The Divergence Theorem - Millersville University of ...banach.millersville.edu/~bob/math311/Divergence/main.pdf · The Divergence Theorem MATH 311, Calculus III J. Robert Buchanan

The Divergence TheoremMATH 311, Calculus III

J. Robert Buchanan

Department of Mathematics

Summer 2011

J. Robert Buchanan The Divergence Theorem

Green’s Theorem Revisited

Green’s Theorem:∮C

M(x , y) dx + N(x , y) dy =

∫∫R

(∂N∂x− ∂M∂y

)dA

R

T

C

n

x

y


Green’s Theorem Vector Form (1 of 3)

Simple closed curve C is described by the vector-valuedfunction

r(t) = 〈x(t), y(t)〉 for a ≤ t ≤ b.

The unit tangent vector and unit (outward) normal vector to Care respectively

T(t) =1

‖r′(t)‖〈x ′(t), y ′(t)〉 and n(t) =

1‖r′(t)‖

〈y ′(t),−x ′(t)〉.



If the vector field F(x , y) = M(x , y)i + N(x , y)j, then along thesimple closed curve C:

F · n = 〈M(x(t), y(t)),N(x(t), y(t))〉 · 1‖r′(t)‖

〈y ′(t),−x ′(t)〉

=(M(x(t), y(t))y ′(t)− N(x(t), y(t))x ′(t)

) 1‖r′(t)‖

.

Now consider the line integral∮C

F · n ds.

Note: this is a line integral with respect to arc length.




F · n = 〈M(x(t), y(t)),N(x(t), y(t))〉 · 1‖r′(t)‖

〈y ′(t),−x ′(t)〉

=(M(x(t), y(t))y ′(t)− N(x(t), y(t))x ′(t)

) 1‖r′(t)‖

.


F · n ds.





F · n = 〈M(x(t), y(t)),N(x(t), y(t))〉 · 1‖r′(t)‖

〈y ′(t),−x ′(t)〉

=(M(x(t), y(t))y ′(t)− N(x(t), y(t))x ′(t)

) 1‖r′(t)‖

.


F · n ds.




∮C

F · n ds =

∫ b

a(F · n)(t) ‖r′(t)‖dt

=

∫ b

a

(M(x(t), y(t))y ′(t)− N(x(t), y(t))x ′(t)

) ‖r′(t)‖‖r′(t)‖

dt

=

∫ b

a

(M(x(t), y(t))y ′(t)− N(x(t), y(t))x ′(t)

)dt

=

∮C

M(x , y) dy − N(x , y) dx

=

∫∫R

(∂M∂x

+∂N∂y

)dA (by Green’s Theorem)

=

∫∫R∇ · F dA


Summary and Objective

Green’s Theorem in vector form states∮C

F · n ds =

∫∫R∇ · F(x , y) dA.

A double integral of the divergence of a two-dimensional vectorfield over a region R equals a line integral around the closedboundary C of R.

The Divergence Theorem (also called Gauss’s Theorem) willextend this result to three-dimensional vector fields.


Summary and Objective

Green’s Theorem in vector form states∮C

F · n ds =

∫∫R∇ · F(x , y) dA.

A double integral of the divergence of a two-dimensional vectorfield over a region R equals a line integral around the closedboundary C of R.

The Divergence Theorem (also called Gauss’s Theorem) willextend this result to three-dimensional vector fields.


Divergence Theorem

Remark: the Divergence Theorem equates surface integralsand volume integrals.

Theorem (Divergence Theorem)

Let Q ⊂ R3 be a region bounded by a closed surface ∂Q andlet n be the unit outward normal to ∂Q. If F is a vector functionthat has continuous first partial derivatives in Q, then∫∫

∂QF · n dS =

∫∫∫Q∇ · F dV .


Proof (1 of 7)

Suppose F(x , y , z) = M(x , y , z)i + N(x , y , z)j + P(x , y , z)k,then the Divergence Theorem can be stated as∫∫

∂QF · n dS

=

∫∫∂Q

M(x , y , z)i · n dS +

∫∫∂Q

N(x , y , z)j · n dS

+

∫∫∂Q

P(x , y , z)k · n dS

=

∫∫∫Q

∂M∂x

dV +

∫∫∫Q

∂N∂y

dV +

∫∫∫Q

∂P∂z

dV

=

∫∫∫Q∇ · F(x , y , z) dV .


Proof (2 of 7)

Thus the theorem will be proved if we can show that∫∫∂Q

M(x , y , z)i · n dS =

∫∫∫Q

∂M∂x

dV∫∫∂Q

N(x , y , z)j · n dS =

∫∫∫Q

∂N∂y

dV∫∫∂Q

P(x , y , z)k · n dS =

∫∫∫Q

∂P∂z

dV .

All of the proofs are similar so we will focus only on the third.


Proof (3 of 7)

Suppose region Q can be described as

Q = {(x , y , z) |g(x , y) ≤ z ≤ h(x , y), for (x , y) ∈ R}

where R is a region in the xy -plane.

Think of Q as being bounded by three surfaces S1 (top), S2(bottom), and S3 (side).


Proof (4 of 7)

S1: z=hHx,yL

S2: z=gHx,yL S3

x

y

z

On surface S3 the the unit outward normal is parallel to thexy -plane and thus∫∫

∂QP(x , y , z) k · n︸︷︷︸

=0

dS =

∫∫∂Q

0 dS = 0.


Proof (5 of 7)

Now we calculate the surface integral over S1.

S1 = {(x , y , z) | z − h(x , y) = 0, for (x , y) ∈ R}

Unit outward normal:

n =∇(z − h(x , y))

‖∇(z − h(x , y))‖=

−hx(x , y)i− hy (x , y)j + k√[−hx(x , y)]2 + [−hy (x , y)]2 + 1

andk · n =

1√[hx(x , y)]2 + [hy (x , y)]2 + 1


Proof (6 of 7)

∫∫S1

P(x , y , z)k · n dS =

∫∫S1

P(x , y , z)√[hx(x , y)]2 + [hy (x , y)]2 + 1

dS

=

∫∫R

P(x , y ,h(x , y)) dA

In a similar way we can show the surface integral over S2 is∫∫S2

P(x , y , z)k · n dS = −∫∫

RP(x , y ,g(x , y)) dA.


Proof (7 of 7)

Finally,∫∫∂Q


=

∫∫S1

P(x , y , z)k · n dS +

∫∫S2


+

∫∫S3


=

∫∫R

P(x , y ,h(x , y)) dA−∫∫

RP(x , y ,g(x , y)) dA

=

∫∫R[P(x , y ,h(x , y))− P(x , y ,g(x , y))] dA

=

∫∫R

P(x , y , z)

∣∣∣∣z=h(x ,y)

z=g(x ,y)

dA

=

∫∫R

∫ h(x ,y)

g(x ,y)

∂P∂z

dz dA =

∫∫∫Q

∂P∂z

dV .


Example (1 of 2)

Let Q be the solid unit sphere centered at the origin. Use theDivergence Theorem to calculate the flux of the vector fieldF(x , y , z) = 〈z, y , x〉 over the surface of the unit sphere.


Example (2 of 2)

F(x , y , z) = 〈z, y , x〉∇ · F = 1

S = {(x , y , z) | x2 + y2 + z2 = 1}Q = {(x , y , z) | x2 + y2 + z2 ≤ 1}

According to the Divergence Theorem,∫∫S

F · n dS =

∫∫∫Q∇ · F dV =

∫∫∫Q

1 dV =4π3.


Example (1 of 3)

Let Q be the solid region bounded by the parabolic cylinderz = 1− x2 and the planes z = 0, y = 0, and y + z = 2.Calculate the flux of the vector field

F(x , y , z) = xy i + (y2 + exz2)j + sin(xy)k

over the boundary of Q.


Example (2 of 3)

Region Q:

−1 ≤ x ≤ 10 ≤ y ≤ 2− z0 ≤ z ≤ 1− x2

-1.0

-0.5

0.0

0.5

1.0

x

0.0

0.5

1.0

1.5

2.0y

0.0

0.5

1.0

z


Example (3 of 3)

F(x , y , z) = 〈xy , y2 + exz2, sin(xy)〉

∇ · F = 3yS = {(x , y , z) | z = 1− x2, z = 0, y = 0, y + z = 2}Q = {(x , y , z) |0 ≤ z ≤ 1− x2, 0 ≤ y ≤ 2− z}


F · n dS =

∫∫∫Q∇ · F dV =

∫∫∫Q

3y dV

=

∫ 1

−1

∫ 1−x2

0

∫ 2−z

03y dy dz dx

=18435


Example (3 of 3)




F · n dS =

∫∫∫Q∇ · F dV =

∫∫∫Q

3y dV

=

∫ 1

−1

∫ 1−x2

0

∫ 2−z

03y dy dz dx

=18435


Example (3 of 3)




F · n dS =

∫∫∫Q∇ · F dV =

∫∫∫Q

3y dV

=

∫ 1

−1

∫ 1−x2

0

∫ 2−z

03y dy dz dx

=18435


Identities (1 of 2)

Show that∫∫

S(∇× F) · n dS = 0.

By the Divergence Theorem∫∫S

(∇× F) · n dS =

∫∫∫Q∇ · (∇× F) dV

=

∫∫∫Q

0 dV

= 0


Identities (1 of 2)

Show that∫∫

S(∇× F) · n dS = 0.

By the Divergence Theorem∫∫S

(∇× F) · n dS =

∫∫∫Q∇ · (∇× F) dV

=

∫∫∫Q

0 dV

= 0


Identities (2 of 2)

Show that∫∫

SDnf (x , y , z) dS =

∫∫∫Q∇2f (x , y , z) dV .

∫∫S

Dnf (x , y , z) dS =

∫∫S∇f (x , y , z) · n dS

=

∫∫∫Q∇ · ∇f (x , y , z) dV (Divergence Th.)

=

∫∫∫Q∇2f (x , y , z) dV


Identities (2 of 2)

Show that∫∫

SDnf (x , y , z) dS =

∫∫∫Q∇2f (x , y , z) dV .

∫∫S

Dnf (x , y , z) dS =

∫∫S∇f (x , y , z) · n dS

=

∫∫∫Q∇ · ∇f (x , y , z) dV (Divergence Th.)

=

∫∫∫Q∇2f (x , y , z) dV


Average Value of a Function

During Calculus I you learned the Integral Mean ValueTheorem for a continuous f (x) defined on [a,b] as

f (c) =1

b − a

∫ b

af (x) dx = favg ,

for some a ≤ c ≤ b.

The analogous result for triple integrals is

f (A) =1V

∫∫∫Q

f (x , y , z) dV

where A is a point in Q and V is the volume of region Q.


Average Value of a Function

During Calculus I you learned the Integral Mean ValueTheorem for a continuous f (x) defined on [a,b] as

f (c) =1

b − a

∫ b

af (x) dx = favg ,

for some a ≤ c ≤ b.

The analogous result for triple integrals is

f (A) =1V

∫∫∫Q

f (x , y , z) dV

where A is a point in Q and V is the volume of region Q.


Interpretation of Divergence of a Vector Field (1 of 3)

[∇ · F]|A =1V

∫∫∫Q∇ · F dV

=1V

∫∫∂Q

F · n dS︸︷︷︸flux per unit volume



Let P be an arbitrary point in the interior of Q (not on ∂Q) thenwe may center a sphere Qa of radius a > 0 at P so that thesphere lies entirely in the interior of Q.

[∇ · F]|A =1

Va

∫∫∂Qa

F · n dS

=1

43πa3

∫∫∂Qa

F · n dS

lima→0+

[∇ · F]|A = lima→0+

143πa3

∫∫∂Qa

F · n dS



Let P be an arbitrary point in the interior of Q (not on ∂Q) thenwe may center a sphere Qa of radius a > 0 at P so that thesphere lies entirely in the interior of Q.

[∇ · F]|A =1

Va

∫∫∂Qa

F · n dS

=1

43πa3

∫∫∂Qa

F · n dS

lima→0+

[∇ · F]|A = lima→0+

143πa3

∫∫∂Qa

F · n dS



Conclusion: the divergence of a vector field at a point is thelimiting value of the flux per unit volume over a sphere centeredat the point as the radius of the sphere approaches zero.

Suppose F represents the flow of a fluid in three dimensions.If ∇ · F < 0 then the divergence represents a net loss offluid (a sink).If ∇ · F > 0 then the divergence represents a net gain offluid (a source).



Conclusion: the divergence of a vector field at a point is thelimiting value of the flux per unit volume over a sphere centeredat the point as the radius of the sphere approaches zero.

Suppose F represents the flow of a fluid in three dimensions.If ∇ · F < 0 then the divergence represents a net loss offluid (a sink).If ∇ · F > 0 then the divergence represents a net gain offluid (a source).


Application (1 of 5)

Suppose a point charge q is located at the origin. ByCoulomb’s Law

F(x , y , z) =c q

‖〈x , y , z〉‖3〈x , y , z〉

where c is constant. Show the electric flux of F over any closedsurface containing the origin is 4πc q.



F is not continuous on any region containing the origin. Think ofQ as the region between two surfaces: (1) Sa a spherecentered at the origin of radius a > 0, and (2) S any surfacecontaining the origin inside it.∫∫∫

Q∇ · F dV =

∫∫Sa

F · n dS +

∫∫S

F · n dS



S

Sa

-2-1

01

2

x

-2

0

2y

-4

-2

0

2

4

z



F(x , y , z) =c q

‖〈x , y , z〉‖3〈x , y , z〉

∇ · F = 0

According to the Divergence Theorem

0 =

∫∫∫Q∇ · F dV =

∫∫Sa

F · n dS +

∫∫S

F · n dS∫∫S

F · n dS = −∫∫

Sa

F · n dS



On surface Sa the unit outward normal points toward the origin.

Sa = {(x , y , z) | x2 + y2 + z2 = a2}

n = −1a〈x , y , z〉

According to the Divergence Theorem∫∫S

F · n dS = −∫∫

Sa

F · n dS

= −∫∫

Sa

c q‖〈x , y , z〉‖3

〈x , y , z〉 ·(−1

a

)〈x , y , z〉dS

=c qa

∫∫Sa

1‖〈x , y , z〉‖3

‖〈x , y , z〉‖2 dS

=c qa

∫∫Sa

1‖〈x , y , z〉‖

dS =c qa

∫∫Sa

1a

dS

=c qa2

∫∫Sa

1 dS = 4πc q


Homework

Read Section 14.7.Exercises: 1–35 odd


The Divergence Theorem - Millersville University of ...banach.millersville.edu/~bob/math311/Divergence/main.pdf · The Divergence Theorem MATH 311, Calculus III J. Robert Buchanan

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