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The Divergence TheoremMATH 311, Calculus III

J. Robert Buchanan

Department of Mathematics

Summer 2011

J. Robert Buchanan The Divergence Theorem

Greens Theorem Revisited

Greens Theorem:C

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

R

(Nx My

)dA

R

T

C

n

x

y

J. Robert Buchanan The Divergence Theorem

Greens 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).

J. Robert Buchanan The Divergence Theorem

Greens Theorem Vector Form (2 of 3)

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)) 1r(t)

y (t),x (t)

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

) 1r(t)

.

Now consider the line integralC

F n ds.

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

J. Robert Buchanan The Divergence Theorem

Greens Theorem Vector Form (2 of 3)

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)) 1r(t)

y (t),x (t)

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

) 1r(t)

.

Now consider the line integralC

F n ds.

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

J. Robert Buchanan The Divergence Theorem

Greens Theorem Vector Form (2 of 3)

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)) 1r(t)

y (t),x (t)

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

) 1r(t)

.

Now consider the line integralC

F n ds.

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

J. Robert Buchanan The Divergence Theorem

Greens Theorem Vector Form (3 of 3)

C

F n ds = b

a(F n)(t) r(t)dt

=

ba

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

) r(t)r(t)

dt

=

ba

(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

(Mx

+Ny

)dA (by Greens Theorem)

=

R F dA

J. Robert Buchanan The Divergence Theorem

Summary and Objective

Greens Theorem in vector form statesC

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 Gausss Theorem) willextend this result to three-dimensional vector fields.

J. Robert Buchanan The Divergence Theorem

Summary and Objective

Greens Theorem in vector form statesC

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 Gausss Theorem) willextend this result to three-dimensional vector fields.

J. Robert Buchanan The Divergence Theorem

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 .

J. Robert Buchanan The Divergence Theorem

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 +

QN(x , y , z)j n dS

+

Q

P(x , y , z)k n dS

=

Q

Mx

dV +

Q

Ny

dV +

Q

Pz

dV

=

Q F(x , y , z) dV .

J. Robert Buchanan The Divergence Theorem

Proof (2 of 7)

Thus the theorem will be proved if we can show thatQ

M(x , y , z)i n dS =

Q

Mx

dVQ

N(x , y , z)j n dS =

Q

Ny

dVQ

P(x , y , z)k n dS =

Q

Pz

dV .

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

J. Robert Buchanan The Divergence Theorem

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).

J. Robert Buchanan The Divergence Theorem

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 =

Q0 dS = 0.

J. Robert Buchanan The Divergence Theorem

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

J. Robert Buchanan The Divergence Theorem

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 isS2

P(x , y , z)k n dS =

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

J. Robert Buchanan The Divergence Theorem

Proof (7 of 7)

Finally,Q

P(x , y , z)k n dS

=

S1

P(x , y , z)k n dS +

S2P(x , y , z)k n dS

+

S3

P(x , y , z)k n dS

=

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)

Pz

dz dA =

Q

Pz

dV .

J. Robert Buchanan The Divergence Theorem

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.

J. Robert Buchanan The Divergence Theorem

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 =43.

J. Robert Buchanan The Divergence Theorem

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.

J. Robert Buchanan The Divergence Theorem

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

J. Robert Buchanan The Divergence Theorem

Example (3 of 3)

F(x , y , z) = xy , y2 + exz2 , sin(xy) F = 3y

S = {(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}

According to the Divergence Theorem,S

F n dS =

Q F dV =

Q

3y dV

=

11

1x20

2z0

3y dy dz dx

=18435

J. Robert Buchanan The Divergence Theorem

Example (3 of 3)

F(x , y , z) = xy , y2 + exz2 , sin(xy) F = 3y

S = {(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}

According to the Divergence Theorem,S

F n dS =

Q F dV =

Q

3y dV

=

11

1x20

2z0

3y dy dz dx

=18435

J. Robert Buchanan The Divergence Theorem

Example (3 of 3)

F(x , y , z) = xy , y2 + exz2 , sin(xy) F = 3y

S = {(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}

According to the Divergence Theorem,S

F n dS =

Q F dV =

Q

3y dV

=

11

1x20

2z0

3y dy dz dx

=18435

J. Robert Buchanan The Divergence Theorem

Identities (1 of 2)

Show that

S( F) n dS = 0.

By the Divergence TheoremS

( F) n dS =

Q ( F) dV

=

Q

0 dV

= 0

J. Robert Buchanan The Divergence Theorem

Identities (1 of 2)

Show that

S( F) n dS = 0.

By the Divergence TheoremS

( F) n dS =

Q ( F) dV

=

Q

0 dV

= 0

J. Robert Buchanan The Divergence Theorem

Identities (2 of 2)

Show that

SDnf (x , y , z) dS =

Q2f (x , y , z) dV .

S

Dnf (x , y , z) dS =

Sf (x , y , z) n dS

=

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

=

Q2f (x , y , z) dV

J. Robert Buchanan The Divergence Theorem

Identities (2 of 2)

Show that

SDnf (x , y , z) dS =

Q2f (x , y , z) dV .

S

Dnf (x , y , z) dS =

Sf (x , y , z) n dS

=

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

=

Q2f (x , y , z) dV

J. Robert Buchanan The Divergence Theorem

Average Value of a Function

During