divergence theorem

1. Divergence Theorem

2. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.

3. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
x
y
σ
z=f(x,y)
D

4. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
x
y
σ
z=f(x,y)
F
D

5. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
x
y
σ
z=f(x,y)
N F
D

6. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
x
y
(xi, yi)
σ
z=f(x,y)
N F
D

7. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
x
y
(xi, yi)
σ
z=f(x,y)
N F
D

8. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
x
y
(xi, yi)
σ
z=f(x,y)
N F
D

9. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
Ni = unit normal
Fi
ΔSi = surface area
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
x
y D
(xi, yi)
σ
z=f(x,y)
N F

10. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
x
y
(xi, yi)
σ
z=f(x,y)
N F
D
Ni = unit normal
Fi
Fi•Ni = height
ΔSi = surface area

11. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
x
y
(xi, yi)
σ
z=f(x,y)
N F
D
Ni = unit normal
Fi Flow Volume
≈ (Fi•Ni) ΔS
Fi•Ni = height

12. ∫ ∫ F•N dS
σ
Divergence Theorem
Recall that the flux integral
which is the limit of the sums Σ(Fi•Ni)Δsi with each term
as the approximate volume of the flow F across a small
patch on the surface σ with N as the unit–normal field.
If the surface σ is defined by z = f(x, y) then
∫D
∫F•Nzx+zy+1 dA
where D is the domain of σ.
2 2
x
y D
(xi, yi)
σ
z=f(x,y)
∫ ∫ F•N dS =
σ
N F
Ni = unit normal
Fi Flow Volume
≈ (Fi•Ni) ΔS
Fi•Ni = height

13. The divergence of a 3D differentiable vector field
F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
is a function and it's written as div(F) ≡ fx + gy + hz.
Divergence Theorem

14. The divergence of a 3D differentiable vector field
F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
is a function and it's written as div(F) ≡ fx + gy + hz.
Divergence Theorem
The Divergence Theorem converts the total outward
flux of the flow F across the surface σ of a “nice” solid
G into a triple integral of the function div(F).

15. The divergence of a 3D differentiable vector field
F(x, y, z) = f(x, y, z) i + g(x, y, z)j + h(x, y, z)k,
is a function and it's written as div(F) ≡ fx + gy + hz.
∫ ∫ F•N dS =
Let G be a “nice” solid enclosed by the
surface σ with the outward unit normal vectors
field N and F = fi + gj + hk be a 3D flow,
then the total outward flux F across σ
∫ ∫ div(F) dV.
σ
Divergence Theorem
The Divergence Theorem converts the total outward
flux of the flow F across the surface σ of a “nice” solid
G into a triple integral of the function div(F).
The Divergence Theorem
∫
G
G
F
σ
the outward
flux =

16. Divergence Theorem
Example A. Given the constant flow F(x, y, z) = 1k,
find the total outward flux across the surface of the
unit cube

17. Divergence Theorem
Example A. Given the constant flow F(x, y, z) = 1k,
find the total outward flux across the surface of the
unit cube
y=1
x=1
z=1
y=1
x=1
z=1
F = 1k = <0, 0, 1>
N

18. Divergence Theorem
Example A. Given the constant flow F(x, y, z) = 1k,
find the total outward flux across the surface of the
unit cube by finding
a. the outward flux across each of the six faces.
b. the divergence–integral over G.
y=1
x=1
z=1
y=1
x=1
z=1
F = 1k = <0, 0, 1>
N

19. Divergence Theorem
Example A. Given the constant flow F(x, y, z) = 1k,
find the total outward flux across the surface of the
unit cube by finding
a. the outward flux across each of the six faces.
The outward normal of the walls are
perpendicular to F hence F•N = 0
so the flux across the walls is 0.
b. the divergence–integral over G.
y=1
x=1
z=1
y=1
x=1
z=1
F = 1k = <0, 0, 1>
N

20. Divergence Theorem
Example A. Given the constant flow F(x, y, z) = 1k,
find the total outward flux across the surface of the
unit cube by finding
a. the outward flux across each of the six faces.
The outward normal of the walls are
perpendicular to F hence F•N = 0
so the flux across the walls is 0.
The upward normal is (0, 0, 1) and
the downward normal is (0, 0, –1)
hence their fluxes are 1 and –1,
therefore the total outward flux is 0.
b. the divergence–integral over G.
y=1
x=1
z=1
y=1
x=1
z=1
F = 1k = <0, 0, 1>
N

21. Divergence Theorem
Example A. Given the constant flow F(x, y, z) = 1k,
find the total outward flux across the surface of the
unit cube by finding
a. the outward flux across each of the six faces.
y=1
x=1
z=1
y=1
x=1
z=1
F = 1k = <0, 0, 1>
The outward normal of the walls are
perpendicular to F hence F•N = 0
so the flux across the walls is 0.
The upward normal is (0, 0, 1) and
the downward normal is (0, 0, –1)
hence their fluxes are 1 and –1,
therefore the total outward flux is 0.
b. the divergence–integral over G.
Since the div(F) = 0 so ∫ ∫ div(F) dV = 0.
∫
G
N

22. Divergence Theorem
Example B. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = xi + yj + zk over the sphere
x2 + y2 + z2 = r2.

23. Divergence Theorem
Example B. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = xi + yj + zk over the sphere
x2 + y2 + z2 = r2.
Div(F) = xx + yy + zz = 1 + 1 + 1 = 3
Solution: Finding the divergence

24. ∫ ∫ F•N dS = ∫ ∫ div(F) dV =
σ
Divergence Theorem
Example B. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = xi + yj + zk over the sphere
x2 + y2 + z2 = r2.
Div(F) = xx + yy + zz = 1 + 1 + 1 = 3
∫
G
∫ ∫ 3 dV
∫
G
Solution: Finding the divergence
Hence

25. ∫ ∫ F•N dS = ∫ ∫ div(F) dV =
σ
Divergence Theorem
Example B. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = xi + yj + zk over the sphere
x2 + y2 + z2 = r2.
Div(F) = xx + yy + zz = 1 + 1 + 1 = 3
∫
G
∫ ∫ 3 dV
∫
G
= 3 * (volume of the sphere)
Solution: Finding the divergence
Hence

26. ∫ ∫ F•N dS = ∫ ∫ div(F) dV =
σ
Divergence Theorem
Example B. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = xi + yj + zk over the sphere
x2 + y2 + z2 = r2.
Div(F) = xx + yy + zz = 1 + 1 + 1 = 3
∫
G
∫ ∫ 3 dV
∫
G
= 3 * (volume of the sphere)
= 3 *
4πr3
3
= 4πr3
Solution: Finding the divergence
Hence

27. Divergence Theorem
Example C. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = x3/3i + y3/3j + zk over the
cylinder x2 + y2 =1 between z = 0 and z = 3.

28. Divergence Theorem
Div(F) = xx + yy + zz = x2 + y2 + 1
Example C. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = x3/3i + y3/3j + zk over the
cylinder x2 + y2 =1 between z = 0 and z = 3.
Solution: Finding the divergence

29. ∫ ∫ F•N dS = ∫ ∫ div(F) dV =
σ
Divergence Theorem
Div(F) = xx + yy + zz = x2 + y2 + 1
∫
G
∫ ∫ x2 + y2 + 1 dV
∫
G
Example C. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = x3/3i + y3/3j + zk over the
cylinder x2 + y2 =1 between z = 0 and z = 3.
Solution: Finding the divergence
Hence

30. ∫ ∫ F•N dS = ∫ ∫ div(F) dV =
σ
Divergence Theorem
Div(F) = xx + yy + zz = x2 + y2 + 1
∫
G
∫ ∫ x2 + y2 + 1 dV
∫
G
change to cylindrical form
= ∫ ∫ (r2 + 1) r dzdrd
∫
z=0
3
r=0
1
=0
2π
Example C. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = x3/3i + y3/3j + zk over the
cylinder x2 + y2 =1 between z = 0 and z = 3.
Solution: Finding the divergence
Hence

31. ∫ ∫ F•N dS = ∫ ∫ div(F) dV =
σ
Divergence Theorem
Div(F) = xx + yy + zz = x2 + y2 + 1
∫
G
∫ ∫ x2 + y2 + 1 dV
∫
G
change to cylindrical form
= ∫ ∫ (r2 + 1) r dzdrd
∫
z=0
3
r=0
1
=0
2π
= ∫ ∫ dz
∫
z=0
3
r=0
1
=0
2π
d * (r2 + 1) r dr *
Example C. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = x3/3i + y3/3j + zk over the
cylinder x2 + y2 =1 between z = 0 and z = 3.
Solution: Finding the divergence
Hence

32. ∫ ∫ F•N dS = ∫ ∫ div(F) dV =
σ
Divergence Theorem
Div(F) = xx + yy + zz = x2 + y2 + 1
∫
G
∫ ∫ x2 + y2 + 1 dV
∫
G
change to cylindrical form
= ∫ ∫ (r2 + 1) r dzdrd
∫
z=0
3
r=0
1
=0
2π
= ∫ ∫ dz = 2π*
∫
z=0
3
r=0
1
=0
2π
d * (r2 + 1) r dr *
3
4 *3 =
9π
2
Example C. Use the Divergence Theorem to find the
outward flux if F(x, y, z) = x3/3i + y3/3j + zk over the
cylinder x2 + y2 =1 between z = 0 and z = 3.
Solution: Finding the divergence
Hence