Andrew Rosen
AT X)
Chapter 14 - Vector Calculus (L
E
14.1 - Vector Fields
A vector eld is written as
(x, y, z)
→
−
F (x, y, z) = hf (x, y, z), g(x, y, z), h(x, y, z)i
because for every point in space
you get a dierent vector
→
−
A vector eld is conservative if it is the gradient ( ∇ ) of a scalar function
There exists a potential function,
→
−
F
φ(x, y, z),
such that
→
−
→
−
F (x, y, z) = ∇φ(x, y, z)
would be considered a gradient eld
14.2 - Line Integrals
If
f
−
C, →
r (t) = hx(t), y(t), z(t)i,
is a continuous function on the smooth curve
1
and
a ≤ t ≤ b,
then the
following is a line integral :
ˆ
ˆ
b
f (x, y, z) ds =
→
f (x(t), y(t), z(t)) |−
r 0 (t)| dt =
a
C
ˆ
b
f (x(t), y(t), z(t))
q
2
2
2
[x0 (t)] + [y 0 (t)] + [z 0 (t)] dt
a
How to Evaluate a Line Integral:
1. Find a parametric description of
2. Compute
−
|→
r 0 (t)| =
q
a
in the form
2
→
−
r (t) = hx(t), y(t), z(t)i
for
a≤t≤b
2
[x0 (t)] + [y 0 (t)] + [z 0 (t)]
3. Make substitutions for
´b
2
C
x
y
and
in the integrand and evaluate an ordinary integral:
´b →
|−
r 0 (t)| dt
a
f (x, y, z) ds =
C
−
f (x(t), y(t), z(t)) |→
r 0 (t)| dt
To nd the length of a curve,
´
→
−
r (t),
a ≤ t ≤ b,
on the bounds
Example:
Set up the following line integral in an ordinary integral :
´
you can do the following line integral:
(2 + x2 y) ds,
where
C
is the upper half of the
C
unit circle
Step 1: Write a parametric description of the curve. A circle can be written in terms of cosine and sine, so
it is
→
−
r (t) = hcos t, sin ti,
f (x(t), y(t), z(t)) = 2 + cos2 t sin t
p
→
−
→
−0
→
−0
Compute the magnitude of the derivative of r (t). r (t) = h− sin t, cos ti, so | r (t)| =
sin2 t + cos2 t =
Step 2: Substitute in for
Step 3:
0≤t≤π
x and y . The function
where
becomes
1
Step 4: Substitute
−
|→
r 0 (t)|
in for
ds.
−
ds = |→
r 0 (t)| dt = 1 dt = dt
This becomes:
Step 5: Put all the pieces together to evaluate an ordinary integral:
ˆ
ˆ
(2 + x2 y) ds =
C
π
(2 + cos2 t + sin t) dt
0
1 If C is not smooth everywhere (piece-wise), the line integral can still be evaluated if the curve is broken up into segments
1
Line Integrals of Vector Fields:
→
−
F (x, y, z) is a continuous
→
−
integral of F on C is:
If
ˆ
vector eld dened on a piece-wise smooth curve,
ˆ
−
→ −
F · d→
r =
C
Let
→
−
F
−
→
( F · T̂ ) ds =
b
from
a ≤ t ≤ b,
the line
−
→ −
F (→
r (t)) · r 0 (t) dt
a
C
be a continuous vector eld and let
ˆ
→
−
r (t),
C
be a closed smooth curve. The circulation is the same as the
previously dened line integral of a vector eld:
˛
ˆ
Circulation =
C
Let
→
−
F
−
→
F · T̂ ds =
F =
ˆ
C
−
→ −
F (→
r (t)) · r 0 (t) dt
a
C
be a continuous vector eld and let
b
be a closed smooth curve with counterclockwise orientation.
The (outward) ux is:
ˆ
ˆ
−
→
F · n̂ ds =
F lux =
b
(f y 0 (t) − g x0 (t)) dt
a
C
n̂ = T̂ × k̂
Note:
Note: A positive answer means a positive outward ux
14.3 - Conservative Vector Fields
A vector eld is said to be conservative on a region if there exists a scalar function,
For a conservative vector eld,
→
− ¸→
− −
F , F · d→
r =0
φ,
such that
on all simple closed smooth oriented curves
→
−
→
−
F = ∇φ
C
C
In
R2 ,
In
R3 ,
∂f
∂y
∂f
=
∂g
for conservative vector elds
∂x
∂g ∂f
∂h
∂g
∂h
=
,
=
, and
=
∂y
∂x ∂z
∂x
∂z
∂y
How to nd
φ
in 2-space:
1. Integrate
φx = f
2. Compute
φy
with respect to
x
to obtain
g
y
g
to obtain an expression for
c(y)
c0 (y)
c(y)
to nd
φ
in 3-space:
1. Integrate
φx = f
2. Compute
φy
with respect to
x
to obtain
φ,
which includes an arbitrary function
via partial dierentation and equate it to
3. Substitute in for
function
which includes an arbitrary function
and integrate with respect to
4. Put it all together for an expression of
φ
φ,
via partial dierentiation and equate it to
3. Substitute in for
How to nd
for conservative vector elds
g
and integrate
cy (y, z)
g
to obtain an expression for
with respect to
y
to obtain
c(y, z),
c(y, z)
cy (y, z)
including an arbitrary
d(z)
4. Compute
φz
equate it to
via partial dierentiation (using
φ
from step 1 and
h
5. Substitute in for
h
and integrate with respect to
2
z
to get
d(z)
d(z)
substituted in from step 3) and
Example:
Determine whether the following vector eld is conservative on
when
→
−
F = h3x, 3yi
→
−
F is conservative
f (x, y) = 3x and g(x, y) = 3y
∂f
∂g
= 0 and
= 0, therefore it is
∂y
∂x
R2 .
If so, determine the potential function
Step 1: Prove if
Step 2: Find
conservative
φ
φx = f = 3x,
´
´
3
φx dx = 3x dx = x2 + c(y)
2
∂ 3 2
x + c(y) = c0 (y) = g
∂y 2
c0 (y) = 3y
´ 0
´
3
c (y) dy = 3y dy = y 2 + C
2
Step 3: Put it together
φ(x, y) =
3 2 3 2
x + y +C
2
2
Example:
Determine whether the following vector eld is conservative on
when
→
−
F = h6y + z, 6x + 7z, x + 7yi
R3 .
If so, determine the potential function
→
−
F is conservative
f (x, y, z) = 6y + z and g(x, y, z) = 6x + 7z and h(x, y, z) = x + 7y
∂g ∂f
∂h
∂g
∂h
∂f
=6=
,
=1=
, and
=7=
, therefore it is conservative
∂y
∂x ∂z
∂x
∂z
∂y
Step 1: Prove if
Step 2: Find
φ
φx = f = 6yz
´
´
φ = φx dx = 6y + z dx = 6xy + xz + c(y, z)
∂
(6xy + xz + c(y, z)) = 6x + cy (y, z) = g
φy =
∂y
6x + cy (y, z) = 6x + 7z → cy (y, z) = 7z
´
´
c(y, z) = cy (y, z) dy = 7z dy = 7yz + d(z)
∂
φz =
(6xy + xz + 7yz + d(z)) = x + 7y + d0 (z) = h
∂z
x + 7y = x + 7y + d0 (z) → d0 (z) = 0
´
´
d(z) = d0 (z) dz = 0 dz = C
Step 3: Put it together
φ(x, y, z) = 6xy + xz + 7yz + C
3
In order to evaluate the line integral
and dot it with
→
−
F
(or
→
−
∇φ
´→
−
−
−
−
∇φ · d→
r given φ(x, y) and →
r (t), you must take the derivative of →
r (t)
C
since they are equivalent), so make sure the gradient of
φ
is taken. Then a simple
integral can be calculated.
Fundamental Theorem for Line Integrals:
ˆ
−
→
F · T̂ ds =
C
ˆ
−
→ −
→
F (→
r (t)) · −
r 0 (t) dt = φ(B) − φ(A)
C
Example:
π
→
−
r (t) = hcos(t), sin(t)i for ≤ t ≤ π and φ(x, y) = xy , use the fundamental theorem for line integrals
2
´→
−
−
evaluate
∇φ · d→
r
If given
to
C
Step 1: Plug each bound into
A : hcos
π
2
, sin
π
2
→
−
r (t)
to nd coordinates A and B
i = h0, 1i ∴ x = 0, y = 1 ∴ A = (0, 1)
B : hcos(π), sin (π)i = h−1, 0i ∴ x = −1, y = 0 ∴ B = (−1, 0)
Step 2: Use the fundamental theorem of line integrals
φ(x, y) = xy →
´→
−
−
∇φ · d→
r = φ(B) − φ(A) = (−1)(0) − (0)(1) = 0
C
4