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Math 2433
Section 26013
MW 1-2:30pm GAR 205
Bekki George
bekki@math.uh.edu
639 PGH
Office Hours:
11:00 - 11:45am MWF or by appointment
Last week’s video went over 17.6 and some of 17.7
We will pick up today with one more topic from 17.7 – Flux of a Vector Field
Suppose S is a smooth surface parameterized by r(u,v) with a unit normal of
n(x,y,z) that is continuous on all of S. If v is a vector field that is continuous on S,
then the surface integral:
∬ (v ⋅ n)dσ
S
gives the flux of v across S in the direction of n.
Example:
Calculate the flux out of the sphere x2 + y2 + z2 = 16 given v = y i − x j
17.8 The Vector Differential Operator
∇
∂
∂
∂
∇ = i + j+ k
∂x ∂y
∂z
So, we are familiar with ∇ from our work with gradients but how does this
work with vector fields?
This is called the divergence of v:
∇ • v = div v =
∂v1 ∂v2 ∂v3
+
+
∂x ∂y ∂z
And the curl of v is:
i
∂
∇ × v = curl v =
∂x
v1
j
∂
∂y
v2
k
∂
∂z
v3
Note that the curl of a gradient is zero
And the divergence of a curl is zero
Example:
(
)
2
2
v(x,
y)
=
x
+
y
i + ( xy ) j
1.Calculate ∇· v and ∇× v given:
2.Calculate ∇· v and ∇× v given:
Popper 22
1 2
v = xyi − y j + zk
1. Find the divergence of
2
17.9 The Divergence Theorem
Recall Green’s Theorem:
⎛ ∂Q ∂P ⎞
∫C h(r) ⋅dr = ∫C P ( x, y ) dx + Q ( x, y ) dy = ∫∫Ω ⎜⎝ ∂x − ∂ y ⎟⎠ dx dy
In vector terms, Green’s Theorem can be written as
∫∫ (∇ • v ) dxdy = ∫ ( v • n) ds
Ω
C
(n is the outer normal)
In higher dimensions, this gives:
∫∫∫ (∇ i v ) dxdydz = ∫∫ ( v i n) dσ
T
S
And the Outward Flux per unit volume is
(∇ i v ) dxdydz = flux of v out of N
∫∫∫
N
∈
Examples:
1.Use the divergence theorem to find the total flux out of the solid
x2 + y2 ≤ 4, 0 ≤ z ≤ 1 given v(x, y, z) = xi + 3y2 j + 2z2 k.
2.Use the divergence theorem to find the total flux out of the solid:
0 ≤ x ≤ 1, 0 ≤ y ≤ 1 - x, 0 ≤ z ≤ 1 - x - y,
given v(x, y, z) = 2x2 i + 4xy j - 4xzk.
3. Use the divergence theorem to find ∫∫S (v i n)dσ for
2
2
v(x, y, z) = (z + 2x)i + (4 y + z 2 )j + (y + 2z)k on the surface S : x + y = 1, 0 ≤ z ≤ 4
2.Set up the integral that uses the divergence theorem to find
where
v = xyi −
1 2
y j + zk
2
∫∫ (v i n)dσ
S
and S is the surface z = 4 − 3x 2 − 3y 2 ; z > 0
17.10 Stoke’s Theorem
Let’s revisit Green’s Theorem:
⎛ ∂Q ∂P ⎞
h(r)
⋅dr
=
P
x,
y
dx
+
Q
x,
y
dy
=
( ) ∫∫Ω ⎜⎝ ∂x − ∂ y ⎟⎠ dx dy
∫C
∫C ( )
If we let v = P i + Q j + R k, then
(∇ × v ) i k =
i
j
k
∂
∂x
∂
∂y
∂
∂Q ∂P
ik =
−
∂z
∂x ∂ y
P
Q
R
So, Green’s Thm can be written as:
∫∫ ⎡⎣(∇ × v ) i k ⎤⎦ dxdy = ∫ v (r ) • dr
Ω
C
And if S is a flat surface in space bounded by a Jordan curve and v is
continuously differentiable on S, we have
∫∫ ⎡⎣(∇ × v ) i n ⎤⎦ dσ = ∫ v (r ) • dr
S
C
This is Stoke’s Theorem.
Example:
1. Let S be the upper half of the unit sphere x2 + y2 + z2 = 1 and take n as
the upper unit normal. Use Stoke's theorem to find ∫∫ ⎡⎣(∇ × v ) i n ⎤⎦ dσ given
v(x, y, z) = 2z2 i + 4x j - 4y3 k.
S
2. Let S be the portion of z = 1- x2 - y2 above the xy-plane. Use Stoke's
theorem to find ∫∫ ⎡⎣(∇ × v ) i n ⎤⎦ dσ given v(x, y, z) = y i - x j + z k.
S
3.Find
∫∫ ⎡⎣(∇ × v ) i n ⎤⎦ dσ where v = -3y i + 3x j + z4 k and S is the portion of
S
1
the ellipsoid 2x + 2y + z = 1 that lies above the plane z = 2 and n is
the upper unit normal.
2
2
2
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