1
The Divergence Theorem
Consider surfaces that are boundaries of finite solids – the surface of a
solid sphere, a solid box, a solid cylinder.
Such surfaces are said to be closed.
A piecewise smooth surface consists of finitely many smooth surfaces joined together at the edges (e.g. a box or a solid cylinder).
We consider only piecewise smooth
surfaces that can be assigned an
inward orientation (towards the interior
of the solid) and an outward orientation
(away from the interior).
The divergence of F~ = f (x, y, z)~i + g(x, y, z) ~j + h(x, y, z) ~k is
divF~ =
∂f ∂g ∂h
+
+
∂x ∂y ∂z
The Divergence or Gauss’s Theorem. Let G be a solid whose
surface σ is oriented outward, and let ~n be the outward unit normal
on σ, then
¨
˚
F~ · ~n dS =
divF~ dV
σ
G
where
F~ = f (x, y, z)~i + g(x, y, z) ~j + h(x, y, z) ~k
and f , g, h have continuous first partial derivatives on some open set
containing G.
1
The flux of a vector field across a closed surface with outward orientation is equal to the triple integral of the divergence over the region
enclosed by the surface.
For G being simultaneously a simple
xy-, yz- and zx-solid the formula splits
¨
˚
∂f
f ~i · ~n dS =
dV
∂x
σ
G
¨
˚
∂g
dV
g ~j · ~n dS =
∂y
σ
G
˚
¨
∂h
h ~k · ~n dS =
dV
∂z
G
σ
We have for a simple xy-solid
˚
G
∂h
dV =
∂z
¨ "ˆ
R
¨
g2 (x,y)
g1 (x,y)
#
∂h
dz dA
∂z
[h(x, y, g2(x, y)) − h(x, y, g1(x, y))] dA
=
R
¨
¨
¨
h ~k · ~n dS =
h ~k · ~n dS +
¨σ1
σ
h ~k · ~n dS
σ2
∂z ~ ~
∂z
h(x, y, g1(x, y)) ~k · ( ~i +
j − k) dA
∂x
∂y
R
¨
∂z
∂z ~ ~
+
h(x, y, g2(x, y)) ~k · (− ~i −
j + k) dA
∂x
∂y
R
where we have taken into account that
¨
~k · ~n = 0 ⇒
h ~k · ~n dS = 0
σ
=
3
σ3
2
Example. Find the flux of the vector field
F(x, y, z) = (−3x+12xy 2+4z 3) i−(2y 2−y+4x2) j+(4+3z+5yz−4y 3) k
across the surface σ: x2 + y 2 + z 2 = 9 with outward orientation.
Answer : flux = 4068
5 π
Divergence viewed as flux density
Let G be small, e.g. a ball of radius << 1 centred at P0.
¨
˚
˚
F~ ·~n dS =
Φ(G) =
σ(G)
divF~ dV ≈ divF~ (P0)
G
G
Thus
divF~ (P0) ≈
where
Φ(G)
vol(G)
dV = divF~ (P0)V
Φ(G)
vol(G)
is called the outward flux density of F~ across G.
Taking the limit vol(G) → 0 ( → 0), we get
Φ(G)
1
divF~ (P0) = lim
= lim
vol(G)→0 vol(G)
vol(G)→0 vol(G)
¨
F~ · ~n dS .
σ(G)
This limit is called the outward flux density of F~ at P0.
It tells us that in a steady-state fluid flow, divF~ can be interpreted as
the limiting flux per unit volume at a point.
3
Sources and sinks
Consider an incompressible fluid which means that its density is constant. Consider a point P0 in fluid, and a small sphere G centred at
P0. If divF~ (P0) > 0 then Φ(G) > 0, and more fluid goes out through
the sphere than comes in. Since the fluid is incompressible this can
only happen if fluid is entering the flow at P0 otherwise the density
would decrees. Similarly, if divF~ (P0) < 0 then Φ(G) < 0, then fluid
is leaving the flow at P0. In an incompressible fluid, points at which
divF~ (P0) > 0 are called sources, and points at which divF~ (P0) < 0
are called sinks. If there are no sources and sinks we have
divF~ (P ) = 0 for every point P
This is the continuity equation for incompressible fluids.
Gauss’s Law for Inverse-square fields
If F~ (~r) = rc3 ~r, r = |~r | is an inverse-square field in 3-space, and if σ is
a closed orientable surface that surrounds the origin, then the outward
flux of F~ across σ is
¨
Φ=
F~ · ~n dS = 4πc .
σ
divF~ (P ) = 0 if P 6= O. Thus
˜
˜
~ · ~n dS +
~ na dS = 0
F
σ
σa F · ~
~na = −~r /r ⇒
˜
˜
~ · ~n dS =
~ ~r
F
σ
σa F · r dS
˜
= ac2 σa dS = 4πc
4
2
Stoke’s Theorem
Let σ be a piecewise smooth oriented surface that is bounded by a
simple, closed, piecewise smooth curve C with positive orientation. If
the components of F~ (x, y, z) = f (x, y, z)~i + g(x, y, z) ~j + h(x, y, z) ~k
are continuous and have continuous first partial derivatives on some
open set containing σ, then
˛
˛
¨
F~ · d~r =
F~ · T~ ds =
(curlF~ ) · ~n dS
C
C
σ
where T~ is the unit tangent vector to C, and
~
~i ~j k ∂h ∂g ~
∂h ∂f ~
∂g ∂f ~ ∂ ∂ ∂ ~ ~
curlF~ =
−
i−
−
j+
−
k = ∂x ∂y ∂z = ∇×F
∂y ∂z
∂x ∂z
∂x ∂y
f g h Example. Let C be the triangle in the plane z = 14 y with vertices (0, 0, 0), (2, 0, 0) and (0, 4, 1) with a counterclockwise orientation
looking
down the positive z-axis. Use Stokes’ Theorem to evaluate
¸
C F · dr.
F(x, y, z) = (2xy + 3x + 4z) i − (2x2 − 5y) j + (y − z) k
Answer : −20
5
Green’s theorem as a particular case of Stoke’s theorem
Let’s regard
F~ (x, y) = f (x, y)~i + g(x, y) ~j
as a vector filed in 3-space
F~ (x, y, z) = f (x, y)~i + g(x, y) ~j + 0 ~k
Let σ = R where R is a region in the xy-plane. Then
‰
¨
F~ · d~r =
(curlF~ ) · ~k dA
C
R
Since
curlF~ =
∂g ∂f ~
−
k
∂x ∂y
we get Green’s theorem
‰
¨ ∂g ∂f
−
f dx + g dy =
dA
∂x
∂y
C
R
6