idea is to consider the curvature of certain curves on a surface through a given point on the surface. It turns out that, for a given point P on the surface, this curvature is always between a certain minimum value k1 and a maximum value k2. Surprisingly enough, the product k1 k2 is the total curvature of the surface at P . · 2 2 For example, for the cylinder x + y = 1 the minimum k1 is attained for the vertical lines (and equals 0), the maximum k2 is attained by the horizontal circles and equals 1. Hence, the total curvature is 0 1 = 0. · 4.5.10 Curvature of surfaces using parametric descriptions The curvature can also be expressed in terms of parametric descriptions of a surface. Sim- ilar computations as above reveal that for the parametric description x(u, v) the absolute value of the curvature at x(u0, v0) is given by

2 (N xuu) (N xvv) (N xuv) • 2 · •2 − •2 . xu xv (xu xv) | | · | | − • 4.6 There is much more on surfaces 111

The denominator is the squared area of the parallellogram spanned by the two tangent vectors xu and xv. In the case of a function, say x(x, y) = (x, y, f(x, y)), the tangent vectors are xx = (1, 0, fx) and xy = (0, 1, fy) and the cross product is ( fx, fy, 1) whose 2 2 − − squared length is 1+fx +fy , a factor which appeared in the denominator for the curvature in (4.5.4).

4.5.11 The curvature of a ruled surface is non–positive We use the expression for the curvature in terms of a parametric description to derive the fact that the total curvature of a ruled surface at any point is 0 (non-positive). Here is why: a ruled surface is always of the form x(u, v) = α(u) + vβ≤(u). (Through the point α(u) a line with direction β(u) is drawn.) Since

xvv(u, v) = 0

the factor N xvv in the formula for the curvature vanishes. But, then the curvature is of the form • 2 (N xuv) 2 − 2• 2 . xu xv (xu xv) | | · | | − • The nominator is obviously always non–positive. Fortunately, the denominator is always positive (if the tangent vectors xu and xv are non–zero)! To see this, we use the fact that xu xv = xu xv cos φ, where φ is the angle between xu and xv. So here is the computation:• | | · | |

2 2 2 2 2 2 2 2 2 2 2 xu xv (xu xv) xu xv xu xv cos φ = xu xv sin φ 0. | | · | | − • | | · | | − | | · | | | | · | | ≥

In fact, the denominator of the fraction is the squared length of the cross product of xu and xv, hence always non–negative. As the nominator is 0 and the denominator is always positive, the curvature is non– ≤ positive. So, any time you construct a surface by moving a line through space, its curvature will be non–positive.

4.6 There is much more on surfaces

4.6.1 There is a lot about surfaces we have not been able to discuss. Here is a short list of related topics.

We have characterized curvature mainly through one number, the total curvature, • but one can analyse the curvature of curves on a surface in more detail with surprising results.

The mean curvature of a surface is a quantity related to the curvature. A surface • whose mean curvature is always 0 is a minimal surface (well, there are a few provisos). Soap films are examples of minimal surfaces. 112 Surfaces

We have not discussed ‘glueing’ surfaces together, like glueing a spherical cap on top • of a cylinder (of finite height).

Numerical approximations for our computations are beyond the scope of this course. • 4.7 Exercises 113

4.7 Exercises

1 The tangent plane to a cylinder a) Compute the (equation or parametric description of the) tangent plane to the cylinder x2 + y2 = 4 at (1, √3, 5).

b) At which points of the cylinder is the tangent plane the same as the one in a)?

c) Show that the intersection of the tangent plane from a) with the cylinder is a line.

2 The intersection of a hyperboloid and a tangent plane In Example (4.3.6) we noticed that the hyperboloid x2 + y2 z2 = 1 and its tangent plane intersect along two lines. Here is the (better: a) computation− that leads to this result. We have to solve the system of two equations x2 + y2 z2 = 1 and x + y z = 1. − − a) Show how to rewrite the equations as (x+z)(x z) = (1+y)(1 y) and x z = 1 y. − − − − b) Suppose for a moment that y = 1. Then show that the system can be rewritten as x + z = 1 + y and x z = 1 6 y. Now argue that this leads to x = 1 and y = z. − − c) Suppose that y = 1. Show that this leads to the line y = 1 and x = z in the intersection.

3 Hyperboloic paraboloids are ruled surfaces Let S be the hyperbolic paraboloid z = xy (by rotating it over 45◦ you get the equation in standard form). a) Show that the intersection of the planes y = z/3 and x = 3 is a line on S.

b) Show that for each t = 0, the intersection of the planes y = z/t and x = t is a line 6 on S. These lines are indicated by Lt.

c) Show that (0, 1, t) is parallel to Lt. Moreover, show by eliminating t and u that (t, 0, 0) + u(0, 1, t) describes S as a ruled surface.

4 On surfaces of revolution In (4.2.2) on page 94 we have set up two equations from which to eliminate t: x2 +y2 +z2 = t2 + t4 + t6 and x = t. Show that if you rotate the curve (t, t2, t3) around the x–axis, you can also use the cylinder y2 + z2 = t4 + t6 and the plane x = t as your starting point.

5 Rotating a line a) Rotate the line (1, 0, 0) + λ(0, 1, 1) around the z–axis. Use a sphere with center O and a horizontal plane to find the−equation of the resulting surface.

b) Now rotate the same line around the line through the point (1, 0, 0) and parallel to the z–axis. What is the equation of the resulting surface? 114 Surfaces

6 Rotating a hyperbola 2 2 Let C be the hyperbola y z = 1 in the y, z–plane (so x = 0). The part C+ where y > 0 can be parametrized by (0,−cosh t, sinh t).

a) Use a rotation matrix to describe the surface that arises if you rotate C+ around the z–axis. What is the equation of this surface?

b) Similar question, but now rotate around the y–axis. The equation you get is also satisfied by points with negative coordinates. Explain!

7 The exterior product

a) Compute the exterior product of the vectors (2, 3, 1) and (1, 0, 2). Same question for the vectors (2, 3, 1) and (4, 6, 2). Can you explain the answer?

b) Use the exterior product to find an equation for the plane λ(2, 3, 1)+µ(1, 0, 2) spanned by (2, 3, 1) and (1, 0, 2).

c) The planes x + y + 2z = 0 and 4x + 6y 2z = 0 intersect along a line. Determine − a parametric description of this line. Compare your result with the exterior product of the normal vectors (1, 1, 2) and (4, 6, 2) of the two planes. −

d) The vectors n1 and n2 are normal vectors of the distinct planes V1 and V2 through the origin, respectively. Argue why n1 n2 spans the line of intersection of the two lines. ×

8 Surface areas In each of the following cases, compute the surface area of the indicated surfaces.

a) The plane 3x + 2y + z = 5, where 1 x 3 and 1 y 5. ≤ ≤ − ≤ ≤ b) The surface given by the graph of f(x, y) = cosh(x), where 0 x 2 and 0 y 4. ≤ ≤ ≤ ≤ c) The surface given by the graph of f(x, y) = xy, where x2 + y2 16. [Hint: use polar coordinates.] ≤

d) The part of the paraboloid z = x2 + y2 inside the cylinder x2 + y2 = 1.

9 Surface area of a plane Over the unit square 0 x 1, 0 y 1 a plane is constructed which passes through (0, 0, 0), (1, 0, a) and (1,≤1, b)≤(for some≤ p≤ositive numbers a and b with b > a). Determine the area of the piece of the roof above the unit square in terms of a and b.

10 Surface area of a piece of a sphere If you intersect the sphere x2 + y2 + z2 = 4 with the plane x + y + z = 3, the intersection is a circle. This circle bounds a ‘cap’ of the sphere. Compute the area of such a cap in the following steps. 4.7 Exercises 115

a) Argue why you can just as well compute the area of the ‘cap’ cut off by the plane z = √3.

b) Describe the upper–half of the sphere as the graph of a function and set up the integral that computes the surface area. What is the region over which to integrate?

c) Use polar coordinates to evaluate the integral.

d) Now we use the parametric description x(u, v) = (2 cos u sin v, 2 sin u sin v, 2 cos v) to compute the surface area. Which u, v–region corresponds to the ‘cap’?

e) Compute xu and xv and find the length xu xv of the exterior product. | × | f) Set up the surface integral and evaluate the integral.

11 Surface area of a cone In this exercise we look at the part of the cone z2 = x2 + y2, where z 0. We consider the piece inside the cylinder x2 + y2 = 4. ≥

a) Describe this part as the graph of a function and set up the corresponding surface integral. Evaluate he integral.

b) Now compute the area using a parametric description like x(r, φ) = (r cos φ, r sin φ, r).

12 The curvature of a cylinder In (4.5.6) on page 109 we computed the curvature of the cylinder x2 + y2 = 1 in (0, 1, 0) by viewing part of the cylinder as the graph of y = √1 x2. In this exercise we lo−ok a bit closer at the cylinder. − −

a) Argue why the curvature of the cylinder will be the same in every point.

b) Argue why you can just as well compute the curvature of the cylinder x2 + z2 = 1.

c) Show why the curvature of cylinder is 0 in every point.

13 The curvature of a hyperboloid In this exercise we compute the curvature of the hyperboloid x2 + y2 z2 = 1 in (1, 0, 0). − a) Find a suitable way of describing a portion of the surface containing (1, 0, 0) as the graph of a function (a function of x and y, or of x and z, or of y and z).

b) Now compute the curvature. Which points of the hyperboloid will have the same curvature?