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## 16.6 Parametric Surfaces and

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### 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.

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### 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.

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### 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.

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### 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

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### 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

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### Example 1 – Solution

So for any point (x, y, z) on the surface, we have x2 + z2 = 4 cos2u + 4 sin2u

= 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

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### 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

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### 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.

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### Parametric Surfaces

If we keep u constant by putting u = u0, then r(u0, v)

becomes a vector function of the single parameter v and defines a curve C1 lying on S. (See Figure 4.)

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### Parametric Surfaces

Similarly, if we keep v constant by putting v = v0, we get a curve C2 given by r(u, v0) 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.

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### 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?

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### Example 2 – Solution

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

Figure 5

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### 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

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### 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 = u0 , then the equation z = u0 + cos v shows that the z-values vary from u0 – 1 to u0 + 1.

cont’d

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### Example 4

Find a parametric representation of the sphere x2 + y2 + z2 = a2

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.

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### 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 j + a cos φ k

cont’d

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### 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

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### 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

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### 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.

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### 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.

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### 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 .

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### Example 8

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

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### 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

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### 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 P0 with position vector r(u0, v0).

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### Tangent Planes

If we keep u constant by putting u = u0, then r(u0, v)

becomes a vector function of the single parameter v and defines a grid curve C1 lying on S. (See Figure 12.) The tangent vector to C1 at P0 is obtained by taking the partial derivative of r with respect to v:

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### Tangent Planes

Similarly, if we keep v constant by putting v = v0, we get a grid curve C2 given by r(u, v0) that lies on S, and its tangent vector at P0 is

If ru × rv 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 ru and rv, and the vector

ru × rv is a normal vector to the tangent plane.

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### Example 9

Find the tangent plane to the surface with parametric equations x = u2, y = v2, z = u + 2v at the point (1, 1, 3).

Solution:

We first compute the tangent vectors:

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### 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

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### 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

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### 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 Rij.

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### Surface Area

Let’s choose to be the lower left corner of Rij. (See Figure 14.)

Figure 14

The image of the subrectangle Rij is the patch Sij.

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### Surface Area

The part Sij of the surface S that corresponds to Rij is called a patch and has the point Pij with position vector as one of its corners.

Let

and

be the tangent vectors at Pij as given by Equations 5 and 4.

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### Surface Area

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

meet at Pij 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.

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### Surface Area

So we approximate Sij 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 Pij.

Approximating a patch by a parallelogram.

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### Surface Area

The area of this parallelogram is

and so an approximation to the area of S is

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### 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

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### Surface Area

This motivates the following definition.

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### 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}

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### Example 10 – Solution

We first compute the cross product of the tangent vectors:

cont’d

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### Example 10 – Solution

= a2 sin2φ cos θ i + a2 sin2 φ sin θ j + a2 sin φ cos φ k Thus

since sin φ ≥ 0 for 0 ≤ φ ≤ π.

cont’d

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### Example 10 – Solution

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

cont’d

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### 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

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### Surface Area of the Graph of a Function

Thus we have

and the surface area formula in Definition 6 becomes

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### Example 11

Find the area of the part of the paraboloid z = x2 + y2 that lies under the plane z = 9.

Solution:

The plane intersects the paraboloid in the circle

x2 + y2 = 9, z = 9. Therefore the given surface lies above the disk D with center the origin and radius 3.

(See Figure 16.)

Figure 16

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### Example 11 – Solution

Using Formula 9, we have

cont’d

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### Example 11 – Solution

Converting to polar coordinates, we obtain

cont’d

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