G.H. Patel College of Engineering and Technology

Subject: Field Theory

Name Enrollment no.

Champaneria Dhvanil J. 150113109004

Chauhan Nisarg D. 150113109005

Jadav Prashant 150113109009

Limbani Milan P. 150113109011

Divergence Theorem

The Divergence Theorem

In this section, we will learn about:

The Divergence Theorem for simple solid regions,

and its applications in electric fields and fluid flow.

INTRODUCTION

• In Section 16.5, we rewrote Green’s Theorem in a vector version as:

• where C is the positively oriented boundary curve of the plane region D.

div ( , )C

D

ds x y dA× =∫ ∫∫F n F

INTRODUCTION

• If we were seeking to extend this theorem to vector fields on, we might make the guess that

• where S is the boundary surface of the solid region E.

div ( , , )S E

dS x y z dV× =∫∫ ∫∫∫F n F ……Equation 1

DIVERGENCE THEOREM• It turns out that Equation 1 is true, under appropriate

hypotheses, and is called the Divergence Theorem.

•Notice its similarity to Green’s Theorem and Stokes’ Theorem in that:• It relates the integral of a derivative of a function (div F in

this case) over a region to the integral of the original function F over the boundary of the region.

SIMPLE SOLID REGION• We state and prove the Divergence Theorem for regions E

that are simultaneously of types 1, 2, and 3.

• We call such regions simple solid regions. For instance, regions bounded by ellipsoids or rectangular boxes are simple solid regions.

•The boundary of E is a closed surface.•That is, the unit normal vector n is directed outward from E.

THE DIVERGENCE THEOREM

• Let:

–E be a simple solid region and let S be the boundary surface of E, given with positive (outward) orientation.

–F be a vector field whose component functions have continuous partial derivatives on an open region that contains E.

• Then, divS E

d dV× =∫∫ ∫∫∫F S F

• Thus, the Divergence Theorem states that:

–Under the given conditions, the flux of F across the boundary surface of E is equal to the triple integral of the divergence of F over E.

THE DIVERGENCE THEOREM

• Let F = P i + Q j + R k

–Then,

–Hence,

divP Q R

x y z

∂ ∂ ∂= + +∂ ∂ ∂

F

Proof

divE

E E E

dV

P Q RdV dV dV

x y z

∂ ∂ ∂= + +∂ ∂ ∂

∫∫∫

∫∫∫ ∫∫∫ ∫∫∫

F

• If n is the unit outward normal of S, then the surface integral on the left side of the Divergence Theorem is:

( )S S

S

S S S

d d

P Q R dS

P dS Q dS R dS

× = ×

= + + ×

= × + × + ×

∫∫ ∫∫

∫∫

∫∫ ∫∫ ∫∫

F S F n S

i j k n

i n j n k n

• So, to prove the theorem, it suffices to prove these equations:

S E

S E

S E

PP dS dV

x

QQ dS dV

y

RR dS dV

z

∂× =∂∂× =∂∂× =∂

∫∫ ∫∫∫

∫∫ ∫∫∫

∫∫ ∫∫∫

i n

j n

k n

• To prove Equation 4, we use the fact that E is a type 1 region:

where D is the projection of E onto the xy-plane.

( ) ( ) ( ) ( ){ }1 2, , , , , ,

E

x y z x y D u x y z u x y

=

∈ ≤ ≤

• By Equation 6 ,we have:

( )( )

( )2

1

,

,, ,

u x y

u x yE D

R RdV x y z dz dA

z z

∂ ∂ = ∂ ∂ ∫∫∫ ∫∫ ∫

•Thus, by the Fundamental Theorem of Calculus,

( )( ) ( )( )2 1, , , , , ,

E

D

RdV

z

R x y u x y R x y u x y dA

∂∂

= −

∫∫∫

∫∫

• The boundary surface S consists of three pieces:

–Bottom surface S1

–Top surface S2

–Possibly a vertical surface S3, which lies above the boundary curve of D(It might happen that S3 doesn’t appear, as in the case of a sphere.)

• Notice that, on S3, we have k ∙ n = 0, because k is vertical and n is horizontal.

–Thus,

3

3

0 0

S

S

R dS

dS

×

= =

∫∫

∫∫

k n

• Thus, regardless of whether there is a vertical surface, we can write:

1 2S S S

R dS R dS R dS× = × + ×∫∫ ∫∫ ∫∫k n k n k n

• The equation of S2 is z = u2(x, y), (x, y) D, and the outward normal n points upward.

–So, from Equation 10 (with F replaced by R k), we have:

( )( )2

2, , ,

S

D

R dS

R x y u x y dA

× =∫∫

∫∫

k n

•On S1, we have z = u1(x, y).

•However, here, n points downward.•So, we multiply

by –1:( )( )

1

1, , ,

S

D

R dS

R x y u x y dA

× =

−

∫∫

∫∫

k n

• Therefore, Equation 6 gives:

( )( ) ( )( )2 1, , , , , ,

S

D

R dS

R x y u x y R x y u x y dA

×

= −

∫∫

∫∫

k n

•Comparison with Equation 5 shows that:

•Equations 2 and 3 are proved in a similar manner using the expressions for E as a type 2 or type 3 region, respectively.

S E

RR dS dV

z

∂× =∂∫∫ ∫∫∫k n

• Find the flux of the vector field F(x, y, z) = z i + y j + x k

over the unit sphere x2 + y2 + z2 = 1

–First, we compute the divergence of F:

Example 1

( ) ( ) ( )div 1z y xx y z

∂ ∂ ∂= + + =∂ ∂ ∂

F

• The unit sphere S is the boundary of the unit ball B given by: x2 + y2 + z2 ≤ 1–So, the Divergence Theorem gives the flux

as:

( ) ( ) 343

div 1

41

3

S B B

F dS dV dV

V Bππ

× = =

= = =

∫∫ ∫∫∫ ∫∫∫F

UNIONS OF SIMPLE SOLID REGIONS

• The Divergence Theorem can also be proved for regions that are finite unions of simple solid regions.

•For example, let’s consider the region E that lies between the closed surfaces S1 and S2, where S1 lies inside S2.

•Let n1 and n2 be outward normal of S1 and S2.

• Then, the boundary surface of E is: S = S1 S2

Its normal n is given by: n = –n1 on S1 n = n2 on S2

• Applying the Divergence Theorem to S, we get:

( )1 2

1 2

1 2

divE S

S

S S

S S

dV d

dS

dS dS

d d

= ×

= ×

= × − + ×

= − × + ×

∫∫∫ ∫∫

∫∫

∫∫ ∫∫

∫∫ ∫∫

F F S

F n

F n F n

F S F S

Thank you…..