integral theorem

8/8/2019 Integral Theorem

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Transformation of Double Integrals into Line Integrals - Green’s theorem in the plane

(Reference, Advanced Engineering Mathematics, Erwin Kreyszig, 1967, Wiley, pg 313Advanced Engineering Mathematics, Grossman/Derrick, 1988, Harper, pg 611

Mathematic of Physics and Modern Engineering, Sokolnikoff/Redheffer, 1966,

McGraw-Hill, pg 370)

Double integrals over a plane region may be transformed into line integrals over the boundary of

the region and conversely. This transformation also known as Green’s (1793-1841) theorem is

of practical as well as theoretical interest. This concept is extended in a tool called planimeter

used to measure the area of a graphically represented planar region.

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The region W can be represented in both of the form

( ) ( ),a x b u x y v x≤ ≤ ≤ ≤ (1)

and

( ) ( ),c y d p y x q y≤ ≤ ≤ ≤ (2)

Consider first

( )

( )( )

( )

( )( ) ( ) , , ,

b v x b b

a u x a a

y v x P P dxdy dy dx P x y dx P x v x P x u x dx

y u x y yΩ

= ∂ ∂= = = − =∂ ∂

∫∫ ∫ ∫ ∫ ∫ (3)

Now

1 2C C Pdx Pdx Pdx

Γ = +∫ ∫ ∫ i (4)

whereG is the entire perimeter (closed - counter clockwise).

From the sketch, it is immediately clear that

( ) ( )1

, ,b

C a P x y dx P x u x dx= ∫ ∫ (5)

Similarly,

( ) ( ) ( )2

, , ,a b

C b a P x y dx P x v x dx P x v x dx= = − ∫ ∫ ∫ (6)

Thus

( ) ( ) ( ) ( ) , , , ,b b b

a a a Pdx P x u x dx P x v x dx P x v x P x u x dx

Γ = − = − − ∫ ∫ ∫ ∫ i (7)

From (3) and (7), it becomes evident that

P dxdy Pdx

y Γ Ω

∂− =

∂∫∫ ∫ i (8)

Similarly, using (2), one obtains immediately

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( )

( )( ) ( ) , ,

d q y d

d p y c

Q Qdxdy dx dy Q q y y Q p y y dy

x xΩ

∂ ∂ = = − ∂ ∂ ∫∫ ∫ ∫ ∫ (9)

which by analogous reasoning yields the following equation:

Qdxdy Qdy

x Γ Ω

∂ =∂∫∫ ∫ i (10)

Adding (8) and (10) completes the Green’s theorem which reads

( )Q P

Pdx Qdy dxdy x yΓ

Ω

∂ ∂+ = − ∂ ∂

∫ ∫∫ i (11)

Consider the area of a plane region as a line integral over the boundary. In (11), let P=0 and

Q=x. Then

dxdy xdyΓ

Ω

=∫∫ ∫ i

The left side of the above integral is the area A of W . Similarly, let P=-y and Q=0; then from

(11)

A dxdy ydxΓ

Ω

= = −∫∫ ∫ i

By adding both areas, one obtains

( )1

2 A xdy ydx

Γ = −∫

This interesting formula expresses the area of W in terms of a line integral over the boundary.

The planimeter is calibrated based on this formula.

Triple Integrals - Divergence Theorem of Gauss (1777-1855)

The triple integral is a generalization of the double integral introduced above. For defining this

integral, one considers a function ( ), , f x y z defined in a bounded closed region T of space. One

subdivides T by planes parallel to the three coordinate planes. Then, one numbers the

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parallelepipeds inside T from 1 to n. In each such parallelepiped, one chooses an arbitrary point,

say ( ), ,k k k x y z in the k th parallelepiped, and form the sum

( )1

, ,

n

n k k k k

k

J f x y z v=

= ∆∑ (1)

where k v∆ is the volume of the k th parallelepiped. One assumes that ( ), , f x y z is continuous in

T , and T is bounded by finitely many smooth surfaces. It can be shown that J n in (1) converges to

a limit which is independent of the choice of subdivisions and corresponding points (x,y,z) as n

approaches infinity. This limit is called the triple integral of ( ), , f x y z over the region T, and is

denoted by

( ) ( ), , or , ,T T

f x y z dxdydz f x y z dv∫∫∫ ∫∫∫ (2)

It is intended to show here that the triple integral of the divergence of a continuously

differentiable vector function u over a region T in space can be transformed into a surface

integral of the normal component of u over the boundary surface S of T . This can be done by a

means of divergence theorem which is the three-dimensional analogue of Green’s theorem in the

plane considered above.

Divergence theorem of Gauss - Let T be a closed bounded region in space whose boundary is a

piecewise smooth orientable surface S . Let ( ), ,u x y z be a vector function which is continuous

and has continuous first partial derivatives in some domain containing T . Then

ˆn

T S

div u dV u dA=∫∫∫ ∫∫ (3)

where ˆˆnu u n= ⋅ (4)

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is the component of ˆ u in the direction of then outer normal of S with respect to T , and ˆ n is the

outer unit normal vector of S . Rewrite ˆ u and ˆ n in terms of components, say

1 2 3ˆˆˆ

ˆˆˆcos cos cos

u u i u j u k

n i j k α β γ

= + +

= + + (5)

so that , , andα β γ are the angles between ˆ n and the positive x, y, and z axes, respectively, the (3)

takes the form

( )31 2

1 2 3cos cos

T S

uu udxdydz u u cos u dA

x y z α β γ

∂∂ ∂+ + = + + ∂ ∂ ∂

∫∫∫ ∫∫ (6)

At this point, it may be useful to review a few elementary vector integral calculus formulas.

Recall the length of a plane curve, ( ) y f x= .

( ) ( )2 2

1 and 1ds f x dx s f x dx ′ ′= + = + ∫ l (7)

Extending (7) (refer to the derivation, Grossman/Derrick, pg 620), the lateral surface area, s of

the graph over the x-y region W is defined by

( ) ( )2 21 , , x y

S

d f x y f x y dAΩ

σ σ = = + +∫∫ ∫∫ (8)

where d s an infinitesimal area in the surface domain S and dA is is an infinitesimal area in the

region W ( x-y plane). Consider a surface ( ), z f x y= as ( ) ( ), , , 0G x y z z f x y= − = . Then

ˆˆˆ ˆˆˆˆ x y z x yG G i G j G k f i f j k ∇ = + + = − − + (9)

and

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2 2

ˆˆˆˆˆ

ˆ 1

x y

x y

f i f j k Gn

G f f

− − +∇= =

∇ + +(10)

Then ˆ n is the outward normal vector and g is the acute angle between ˆ n and the positive z-axis.

Hence, by definition

2 2

2 2

ˆˆ 1cos and sec 1

ˆˆ 1 x y

x y

n k f f

n k f f γ γ

⋅= = = + +

+ +(11)

and

( ) ( ) ( ), , , , , sec ,S

F x y z d F x y f x y x y dxdyΩ

σ γ = ∫∫ ∫∫ (12)

Rewriting (6)

ˆˆˆˆS T T

F nd div Fdxdydz div Fdvσ ⋅ = =∫∫ ∫∫∫ ∫∫∫ (6)

Just as Green’s theorem equates a line integral to an ordinary double integral, the divergence

theorem equates a surface integral to an ordinary triple integral.

If

( ) ( ) ˆˆ ˆˆ, , , , x y z F u u u P Q R Pi Qj Rk = = = + + (13)

then

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( )cos cos cosS T

P Q R P Q R d dxdydz

x y z α β γ σ

∂ ∂ ∂+ + = + + ∂ ∂ ∂

∫∫ ∫∫∫ (6)

where , , andα β γ are the angles that ˆ n makes with the positive coordinate axes. From (6), it is

sufficient to prove that

cosS T

P P d dxdydz

xα σ

∂=

∂∫∫ ∫∫∫ (14)

cosS T

QQ d dxdydz

yβ σ

∂=

∂∫∫ ∫∫∫ (15)

and

cosS T

R R d dxdydz

z γ σ

∂=

∂∫∫ ∫∫∫ (16)

The proofs of (14),(15), and (16) are virtually identical, hence only proof of (16) is illustrated.

Consider S as “walnut shaped” as shown below.

That is, S consists of two surfaces: an upper surface S 2: ( )2, z f x y= and a lower surface S 1:

( )1, z f x y= . Then

1 2

cos cos cosS S S

R d R d R d γ σ γ σ γ σ = +∫∫ ∫∫ ∫∫ (17)

From (12)

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( ) ( ) ( )2

2 2, , cos , , , cos sec , , ,

S

R x y z d R x y f x y dxdy R x y f x y dxdyΩ Ω

γ σ γ γ = = ∫∫ ∫∫ ∫∫ (18)

where W is the projection of S 2 into xy-plane. On S 1 the situation is a bit different because the

outer normal ˆ n on S 1 points downward and it was assumed that the angle g is an acute angle

between the positive z-axis and ˆ n . This situation can be readily rectified if one uses - ˆ n in S 1.

( ) ( ) ( )1

1 1, , cos , , , cos sec , , ,

S

R x y z d R x y f x y dxdy R x y f x y dxdyΩ Ω

γ σ γ γ = − = − ∫∫ ∫∫ ∫∫ (19)

thus

( ) ( )

( )

( )2

1

2 1

,

,

cos , , , , , ,

S

f x y

f x yT

R d R x y f x y R x y f x y dxdy

R Rdz dxdy dzdxdy

z z

Ω

Ω

γ σ = −

∂ ∂ = = ∂ ∂

∫∫ ∫∫

∫∫ ∫ ∫∫∫ (20)

Similarly (14) and (15) can be proved following exactly the same procedure outlined above.

This completes the proof.

Since (6) contains no reference to any special coordinate system, the result is true in all

coordinate systems. This integral transformation theorem, known as Gauss, or divergence,

theorem, is fundamental to all developments in mechanics of continuous media (continuum

mechanics).

Example 1

Computeˆ ˆ ,

S

F n d σ ⋅∫∫ where ( ) 2 2 ˆˆ ˆˆ, , 2 4 F x y z x i yj z k = + + and S is the surface of the cylinder

2 24, 0 2 x y z + = ≤ ≤ .

Solution

div ˆ 2 2 8 F x z = + + , so

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( ) ( )

( ) ( )

2 2

2 2 2 2

2

04

2

4 4

ˆ ˆ 2 2 8 2 2 8

z=22 2 4 4 20

z=0

S T x y

x y x y

F n x z dzdxdy x z dzdxdy

xz z z dxdy x dxdy

+ ≤

+ ≤ + ≤

⋅ = + + = + +

= + + = +

∫∫ ∫∫∫ ∫∫ ∫

∫∫ ∫∫

Let cos x r θ = and sin y r θ = , then2 2 2 2 2 2 2

4 cos sin 2 x y r r r r θ θ + = = + = ⇒ = .

The differential area in the polar coordinate system becomes dA dxdy rdrd θ = = .)

( ) ( )2 2 2 2

2

0 0 0 0

2

0

4 cos 20) 4 cos 5

84 cos 10 80

3

r rdrd r r drd

d

π π

π

θ θ θ θ

θ θ π

= + = +

= + =

∫ ∫ ∫ ∫

∫ Example 2

By transforming to a trip[le integral, evaluate

( )3 2 2

S

I x dydz x y dxdz x z dxdy= + +∫∫

where S is the closed surface consisting of the cylinder ( )2 2 20 z b x y a+ = ≤ ≤ and the dicks

z=0 and z=b. From (6), it is evident that

3 2 2, , P x Q x y R x z = = =

Hence, by taking advantage of the symmetry of the circle (evaluate only 1/4 of the circle), the

corresponding triple integral takes the from by (6)

( ) ( )2 2

3/ 22 2 2 2 2 2

0 0 0 0 0

13 4 5 20

3

b a a y b a

T

I x x x dxdydz x dxdydz a y dydz − = + + = × = − ∫∫∫ ∫ ∫ ∫ ∫ ∫

Let sin , cos y a dy a d θ θ θ = = , then

( ) ( ) ( )3/ 23/ 2 3/ 2

2 2 2 2 2 2 3 31 sin cos cosa y a a aθ θ θ − = − = =

and

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( )43/ 2 / 2

2 2 4 4 4

0 0

1 1 1 3cos

3 3 3 8 2 16

a aa y dy a d a

π π π θ θ − = = =∫ ∫

Hence, the integral I becomes

44

0

520

16 4

b a I dz a b

π π = × =∫

Example 3

If ˆ F and ˆ n are expressed in terms of their Cartesian components such that

( ) ( ) ( ) ˆˆˆ ˆˆˆˆˆ cos , cos , cos ,x y z

F n x n i y n j z n k i j k r r r

= = + + = + +

where

2 2 2r x y z = + + and the region T is the sphere2 2 2 2 x y z a+ + ≤

Evaluate the triple integral of (6).

Solution

Assuming that 0r ≠

2 2 2 2 2 2 2 2 2

2 2 2 32 2 2/ x y z x x y z P x r x

x x x y z r x y z + + − + +∂ ∂ − = = = ∂ ∂ + ++ +

Likewise,

2 2

3

Q r y

y r

∂ −=

∂and

2 2

3

R r z

z r

∂ −=

∂

Hence,

2 2

3

3 20

r r div F if r r r −= = ≠ and 2

T

dxdydz r ∫∫∫

Since the triple integral is discontinuous when r=0, care must be exercised. Returning back to

the equivalent surface integral reveals that

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2 2 2

2

2 2 2ˆ ˆ 1 4

S S S

x y z F nd d d a

r r r σ σ σ π

⋅ = + + = =

∫∫ ∫∫ ∫∫

The integral is simply the surface area of a sphere of radius a, thus avoiding an integral

containing a discontinuity at r=0.

The triple integral may be evaluated by transforming the Cartesian coordinates into the spherical

coordinates. From the sketch, it is clear that

In this case, r ρ = .2sindv dxdydz r drd d φ φ θ = =

( ) ( )

2 22 2

0 0 0 0 0

22 2 2

0

2 12 sin sin

cos 2 1 1 40

a

T

dxdydz r drd d a d d r r

a d a a

π π π π

π

φ φ θ φ φ θ

π φ θ π π

= =

= − = + =

∫∫∫ ∫ ∫ ∫ ∫ ∫

∫

As expected, it yields the same result.

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integral theorem

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