Review
Recitation Problems
AMS261 Recitation Course 13
Recitation Instructor: Yiyang Yang
Department of Applied Mathematics and Statistics
State University of New York at Stony Brook
Spring 2013
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Outline
1
Review
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
2
Recitation Problems
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Vector Fields
Deffinition of Vector Field
~ that assigns a
A vector field over a plane region R is a function F
~
vector F (x, y ) to each point in R.
~ that
A vector field over a solid region Q in space is a function F
~
assigns a vector F (x, y , z) to each point in Q.
~ (x, y , z) = M (x, y , z)~i + N (x, y , z)~j + P (x, y , z) ~k
A vector field F
is continuous at a point if and only if each of its component
functions M, N and P is continuous at that point.
Some physical examples of vector fields are velocity fields,
gravitational fields and electric force fields.
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Conservative Vector Fields
Definition of Conservative Vector Field
~ is called conservative if there exists a differentiable
A vector field F
~ = ∇f . The function f is called the potential
function f such that F
~.
function for F
Test for Conservative Vector Field in the Plane
Let M and N have continuous first partial derivatives on an open disk R.
~ (x, y ) = M~i + N~j is conservative if and only if
The vector field given by F
∂N
∂M
=
.
∂x
∂y
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Curl of a Vector Field
Definition of Curl of a Vector Field
~ (x, y , z) = M~i + N~j + P ~k is
The curl of F
~ (x, y , z) = ∇ × F
~ (x, y , z)
curl F
=
∂P
∂N
−
∂y
∂z
~i −
∂P
∂M
−
∂x
∂z
~j +
∂N
∂M
−
∂x
∂y
~k.
Test for Conservative Vector Field in Space
Suppose that M, N and P have continuous first partial derivatives in an open sphere
~ (x, y ) = M~i + N~j + P ~k is conservative if and
Q in space. The vector field given by F
only if
~ (x, y , z) = ~0.
curl F
~ is conservative if and only if
That is, F
∂P
∂N ∂P
∂M ∂N
∂M
=
,
=
,
=
.
∂y
∂z ∂x
∂z
∂x
∂y
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Divergence of a Vector Field
Definition of Divergence of a Vector Field
~ (x, y ) = M~i + N~j is
The divergence of F
~ (x, y ) = ∇ · F
~ (x, y ) = ∂M + ∂N .
div F
∂x
∂y
~ (x, y ) = M~i + N~j + P ~k is
The divergence of F
~ (x, y , z) = ∇ · F
~ (x, y , z) = ∂M + ∂N + ∂P .
div F
∂x
∂y
∂z
~ = 0, then F
~ is said to be divergent free.
If div F
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Piecewise Smooth Curves
Deffinition of Smooth and Piecewise Smooth
A plane curve C : ~r (t) = x (t)~i + y (t)~j, a ≤ t ≤ b is smooth if
dx dy
,
dt dt
are continuous on [a, b] and not simultaneously 0 on (a, b).
A space curve C : ~r (t) = x (t)~i + y (t)~j + z (t) ~k, a ≤ t ≤ b is
smooth if
dx dy dz
,
,
dt dt dt
are continuous on [a, b] and not simultaneously 0 on (a, b).
A curve C is piecewise smooth if the interval [a, b] can be
partitioned into a finite number of subintervals, on each of which C
is smooth.
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Line Integrals
Deffinition of Line Integrals
If f is defined in a region containing a smooth curve C of finite length,
then the line integral of f along C is given by
ˆ
n
X
f (x, y ) ds = lim
C
or
k4k→0
ˆ
f (x, y , z) ds = lim
C
n
X
k4k→0
f (xi , yi ) 4si
i=1
f (xi , yi , zi ) 4si
i=1
provided this limit exists.
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Line Integrals (Cont.)
Evaluation of a Line Integral as a Definite Integral
Let f be continuous in a region containing a smooth curve C . If C is
given by ~r (t) = x (t)~i + y (t)~j, where a ≤ t ≤ b then
ˆ
ˆ
f (x, y ) ds =
C
b
q
2
2
f (x (t) , y (t)) [x 0 (t)] + [y 0 (t)] dt.
a
If C is given by ~r (t) = x (t)~i + y (t)~j + z (t) ~k, where a ≤ t ≤ b then
ˆ
f (x, y , z) ds
C
ˆ
=
b
q
f (x (t) , y (t) , z (t))
2
2
a
Yiyang Yang
2
[x 0 (t)] + [y 0 (t)] + [z 0 (t)] dt.
AMS261 Recitation Course 13
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Review
Recitation Problems
Line Integrals of Vector Fields
Deffinition of the Line Integral of a Vector Field
~ be a continuous vector field defined on a smooth curve C given by
Let F
~ on C is given by
~r (t), where a ≤ t ≤ b. The line integral of F
ˆ
ˆ
~ · d~r =
F
C
ˆ
~ (x (t) , y (t) , z (t)) · ~r 0 (t) dt.
F
~ ·T
~ ds =
F
C
C
ˆ
ˆ
~ · d~r =
F
~ · d~r dt
F
dt
ˆ =
M~i + N~j · x 0 (t)~i + y 0 (t)~j dt
ˆ C
ˆ
dy
dx
=
+N
dt =
Mdx + Ndy
M
dt
dt
C
C
C
C
This differential form can
be extended to three variables as
ˆ
Mdx + Ndy + Pdz
C
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Fundamental Theorem of Line Integrals
Deffinition of the Line Integral of a Vector Field
Let C be a piecewise smooth curve lying in an open region R and given by
~r (t) = x (t)~i + y (t)~j, a ≤ t ≤ b.
~ (x, y ) = M~i + N~j is conservative in R, and M and N are continuous
If F
in R, then
ˆ
ˆ
~
F · d~r =
∇f · d~r = f (x (b) , y (b)) − f (x (a) , y (a))
C
C
~ . That is F
~ (x, y ) = ∇f (x, y ).
where f is a potential function of F
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Vector Fields
Line Integrals
Conservative Vector Fields and Independence of Path
Independence of Path
Equivalent Conditions
~ (x, y , z) = M~i + N~j + P ~k have continuous first partial derivatives
Let F
in an open connected region R, and let C be a piecewise smooth curve in
R. The following conditions are equivalent.
~ is conservative. That is F
~ = ∇f for some function f .
F
´
~ · d~r is independent of path.
F
C
´
~ · d~r = 0 for every closed curve C in R.
F
C
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Textbook Problems
15.1.81 on p.1068
~) = ∇· ∇×F
~ where F
~ (x, y , z) = xyz~i + y~j + z ~k.
Find div (curl F
Solution:
~i
∂
~
curl F = ∂x
xyz
~j
∂
∂y
y
~k ∂ = 0 · ~i − xy~j + xz ~k
∂z z
~ = div 0 · ~i − xy~j + xz ~k = 0 − x + x = 0
div curl F
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Textbook Problems (Cont.)
15.2.73 on p.1082
Find the moments of inertia for the wire of density ρ. A wire lies along
~r (t) = a cos t~i + a sin t~j, 0 ≤ t ≤ 2π and a > 0, with density function
ρ (x, y ) = 1.
Solution:
0
Since ~r (t) = −a sin t~i + a cos t~j
q
p
2
2
[x 0 (t)] + [y 0 (t)] = a2 sin2 t + a2 cos2 t = a
The moments of inertia are
ˆ
ˆ
y 2 ρ (x, y ) ds =
Ix =
C
ˆ
2π
y 2 ρ (x, y )
q
2
2
[x 0 (t)] + [y 0 (t)] dt
0
ˆ 2π
(a sin t)2 · adt = a3
sin2 tdt = a3 π
0
0
ˆ
ˆ 2π
Iy =
x 2 ρ (x, y ) ds = a3
cos2 tdt = a3 π
2π
=
C
Yiyang Yang
0
AMS261 Recitation Course 13
Review
Recitation Problems
Textbook Problems (Cont.)
15.3.35 on p.1091
~ (x, y ) = 9x 2 y 2~i + 6x 3 y − 1 ~j in
Find the work done by the force field F
moving an object from P (0, 0) to Q (5, 9).
Solution:
Since
∂ (9x 2 y 2 )
∂y
= 18x 2 y =
∂ (6x 3 y −1)
,
∂x
~ (x, y ) is conservative.
F
The work done by the force field is
ˆ
ˆ
~
F · d~r =
∇f · d~r = f (x (b) , y (b)) − f (x (a) , y (a))
C
C
(5,9)
= 3x 3 y 2 − y |(0,0) = 30366.
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Textbook Problems (Cont.)
15.4.31 on p.1100
Use Green’s theorem to verify the line integral formulas. The centroid of
the region
area A bounded
´ have
´ by2 the simple closed path C is
1
1
2
x̄ = 2A
x
dy
and
ȳ
=
−
2A C y dx.
C
Proof:
Based on Green’s formula
´
Mdx + Ndy =
C
´ ´ ∂N
R
∂x
−
∂M
∂y
´ ´
xdA
2xdA = ´R ´
= x̄
dA
R
R
´ ´
ˆ
ˆ ˆ
ydA
1
1
2
−
y dx = −
(−2y ) dA = ´R ´
= ȳ
2A C
2A R
dA
R
1
2A
ˆ
1
x dy =
2A
C
ˆ ˆ
2
conclusion is proved.
Yiyang Yang
AMS261 Recitation Course 13
dA
Review
Recitation Problems
Textbook Problems (Cont.)
15.6.11 on p.1122
Find the mass of the surface lamina S : 2x + 3y + 6z = 12 first octant
with density ρ (x, y , z) = x 2 + y 2 .
Solution:
The projection of S in first octant on xy -plane is
x ≥ 0, y ≥ 0, 2x + 3y ≤ 12, then the integration domain is
2
0 ≤ x ≤ 6, 0 ≤ y ≤ 4 − x
3
Yiyang Yang
AMS261 Recitation Course 13
Review
Recitation Problems
Textbook Problems (Cont.)
The mass of the surface z = f (x, y ) = 2 − 13 x − 21 y is
ˆ ˆ
ˆ ˆ q
m=
ρds =
ρ 1 + fx2 + fy2 dA
ˆ ˆ
r
=
=
2
2
ˆ
4− 23 x
x +y
7
6
ˆ
0
6
1+
1 1
+ dA
9 4
x 2 + y 2 dydx
0
"
4 2 1 4 1
=
x − x −
3
6
8
Yiyang Yang
2
4− x
3
4 #
|60 =
364
.
3
AMS261 Recitation Course 13