Divergence Theorem Examples

Gauss' divergence theorem relates triple integrals and surface integrals.

GAUSS' DIVERGENCE THEOREM

Let be a vector field. Let be a closed surface, and let be the region inside of . Then:F W We

(( ((( a bW

F A F† . œ .Ze

div

EXAMPLE 1 Evaluate , where is the sphere .(( a bW

# # #$B #C † . W B C D œ *i j A

SOLUTION We could parameterize the surface and evaluate the surface integral, but it is much faster touse the divergence theorem. Since:

diva b a b a b a b$B #C œ $B #C ! œ &` ` `

`B `C `Di j

the divergence theorem gives:

(( (((a b a bW

$B #C † . œ & .Z œ & ‚ œ ")!i j Ae

the volume of the sphere 1 è

EXAMPLE 2 Evaluate , where is the boundary of the cube defined by(( ˆ ‰W

# $C D C BD † . Wi j k A

" Ÿ B Ÿ " " Ÿ C Ÿ " ! Ÿ D Ÿ #, , and .

SOLUTION Since:

divˆ ‰ ˆ ‰ ˆ ‰ a bC D C BD# $i j k œ C D C BD œ $C B` ` `

`B `C `D# $ #

the divergence theorem gives:

(( (((ˆ ‰ ˆ ‰

( ( ( ˆ ‰

(

W

# $ #

! " "

# " "#

"

"#

C D C BD † . œ $C B .Z

œ $C B .B .C .D

œ # 'C .C œ )

i j k Ae

è

EXAMPLE 3 Let be the region in bounded by the paraboloid and the plane ,e ‘$ # #D œ B C D œ "

and let be the boundary of the region . Evaluate .W C B D † .e (( ˆ ‰W

#i j k A

SOLUTION Here is a sketch of the region in question:

r

z (1, 1)

z = r2

z = 1

Since:

divˆ ‰ ˆ ‰a b a bC B D œ C B D œ #D` ` `

`B `C `Di j k# #

the divergence theorem gives:

(( (((W

#D † . œ #D .Zk Ae

It is easiest to set up the triple integral in cylindrical coordinates:

((( ( ( (

( ’ “

( ˆ ‰

Œ

e

1

#D .Z œ #D < .D .< .

œ # D < .<

œ # < < .<

œ # œ" " #

# ' $

! ! <

# " "

!

"#

Dœ<

"

!

"&

#

#

)

1

1

11

è

In general, you should probably use the divergence theorem whenever you wish to evaluate a vectorsurface integral over a closed surface.

The divergence theorem can also be used to evaluate triple integrals by turning them into surfaceintegrals. This depends on finding a vector field whose divergence is equal to the given function.

EXAMPLE 4 Find a vector field whose divergence is the given function .F 0 Ba b (a) (b) (c)0 B œ " 0 B œ B C 0 B œ B Da b a b a b È# # #

SOLUTION The formula for the divergence is:

diva bF Fœ f † œ `J `J

`B `C `D

`JB DC

We get to choose , , and , so there are several possible vector fields with a given divergence.J J JB C D

(This is similar to the freedom enjoyed when finding a vector field with a given rotation.)

(a) works, as does , , and so forth.F i F j F kœ B œ C œ D

(b) Three possible solutions are , , and .F i F j F kœ B C œ B C œ B CD" "

$ #$ # # #

(c) It is difficult to integrate with respect to or , but we can integrate with respect to ÈB D B D C# #

to get .F jœ C B DÈ # # è

EXAMPLE 5 Let be the region defined by . Use the divergence theorem toe B C D Ÿ "# # #

evaluate .(((e

D .Z#

SOLUTION Let be the unit sphere . By the divergence theorem:W B C D œ "# # #

((( ((e

D .Z œ † .#

W

F A

where is any vector field whose divergence is . One possible choice is :F F kD œ D"

$# $

((( ((e

D .Z œ D † ."

$# $

W

k A

All that remains is to compute the surface integral .((W

$"

$D † .k A

We have parameterized the sphere many times by now:

r

z

> œ B œ ? >

< œ ? C œ ? >

D œ ? D œ ?

! Ÿ > Ÿ # Ÿ ? Ÿ# #

)

11 1

cos cos sin

sin sin

cos cos

so

and

This gives:

. œ .> .? œ .> .? ? > ? > !

? > ? > ?

œ ? >ß

A

i j k i j kâ â â ââ â â ââ â â ââ â â ââ â â ââ â â ââ â â ââ â â â

ˆ ‰

`B `D`> `> `>

`C

`B `D`? `? `?

`C

#

cos sin cos cos

sin cos sin sin cos

cos cos cos sin cos sin#? >ß ? ? .> .?

so:

(( ( ( a b

(

” •

W ! Î#

$# Î#

$

Î#

Î#%

&

Î#

Î#

" "

$ $D † . œ ? ? ? .? .>

œ ? ? .?#

$

œ ?# "

$ &

œ%

"&

k A1 1

1

1

1

1

1

sin cos sin

sin cos

sin

1

1

1è

Of course, in the last example it would have been faster to simply compute the triple integral. In reality,the divergence theorem is only used to compute triple integrals that would otherwise be difficult to setup:

EXAMPLE 6 Let be the surface obtained by rotating the curveW

< œ ?

D œ #? Ÿ ? Ÿ# #

cos

sin

1 1

around the -axis:D

r

z

Use the divergence theorem to find the volume of the region inside of .W

SOLUTION We wish to evaluate the integral , where is the region inside of . By the(((e

.Z We

divergence theorem:

((( ((e

.Z œ † .W

F A

where is any vector field whose divergence is . Because of the cylindrical symmetry, and areF i j" B C

poor choices for . We therefore let :F F kœ D

((( ((e

.Z œ D † .W

k A

All that remains is to evaluate the surface integral .((W

D † .k A

We were essentially given the parameterization of the surface:

r

z

> œ B œ ? >

< œ ? C œ ? >

D œ #? D œ #?

! Ÿ > Ÿ # Ÿ ? Ÿ# #

)

11 1

cos cos sin

sin sin

cos cos

so

and

Thus:

. œ .> .? œ .> .? ? > ? > !

? > ? > # #?

œ # ? #

A

i j k i j kâ â â ââ â â ââ â â ââ â â ââ â â ââ â â ââ â â ââ â â â

a b

`B `D`> `> `>

`C

`B `D`? `? `?

`C

cos sin cos cos

sin cos sin sin cos

cos cos ? >ß # ? #? >ß ? ? .> .?cos cos cos sin cos sin

so:

(( ( ( a b

(

” •a b

W ! Î#

# Î#

Î#

Î#

Î#

Î#

#

D † . œ #? ? ? .? .>

œ # #? ? ? .?

œ # %? %?"

"'

œ#

k A1 1

1

1

1

1

1

sin cos sin

sin cos sin

sin

1

1

1

(according to my calculator)

è