**16.6** Parametric Surfaces and

### Their Areas

### Parametric Surfaces and Their Areas

Here we use vector functions to describe more general
*surfaces, called parametric surfaces, and compute their *
areas.

Then we take the general surface area formula and see how it applies to special surfaces.

### Parametric Surfaces

### Parametric Surfaces

In much the same way that we describe a space curve by a
**vector function r(t) of a single parameter t, we can describe ****a surface by a vector function r(u, v) of two parameters ***u and v.*

We suppose that

**r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k**

*is a vector-valued function defined on a region D in the *
*uv-plane.*

### Parametric Surfaces

**So x, y, and z, the component functions of r, are functions ***of the two variables u and v with domain D. *

*The set of all points (x, y, z) in such that*

*x = x(u, v) y = y(u, v)* *z = z(u, v) *
**and (u, v) varies throughout D, is called a parametric ****surface S and Equations 2 are called parametric **

**equations of S. **

### Parametric Surfaces

*Each choice of u and v gives a point on S; by making all *
*choices, we get all of S. *

*In other words, the surface S is traced out by the tip of the *
**position vector r(u, v) as (u, v) moves throughout the **

*region D. (See Figure 1.)*

**Figure 1**

A parametric surface

### Example 1

Identify and sketch the surface with vector equation
**r(u, v) = 2 cos u i + v j + 2 sin u k**

Solution:

The parametric equations for this surface are
*x = 2 cos u y = v z = 2 sin u*

*Example 1 – Solution*

*So for any point (x, y, z) on the surface, we have*
*x*^{2} *+ z*^{2} = 4 cos^{2}*u + 4 sin*^{2}*u *

= 4

This means that vertical cross-sections parallel to the
*xz-plane (that is, with y constant) are all circles with *
radius 2.

cont’d

*Example 1 – Solution*

*Since y = v and no restriction is placed on v, the surface is *
*a circular cylinder with radius 2 whose axis is the y-axis *
(see Figure 2).

cont’d

### Parametric Surfaces

*If a parametric surface S is given by a vector function *

**r(u, v), then there are two useful families of curves that lie ***on S, one family with u constant and the other with v*

constant.

These families correspond to vertical and horizontal lines in
*the uv-plane.*

### Parametric Surfaces

*If we keep u constant by putting u = u*_{0}**, then r(u**_{0}*, v) *

*becomes a vector function of the single parameter v and *
*defines a curve C*_{1 }*lying on S. (See Figure 4.)*

### Parametric Surfaces

*Similarly, if we keep v constant by putting v = v*_{0}, we get a
*curve C*_{2} **given by r(u, v**_{0}*) that lies on S. *

**We call these curves grid curves. (In Example 1, for **

*instance, the grid curves obtained by letting u be constant *
*are horizontal lines whereas the grid curves with v constant *
are circles.)

In fact, when a computer graphs a parametric surface, it

usually depicts the surface by plotting these grid curves, as we see in the next example.

### Example 2

Use a computer algebra system to graph the surface

**r(u, v) = 〈(2 + sin v) cos u, (2 + sin v) sin u, u + cos v〉**

*Which grid curves have u constant? Which have v constant?*

*Example 2 – Solution*

We graph the portion of the surface with parameter domain 0 ≤ u ≤ 4π, 0 ≤ v ≤ 2π in Figure 5.

**Figure 5**

*Example 2 – Solution*

It has the appearance of a spiral tube.

To identify the grid curves, we write the corresponding parametric equations:

*x = (2 + sin v) cos u y = (2 + sin v) sin u z = u + cos v*
*If v is constant, then sin v and cos v are constant, so the *
parametric equations resemble those of the helix.

*Thus the grid curves with v constant are the spiral curves in *
Figure 5.

cont’d

*Example 2 – Solution*

*We deduce that the grid curves with u constant must be the *
curves that look like circles in the figure.

*Further evidence for this assertion is that if u is kept *

*constant, u = u*_{0} *, then the equation z = u*_{0} *+ cos v shows *
*that the z-values vary from u*_{0} *– 1 to u*_{0} + 1.

cont’d

### Example 4

Find a parametric representation of the sphere
*x*^{2} *+ y*^{2} *+ z*^{2} *= a*^{2}

Solution:

The sphere has a simple representation ρ *= a in spherical *
coordinates, so let’s choose the angles φ and θ in spherical
coordinates as the parameters.

*Example 4 – Solution*

Then, putting ρ *= a in the equations for conversion from *
spherical to rectangular coordinates, we obtain

*x = a sin *φ cos θ *y = a sin *φ sin θ *z = a cos *φ
as the parametric equations of the sphere.

The corresponding vector equation is

**r(**φ, θ*) = a sin *φ cos θ * i + a sin *φ sin θ

*φ*

**j + a cos****k**

cont’d

*Example 4 – Solution*

We have 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π, so the parameter
*domain is the rectangle D = [0, *π] × [0, 2π].

The grid curves with φ constant are the circles of constant latitude (including the equator).

cont’d

*Example 4 – Solution*

The grid curves with θ constant are the meridians

(semi-circles), which connect the north and south poles (see Figure 7).

**Figure 7**

cont’d

### Parametric Surfaces

**Note**

We saw in Example 4 that the grid curves for a sphere are curves of constant latitude or constant longitude.

For a general parametric surface we are really making a map and the grid curves are similar to lines of latitude and longitude.

Describing a point on a parametric surface (like the one in
*Figure 5) by giving specific values of u and v is like giving *
the latitude and longitude of a point.

### Surfaces of Revolution

### Surfaces of Revolution

Surfaces of revolution can be represented parametrically
and thus graphed using a computer. For instance, let’s
*consider the surface S obtained by rotating the curve *
*y = f(x), a* *≤ x ≤ b, about the x-axis, where f(x) ≥ 0. *

Let θ be the angle of rotation as shown in Figure 10.

### Surfaces of Revolution

*If (x, y, z) is a point on S, then*

*x = x y = f(x) cos *θ *z = f(x) sin *θ

*Therefore we take x and *θ as parameters and regard
*Equations 3 as parametric equations of S. *

*The parameter domain is given by a* *≤ x ≤ b, 0 ≤* θ ≤ 2π .

### Example 8

Find parametric equations for the surface generated by
*rotating the curve y = sin x, 0 ≤ x ≤ 2*π*, about the x-axis. *

Use these equations to graph the surface of revolution.

Solution:

From Equations 3, the parametric equations are

*x = x y = sin x cos *θ *z = sin x sin *θ
and the parameter domain is 0 ≤ x ≤ 2π , 0 ≤ θ ≤ 2π*.*

*Example 8 – Solution*

Using a computer to plot these equations and then rotating the image, we obtain the graph in Figure 11.

**Figure 11**

cont’d

### Tangent Planes

### Tangent Planes

*We now find the tangent plane to a parametric surface S *
traced out by a vector function

**r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k***at a point P*_{0} **with position vector r(u**_{0}*, v*_{0}).

### Tangent Planes

*If we keep u constant by putting u = u*_{0}**, then r(u**_{0}*, v) *

*becomes a vector function of the single parameter v and *
*defines a grid curve C*_{1} *lying on S. (See Figure 12.) The *
*tangent vector to C*_{1} *at P*_{0} is obtained by taking the partial
**derivative of r with respect to v:**

### Tangent Planes

*Similarly, if we keep v constant by putting v = v*_{0}, we get a
*grid curve C*_{2 }**given by r(u, v**_{0}*) that lies on S, and its tangent *
*vector at P*_{0} is

**If r**_{u}**× r**_{v}* is not 0, then the surface S is called smooth *
(it has no “corners”).

**For a smooth surface, the tangent plane is the plane that **
**contains the tangent vectors r**_{u}**and r*** _{v}*, and the vector

**r**_{u}**× r*** _{v}* is a normal vector to the tangent plane.

### Example 9

Find the tangent plane to the surface with parametric
*equations x = u*^{2}*, y = v*^{2}*, z = u + 2v at the point (1, 1, 3).*

Solution:

We first compute the tangent vectors:

*Example 9 – Solution*

Thus a normal vector to the tangent plane is

Notice that the point (1, 1, 3) corresponds to the parameter
*values u = 1 and v = 1, so the normal vector there is*

**–2 i – 4 j + 4 k**

cont’d

*Example 9 – Solution*

Therefore an equation of the tangent plane at (1, 1, 3) is
*–2(x – 1) – 4(y – 1) + 4(z – 3) = 0*

or

*x + 2y – 2z + 3 = 0*

cont’d

### Surface Area

### Surface Area

Now we define the surface area of a general parametric surface given by Equation 1.

For simplicity we start by considering a surface whose
*parameter domain D is a rectangle, and we divide it into *
*subrectangles R** _{ij}*.

### Surface Area

Let’s choose to be the lower left corner of R* _{ij}*.
(See Figure 14.)

**Figure 14**

*The image of the subrectangle R*_{ij}*is the patch S** _{ij}*.

### Surface Area

*The part S*_{ij}*of the surface S that corresponds to R** _{ij}* is called

*a patch and has the point P*

*with position vector as one of its corners.*

_{ij}Let

and

*be the tangent vectors at P** _{ij}* as given by Equations 5 and 4.

### Surface Area

Figure 15(a) shows how the two edges of the patch that

*meet at P** _{ij}* can be approximated by vectors. These vectors,
in turn, can be approximated by the vectors and
because partial derivatives can be approximated by

difference quotients.

**Figure 15(a)**

Approximating a patch by a parallelogram.

### Surface Area

*So we approximate S** _{ij}* by the parallelogram determined by
the vectors and .

This parallelogram is shown in Figure 15(b) and lies in the
*tangent plane to S at P*_{ij}.

Approximating a patch by a parallelogram.

### Surface Area

The area of this parallelogram is

*and so an approximation to the area of S is*

### Surface Area

Our intuition tells us that this approximation gets better as we increase the number of subrectangles, and we

recognize the double sum as a Riemann sum for the double integral

### Surface Area

This motivates the following definition.

### Example 10

*Find the surface area of a sphere of radius a.*

Solution:

In Example 4 we found the parametric representation
*x = a sin *φ cosθ *y = a sin *φ sin θ *z = a cos *φ
where the parameter domain is

*D = {(*φ, θ*) | 0 ≤* φ ≤ π, 0 ≤ θ ≤ 2π}

*Example 10 – Solution*

We first compute the cross product of the tangent vectors:

cont’d

*Example 10 – Solution*

*= a*^{2} sin^{2}φ cos θ **i + a**^{2 }sin^{2 }φ sin θ **j + a**^{2} sin φ cos φ **k**
Thus

since sin φ ≥ 0 for 0 ≤ φ ≤ π.

cont’d

*Example 10 – Solution*

Therefore, by Definition 6, the area of the sphere is

cont’d

### Surface Area of the Graph of a function

### Surface Area of the Graph of a Function

*For the special case of a surface S with equation z = f(x, y), *
*where (x, y) lies in D and f has continuous partial *

*derivatives, we take x and y as parameters. *

The parametric equations are

*x = x y = y z = f(x, y)*
so

and

### Surface Area of the Graph of a Function

Thus we have

and the surface area formula in Definition 6 becomes

### Example 11

*Find the area of the part of the paraboloid z = x*^{2} *+ y*^{2 } that
*lies under the plane z = 9.*

Solution:

The plane intersects the paraboloid in the circle

*x*^{2} *+ y*^{2} = 9, z = 9. Therefore the given surface lies above
*the disk D with center the origin and radius 3. *

(See Figure 16.)

**Figure 16**

*Example 11 – Solution*

Using Formula 9, we have

cont’d

*Example 11 – Solution*

Converting to polar coordinates, we obtain

cont’d