The Divergence Theorem - MATH 311, Calculus IIIbanach.millersville.edu/~bob/math311/Divergence/main.pdf · Green’s Theorem Vector Form (1 of 3) Simple closed curve C is described

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The Divergence TheoremMATH 311, Calculus III

J. Robert Buchanan

Department of Mathematics

Spring Summer 2019

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,∫∫

∂QP(x , y , z)k · n dS

=

∫∫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

. -1.0-0.50.00.51.0

x

0.0 0.5 1.0 1.5 2.0y

0.00.51.0z

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.I If ∇ · F < 0 then the divergence represents a net loss of

fluid (a sink).I If ∇ · F > 0 then the divergence represents a net gain of

fluid (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.I If ∇ · F < 0 then the divergence represents a net loss of

fluid (a sink).I If ∇ · F > 0 then the divergence represents a net gain of

fluid (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 0 1 2x

-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

Application (5 of 5)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

I Read Section 14.7.I Exercises: 1–35 odd

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The Divergence Theorem - MATH 311, Calculus IIIbanach.millersville.edu/~bob/math311/Divergence/main.pdf · Green’s Theorem Vector Form (1 of 3) Simple closed curve C is described