Multiple Integrals 12

Surface Area12.6

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Surface Area

In this section we apply double integrals to the problem of

computing the area of a surface.

We start by finding a formula for the area of a parametric

surface and then, as a special case, we deduce a formula

for the surface area of the graph of a function of two

variables.

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Surface AreaRecall that a parametric surface S is defined by a vector-valued function of two parameters

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

or, equivalently, by parametric equations

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

where (u, v) varies throughout a region D in the uv-plane.

We will find the area of S by dividing S into patches and approximating the area of each patch by the area of a piece of a tangent plane. So first let’s recall how to find tangent planes to parametric surfaces.

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Surface AreaLet P0 be a point on S with position vector r(u0, v0).

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 1.)

Figure 1

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Surface AreaThe tangent vector to C1 at P0 is obtained by taking the partial derivative of r with respect to v:

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

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Surface AreaIf the normal vector ru rv is not 0, then the surface S is

called smooth. (It has no “corners”.) In this case the tangent

plane to S at P0 exists and can be found using the normal

vector.

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 AreaLet’s choose to be the lower left corner of Rij. (See Figure 2.)

Figure 2

The image of the subrectangle Rij is the patch Sij.

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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

be the tangent vectors at Pij as given by Equations 3 and 2.

Surface Area

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Surface AreaFigure 3(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.

So we approximate Sij by the parallelogram determined by the vectors and .

Figure 3(a)

Approximating a patch by a parallelogram

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Surface AreaAs shown in Figure 3(b), this parallelogram lies in the tangent plane to S at Pij.

The area of this parallelogram is

and so an approximation to the area of S is Figure 3(b)

Approximating a patch by a parallelogram

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Surface AreaOur 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

D | ru rv | du dv.

This motivates the following definition.

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Example 1 – Area of a Parametric Surface Find the surface area of a sphere of radius a.

Solution:

We have found the parametric representation

x = a sin cos y = a sin sin z = a cos where the parameter domain is

D = {(, ) | 0 , 0 2}

We first compute the cross product of the tangent vectors:

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Example 1 – Solution

Thus

since sin 0 for 0 .

cont’d

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Example 1 – SolutionTherefore, by Definition 4, the area of the sphere is

cont’d

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Surface Area of a Graph

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Surface Area of a GraphFor 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 a GraphThus the surface area formula in Definition 4 becomes

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Example 2 – Surface Area of a GraphFind 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 4.)

Figure 4

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Example 2 – SolutionUsing Formula 6, we have

cont’d

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Example 2 – SolutionConverting to polar coordinates, we obtain

cont’d

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Surface Area of a Graph

A common type of surface is a surface of revolution S

obtained by rotating the curve y = f (x), a x b, about the

x-axis, where f (x) 0 and f is continuous.