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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
J. Robert Buchanan The Divergence Theorem
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)〉.
J. Robert Buchanan The Divergence Theorem
Green’s 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))〉 · 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.
J. Robert Buchanan The Divergence Theorem
Green’s 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))〉 · 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.
J. Robert Buchanan The Divergence Theorem
Green’s 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))〉 · 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.
J. Robert Buchanan The Divergence Theorem
Green’s Theorem Vector Form (3 of 3)
∮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
J. Robert Buchanan The Divergence Theorem
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.
J. Robert Buchanan The Divergence Theorem
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.
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 +
∫∫∂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 .
J. Robert Buchanan The Divergence Theorem
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.
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 =
∫∫∂Q
0 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 is∫∫S2
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 +
∫∫S2
P(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)
∂P∂z
dz dA =
∫∫∫Q
∂P∂z
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 =4π3.
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 = 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}
According to the Divergence Theorem,∫∫S
F · n dS =
∫∫∫Q∇ · F dV =
∫∫∫Q
3y dV
=
∫ 1
−1
∫ 1−x2
0
∫ 2−z
03y dy dz dx
=18435
J. Robert Buchanan The Divergence Theorem
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}
According to the Divergence Theorem,∫∫S
F · n dS =
∫∫∫Q∇ · F dV =
∫∫∫Q
3y dV
=
∫ 1
−1
∫ 1−x2
0
∫ 2−z
03y dy dz dx
=18435
J. Robert Buchanan The Divergence Theorem
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}
According to the Divergence Theorem,∫∫S
F · n dS =
∫∫∫Q∇ · F dV =
∫∫∫Q
3y dV
=
∫ 1
−1
∫ 1−x2
0
∫ 2−z
03y dy dz dx
=18435
J. Robert Buchanan The Divergence Theorem
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
J. Robert Buchanan The Divergence Theorem
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
J. Robert Buchanan The Divergence Theorem
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
J. Robert Buchanan The Divergence Theorem
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
J. Robert Buchanan The Divergence Theorem
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.
J. Robert Buchanan The Divergence Theorem
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.
J. Robert Buchanan The Divergence Theorem
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
J. Robert Buchanan The Divergence Theorem
Interpretation of Divergence of a Vector Field (2 of 3)
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
J. Robert Buchanan The Divergence Theorem
Interpretation of Divergence of a Vector Field (2 of 3)
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
J. Robert Buchanan The Divergence Theorem
Interpretation of Divergence of a Vector Field (3 of 3)
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).
J. Robert Buchanan The Divergence Theorem
Interpretation of Divergence of a Vector Field (3 of 3)
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).
J. Robert Buchanan The Divergence Theorem
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.
J. Robert Buchanan The Divergence Theorem
Application (2 of 5)
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
J. Robert Buchanan The Divergence Theorem
Application (4 of 5)
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
J. Robert Buchanan The Divergence Theorem
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
J. Robert Buchanan The Divergence Theorem
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