surface integration with respect to area

surface integration with respect to area

.

1

Introduction

In pure and applied math, one frequently encounters integrals over surfaces with respect to

surface area

. The simplest

instance

occurs when computing the area of a surface. More complicated instances occur when one is computing, say the moments of a shell or the total force acting on a surface. In this entry, methods of computing such

integrals

together with their theoretical justifications will be discussed.

The material to be found in this entry varies greatly in the level of mathematical sophistication. At the low end, there are

formulas

for computing such integrals which can be understood and used profitably by a

calculus

student. At the high end, a deeper discussion of this topic requires such relatively advanced mathematical tools as

exterior

calculus, Riemannian geometry, and

measure theory

. On the one hand, because this is an encyclopaedia, the coverage should be as thorough as possible. On the other hand, discussing the material at a high level of mathematical sophistication would make it

inaccessible

to many readers who could benefit from it. As a way out of this

dilemma

, we have chosen to write this entry in several

sections

with the level of mathematical sophistication and rigor incerasing from section to section. In the beginning, we will only assume familiarity with integral calculus and derive formulas using heuristic

arguments

.

Rigorous

proofs will only come in later sections because they require mathematical tools which may not yet be available to the typical calculus student. Likewise, in the earlier sections we shall treat the three-dimensional case separately from the more general formailsm that applies in any number of

dimensions

.

2

Calculus

2.1

Basic idea and definition

Let

S

be a surface in three-dimensional space (

ℝ

3

) and let

f

be a

function

defined on this surface. To describe the surface, we will choose a parameterization of the surface by two variables which we shall call

u

and

v

. (For instance, if

S

is a sphere,

u

and

v

might be the

latitude

and

longitude

.) In terms of this parameterization, the function

f

can be expressed as a function of

u

and

v

. Then the

integral of

f

with respect to the surface area

is usually notated as

∫

S

f

⁢

(

u

,

v

)

⁢

d

2

⁢

A

.

Just as with the ordinary integral, this quantity may be understood as a limit of sums.

1

1

Note on rigour, or lack thereof:

For our definition of integral to be sound, we need to show that the limit exists and depends only on the surface

S

and the function

f

and not on the details of how we choose to subdivide the surface. Throughout this section, we shall sweep such questions of mathematical propriety under the rug for three reasons: 1) In this section, our main interest is in deriving practical formulas and procedures which are of use in computing

surface integrals

that arise in practise. 2) This section is written for the benefit of beginners who may not be familiar with the techniques necessary to properly justify the manipulations presented here; the more sophisticated reader may skim through the formulas and examples and proceed to the later sections. 3) “The physicist’s excuse” — As long as we restrict ourselves to rather familiar and well-behaved surfaces and functions, our intuition should save us from making really serious mistakes. (Along the same lines, it is worth pointing out that we shall also adopt the naive point of view that

differentials

like

d

⁢

x

represent tiny displacements rather than a more sophisticated

interpretation

such as differential forms.)

To form an approximating sum, we subdivide the surface

S

into a miniscule pieces, multiply the area of each piece by the value of

f

at a point located inside that piece, sum over all the miniscule pieces into which we have subdivided the surface. The limiting value of these sums as the size of the pieces shrinks to zero is the integral with respect to surface area.

To compute this quantity, one may use the following formula to convert it to a

http://planetmath.org/node/11658

double integral

:

∫

S

f

⁢

(

u

,

v

)

⁢

d

2

⁢

A

=

∫

f

⁢

(

u

,

v

)

⁢

(

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

y

,

z

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

z

,

x

)

∂

⁡

(

u

,

v

)

)

2

⁢

𝑑

u

⁢

𝑑

v

.

If the surface is described as the graph of a function

g

, then we may also use the following formula:

∫

S

f

⁢

(

x

,

y

)

⁢

d

2

⁢

A

=

∫

f

⁢

(

x

,

y

)

⁢

1

+

(

∂

⁡

g

∂

⁡

x

)

2

+

(

∂

⁡

g

∂

⁡

y

)

2

⁢

𝑑

x

⁢

𝑑

y

.

2.2

Examples

To explain the use of these formulae, several worked examples have been presented in the forms of

supplements

to this entry. The first four examples illustrate the formula for integrals over parameterized surfaces and the latter four examples deal with surfaces presened as graphs of functions.

http://planetmath.org/node/6664

Example 1. This example shows how integrals over spheres with respect to surface area may be rewriten as integrals with respect to the

spherical coordinates

.

http://planetmath.org/node/6665

Example 2. This example builds on the previous example by carrying out the computation of an integral over the sphere using the formula derived in example 1.

http://planetmath.org/node/6666

Example 3. This example shows how to re-express integrals over

helicoids

as integrals over the parameters.

http://planetmath.org/node/6667

Example 4. This example uses the result of example 3 to compute the area of a portion of a helicoid.

http://planetmath.org/node/6668

Example 5. In this example, we revisit the sphere and consider it as the graph of the function

g

⁢

(

x

,

y

)

=

x

2

+

y

2

.

http://planetmath.org/node/6669

Example 6. We use the result of example 5 to compute the

moment of inertia

of a

spherical shell

.

http://planetmath.org/node/6672

Example 7. In this example, we consider integration on the paraboloid described by the equation

z

=

x

2

+

3

⁢

y

2

.

http://planetmath.org/node/6673

Example 8. We use the result of example 7 to compute the area of a portion of the paraboloid. From a purely technical point of view, this example is the longest and most complicated of the examples. Evaluating the integral is a long process and the answer involves elliptic integrals.

2.3

A common mistake

Before proceeding further, a caveat may be in order. Beginners are often apt to make the mistake that

∫

S

f

⁢

(

u

,

v

)

⁢

d

2

⁢

A

=

∫

S

f

⁢

(

u

,

v

)

⁢

𝑑

u

⁢

𝑑

v

.

By a fluke, this may give the right answer in certain cases, but as a general principle,

IT IS WRONG

. (This is a lot like the fact that in elementary arithmetic, one cannot reduce a fraction to lowest terms by striking out digits that appear in both the numerator and denominator, even if this happens to give the right answer in a few cases.) Since this mistake is so common and easy to make, it might not be inappropriate if we take out some time to explain why the above formula is wrong. The reason is that

d

2

⁢

A

refers to the area of a small portion of surface whilst

d

⁢

u

⁢

d

⁢

v

refers to the change in

u

multiplied by the change in

v

. In general, these two quantity are not equal for two reasons.

First, to compute an area, we need to multiply two lengths, but

u

and

v

may not always

measure

length. The following example should help clarify this problem. Suppose that

S

is a sphere and that the parameters

u

and

v

are latitude and longitude, respectively. Then, as every navigator knows, if we travel due west a degree of longitude starting in Florida (

d

⁢

v

=

1

), we will have travelled a

greater

distance

than if we travel due west a degree of longitude starting in Sweden (also

d

⁢

v

=

1

, but different values for

u

and

v

). In fact, to obtain the distance travelled, we must multiply the change in longitude by the cosine of the latitude. In fact, this is why, as we shall see, on a

unit sphere

the correct formula is

d

2

⁢

A

=

cos

⁡

u

⁢

d

⁢

u

⁢

d

⁢

v

,

which contains the sine to account for this fact.

Second, even when

u

and

v

measure length, they may not be

perpendicular

. If so, then the portion of surface which is traced out by letting the one parameter vary between

u

and

u

+

d

⁢

u

and letting the other parameter vary between

v

and

v

+

d

⁢

v

will look more like a

parallelogram

than a

rectangle

. As every student of

plane geometry

knows, the

product

of the length of the sides will not equal the area of the parallelogram except in the special case when it happens to be a rectangle.

2.4

Derivation of formulas for area integration

Having discoursed at some length about what not to do, let us now derive a correct formula for

d

2

⁢

A

. We shall show how to do this in two different ways, not only for the sake of completeness, but because one method is better adapted to deriving one formula and the other method is better adapted to deriving the other formula.

Also, it is worth checking explicitly that the answer depends only on the surface

S

and the function

f

but not on the choice of parameterization. This check is a routine application of the

chain rule

and the rule for change of variables in

multiple integrals

.

Title

surface integration with respect to area

Canonical name

SurfaceIntegrationWithRespectToArea

Date of creation

2013-03-22 14:57:44

Last modified on

2013-03-22 14:57:44

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

26

Author

rspuzio (6075)

Entry type

Topic

Classification

msc 28A75

Related topic

GaussGreenTheorem

Related topic

VectorPotential

Generated on Fri Feb 9 20:10:28 2018 by

LaTeXML