Divergence Theorem Examples

Gauss' divergence theorem relates triple integrals and surface integrals.

GAUSS' DIVERGENCE THEOREM Let F be a vector field. Let W be a closed surface, and let e be the region inside of W . Then:

F†.œ Adiv F .Z ((W ((( e a b

EXAMPLE 1 Evaluate $B#C†.i j A , where W is the sphere BCDœ*# # # . (( W a b SOLUTION We could parameterize the surface and evaluate the surface integral, but it is much faster to use the divergence theorem. Since: ` ` ` div $B#Ci j œ $B #C !œ& a b`B ab `C ab `D ab

the divergence theorem gives:

$B#C†.œi j A &.Zœ&‚the volume of the sphere œ")! 1 è ((W ab ((( e a b

# $ EXAMPLE 2 Evaluate ˆCDCBDi j k ‰ †. A , where W is the boundary of the cube defined by (( W "ŸBŸ""ŸCŸ", , and !ŸDŸ# .

SOLUTION Since: ` ` ` div CD#i C $ j BD k œ CD# C $  BDœ$CB # ˆ ‰`B ˆ‰ˆ‰ `C `D a b the divergence theorem gives:

# $ # ˆCDCi j BD k ‰ †. A œ ˆ‰ $CB.Z ((W ((( e

# " " # œˆ $CB.B.C.D ‰ (! ( " ( "

" œ# 'C.Cœ)# è (" EXAMPLE 3 Let e be the region in ‘ $ bounded by the paraboloid DœBC# # and the plane Dœ" , e # and let W be the boundary of the region . Evaluate ˆCBDi j k ‰ †. A . (( W SOLUTION Here is a sketch of the region in question: z z = 1 (1, 1)

z = r2

r

Since: ` ` ` div CBDi j# k œ C B D# œ#D ˆ ‰`Bab `C ab `D ˆ‰ the divergence theorem gives:

D#k †. A œ #D.Z ((W ((( e It is easiest to set up the triple integral in cylindrical coordinates:

#1 " " #D.Zœ #D<.D.<. ) (((e (((! ! < #

" " œ#1 D<# .< # (! ’ “ Dœ<

" 1 & œ#ˆ << ‰ .< (! " " # 1 œ#1  œ è Œ# '  $

In general, you should probably use the divergence theorem whenever you wish to evaluate a vector surface integral over a closed surface. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. This depends on finding a vector field whose divergence is equal to the given function.

EXAMPLE 4 Find a vector field F whose divergence is the given function 0 B . a b (a)0Bœ" (b) 0BœBC# (c) 0BœBD# # ab ab ab È SOLUTION The formula for the divergence is: `J`J `J div Fœf†œ F B C  D a b `B `C `D

We get to choose JB , J C , and J D , so there are several possible vector fields with a given divergence. (This is similar to the freedom enjoyed when finding a vector field with a given rotation.) (a) FiœB works, as does FjFk œCœD , , and so forth. " " (b) Three possible solutions are FœBC$ iF , œBC ## jF , and œBCD # k . $ # (c) It is difficult to integrate ÈBD# # with respect to BD or , but we can integrate with respect to C to get FœCÈ B# D # j . è

EXAMPLE 5 Let e be the region defined by B# C # D # Ÿ" . Use the divergence theorem to evaluate D# .Z . ((( e SOLUTION Let W be the unit sphere BCDœ"# # # . By the divergence theorem:

D.Zœ# F †. A (((e (( W " where F is any vector field whose divergence is D# . One possible choice is F œ D $ k : $ " D.Zœ# D $ k †. A (((e (( W $

" All that remains is to compute the surface integral D$k † . A . (( W $ We have parameterized the sphere many times by now: z >œ) Bœ?>cos cos <œ?cosso Cœ?> cossin r Dœ?sin Dœ? sin 1 1 !Ÿ>Ÿ#1 and  Ÿ?Ÿ # # This gives: âââi j k â âââi j k â âââ â .œA âââ`B`C `D .>.?œ?>cos sin cos ?> cos ! â .>.? âââ`> `> `> â âââ â âââ`B`C `D ?sin cos >?> sin sin cos ? â âââ`? `? `? â # # œˆcos ? cos >ß cos? sin >ß cos ? sin ?.>.? ‰ so:

" " #1 1 Î# D†.œ$k A sin ?$ cos ??.?.> sin ((W$ $ (( !Î#1 a b

#1 1Î# œsin% ??.? cos $ (1 Î#

1 #1 " Î# œsin & ? $” & • 1 Î#

%1 œ è "&

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 set up:

EXAMPLE 6 Let W be the surface obtained by rotating the curve < œcos ? 1 1  Ÿ?Ÿ Dœsin #? # #

around the D -axis:

z

r

Use the divergence theorem to find the volume of the region inside of W . SOLUTION We wish to evaluate the integral .Z , where e is the region inside of W . By the ((( e divergence theorem:

.ZœF †. A (((e (( W where F is any vector field whose divergence is " . Because of the cylindrical symmetry, Bi and C j are poor choices for F . We therefore let Fœ D k :

.Zœ D†.k A (((e (( W

All that remains is to evaluate the surface integral Dk † . A . (( W We were essentially given the parameterization of the surface: z >œ) Bœ?>cos cos <œ?cosso Cœ?> cossin r Dœsin #? Dœ sin #? 1 1 !Ÿ>Ÿ#1 and  Ÿ?Ÿ # #

Thus: âââi j k â âââi j k â âââ â .œA âââ`B`C `D .>.?œ?>cos sin cos ?> cos ! â .>.? âââ`> `> `> â âââ â âââ`B`C `D sin ? cos > sin ? sin ># cos #? â âââ`? `? `? â œ#cos ? cos # ?cos >ß# cos ? cos #? sin >ß cos ? sin ?.>.? a b so:

#1 1 Î# D†.œk A sin #? cos ? sin ?.?.> ((W (( !Î#1 a b

1Î# œ#1 sin #? cos ? sin ?.? (1 Î#

1 " Î# œ#1 %?sin %? (according to my calculator) a b ”"' • 1 Î#

1# œ è #