I love Greens Theorem!
1
Use the two formulas to nd the two line integrals
(9)
(10)
C
F(x, y ) dr =
C
D
F(x, y ) n ds =
(curl F(x, y ) k dA
D
(div F(x, y ) dA
1) Let C be the positively oriented circle with center (0, 0), and radius 2.
That is, C : r(t) =
I love line integrals!
1
1) Let C be the positively oriented circle with center (0, 0), and radius 2.
That is, C : r(t) = 2 cos(t), 2 sin(t) , 0 t 2 . Let F(x, y ) = xy 2 y, x2 y + 2x and let
T be the unit tangent to the circle.
a) Find T(t)
b) Find F (r(
(11)
A (S ) =
(12)
A (S ) =
(13)
(14)
(15)
(16)
(17)
(18)
(19)
S
rs rt dA where r(s, t) = x(s, t), y (s, t), z (s, t)
z 2
x
D
S
C
f (x, y, z ) dS =
F dS =
F dr =
S
D
F dS =
S
S
z
y
D
2
+ 1 dA
z 2
x
+
z
y
x dS ; y =
1
m
S
f (x, y, g (x, y )
x=
(x, y, z ) d
For the formulas below, C is a curve in R2 (or R3 ), and x(t), y (t), z (t) , a t b
is a parametrization of C . Also
dy
dz
dx
,y= ,z=
dt
dt
dt
x=
(1)
in R2 : ds =
and in R3 : ds =
x2 + y 2 dt
(2)
C
f (x, y ) ds =
b
a
f (x(t), y (t) ds and
(3)
C
f (x, y )
Integrals
1
Integrals you get by reversing the derivatives of basic functions:
1
du = ln(u)
u
eu du = eu
sin(u) du = cos(u)
csc2 (u) du = cot(u)
up du =
1
du = arcsin(u)
1 u2
Trigonometric integrals (that follow from a basic substitution):
cot(u) du = ln(
Richard Reynolds - MAT 272
1
Summary of Grad f =
f (Del f )
Denition:
f (x, y ) = fx (x, y ), fy (x, y )
f (x, y, z ) = fx (x, y, z ), fy (x, y, z ), fz (x, y, z )
Directional Derivative
(Instantaneous rate of change of f in the direction u at the point (
Divergence and Stokes Theorem
1
Suppose that Q R3 is bounded by the closed surface Q and that n(x, y, z ) denotes
the exterior unit normal vector to Q. Then if the components of the vector eld F(x, y, z )
have continuous partial derivatives in Q, we have
1) Let C be the positively oriented circle with center (0, 0), and radius 2.
That is,
. Let
.
a)
b)
c)
d)
e)
By Greens Theorem (9), the circulation integral is
2) Let C be the positively oriented circle with center (0, 0), and radius 2.
That is,
. Let
.
(