3. 2. Line integrals
◦
integrate over a curve in #% =
line integrals
1123
I. Stds if f- real-valued
¢ function defined on
the curve @
p vector field
2. J F. di if É is a vector field
on the curve c
c
'
have to first
' describe
To determine a line integral we
the curve that is find a vector valued function
-
#
Parametrize
if can 't parametrize
FCH E- C- [q ,b ] you a
, curve =
can't do line integral
C :
Fct ) cxct), > 1- C- [ 9lb ] AY
yctl 2- It )
=
y☒•Ñ•
, ,
Ifaf initial point of curve
=
•
F (b) =
terminal (end) point of curve Feb)
r
•
>n
study guide
~
Standard Parametrization s
from A Cal / 92,93)
① Line
segment to B Cbi bi bs
, , ) in 1122 or /Rˢ
r (f) =
< di
, di d} > + t < bi -
91 , bz -92 bs -93 > t c- [ 91 ]
,
,
=
[ die tlbi -91 ) ] it [ 92 Case 1- Cbs -93)]K
1- ( bi 92 )
)j TE cost ]
'
- -
+
◦ B
7 7
◦
A Direction of parametrization
is from A to B
, ② circle with Center ( a / b) and radius C
F) rlt) Cd b) CC cost Sint > 1- C- [0,21T ]
=
→
/ ,
[0,21T ]
:b (94 Ccostlit Cbi csint
)j 1- C-
=
•
ca )
anti-clockwise
of parametrization
=
Direction
③ Function )
y the
H
'
✗ C- [a b) in IR
-
_
, ,
let × t then ,
-
-
,
i ◦
rlt ) =
< t
,
Flt ) > ttcq, b) ^↳ ◦
tie
flttj
=
I 1
a b
Direction of parametrization
is from la, f- (d) → ( b f- (b) )
,
Reversal of parametrization
Tfg C has orientation
curve d
given ,
then -
C is the
curve with the same
points as C but with opposite
orientation
[ → ( t) ⇐
< 11-1 ,
yltl / b)
r ✗ 2- ( t ) > 1- C- [ a
,
C r.lt ) rlzt )
:
- →
=
< ✗
1--1-7 , yczt) zczt) > ,
1- C- C- b , -
9)