derivation of first formula for surface integration with respect to area

derivation of first formula for surface integration with respect to area

Since the

integral

is defined as a limit, we may make life simpler by making approximations as long as the errors incurred in the approximation go zero in the limit of fine sudivisions. In particular, we have in mind approximating small portions of our surface by planes. A systematic way of carrying out such an approximation is to approximate the

neighborhood

of a surface near a point by a portion of the

tangent plane

to that surface through that point. Again, one needs to prove that such an approximation is mathematically justified and that the error really does tend to zero, but we are deferring such considerations to a later

section

.

The approach presented here can be seen as avoiding the pitfalls pointed out in the last section of the main entry to obtain a correct answer. We pick a

sequence

of closely spaced values of

u

and

v

and draw a grid on our surface. This subdivides the surface into many small curved

quadrilaterals

. If we pick this grid sufficiently fine, then we can approximate these quadrilaterals by plane

parallelograms

.

Having been warned that

u

and

v

may not necessarily

measure

length, we shall be more careful. Rather, to measure length, we should employ the

rectangular coordinates

x

,

y

, and

z

. Using the

first order

Taylor expansion

, we see that the

components

of the vector

a

→

connecting the point on

S

parameterized by

(

u

,

v

)

to the point

(

u

+

d

⁢

u

,

v

)

is

a

→

=

(

∂

⁡

x

∂

⁡

u

⁢

d

⁢

u

,

∂

⁡

y

∂

⁡

u

⁢

d

⁢

u

,

∂

⁡

z

∂

⁡

u

⁢

d

⁢

u

)

and, likewise, the vector connecting the point

(

u

,

v

)

to the point

(

u

,

v

+

d

⁢

v

)

is

b

→

=

(

∂

⁡

x

∂

⁡

v

⁢

d

⁢

v

,

∂

⁡

y

∂

⁡

v

⁢

d

⁢

v

,

∂

⁡

z

∂

⁡

v

⁢

d

⁢

v

)

.

As we said, the curved portion of quadrilateral may be approximated by a parallelogram having these two vectors as sides. Recall that the area of a parallelogram

spanned by

vectors

a

→

and

b

→

is given as

A

⁢

(

a

→

,

b

→

)

=

|

a

→

|

⁢

|

b

→

|

⁢

sin

⁡

θ

a

⁢

b

where

θ

a

⁢

b

is the angle between

a

→

and

b

→

. Using the fact that

a

→

⋅

b

→

=

|

a

→

|

⁢

|

b

→

|

⁢

cos

⁡

θ

a

⁢

b

and the famous

trigonometric identity

sin

2

⁡

θ

+

cos

2

⁡

θ

=

1

, we can rewrite our expression for the area as follows:

A

⁢

(

a

→

,

b

→

)

=

|

a

→

|

2

⁢

|

b

→

|

2

-

(

a

→

⋅

b

→

)

2

Remembering what the vectors

a

→

and

b

→

happen to be for our problem, we obtain

d

2

⁢

A

=

(

∂

⁡

x

∂

⁡

u

⁢

∂

⁡

y

∂

⁡

v

-

∂

⁡

y

∂

⁡

u

⁢

∂

⁡

x

∂

⁡

v

)

2

+

(

∂

⁡

x

∂

⁡

u

⁢

∂

⁡

z

∂

⁡

v

-

∂

⁡

z

∂

⁡

u

⁢

∂

⁡

x

∂

⁡

v

)

2

+

(

∂

⁡

y

∂

⁡

u

⁢

∂

⁡

z

∂

⁡

v

-

∂

⁡

z

∂

⁡

u

⁢

∂

⁡

y

∂

⁡

v

)

2

⁢

d

⁢

u

⁢

d

⁢

v

.

Note that the quantities which

occur in

the parentheses are Jacobi determinants, so we can rewrite this as

d

2

⁢

A

=

(

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

y

,

z

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

z

,

x

)

∂

⁡

(

u

,

v

)

)

2

Hence, we can express our integral as

∫

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

.

Title

derivation of first formula for surface integration with respect to area

Canonical name

DerivationOfFirstFormulaForSurfaceIntegrationWithRespectToArea

Date of creation

2013-03-22 15:07:23

Last modified on

2013-03-22 15:07:23

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

6

Author

rspuzio (6075)

Entry type

Derivation

Classification

msc 28A75

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