Section 9.8 Line Integrals
Parametric Curve
x = f (t ) , y = g (t )
f ′ and g ′ continuous
for t in [ a, b ]
Consists of a finite
number of smooth curves
Starts and ends at the same
point and doesn't cross itself
Starts and ends
at the same pt.
Orientation of the curve
Geometric Interpretation of a Line Integral
9.8
Definite Integral
Line Integral
1
9.8
b
∫ P ( x, y ) dx + Q ( x, y ) dy = ∫ U ( t ) f ′ ( t ) dt + V ( t ) g ′ ( t ) dt
a
C
Parametrize the path
Substitute everything
x = f (t ) , y = g ( t ) a ≤ t ≤ b
P ( x, y ) = P ( f ( t ) , g ( t ) ) = U ( t )
dx = f ′ ( t ) dt , dy = g ′ ( t ) dt
Q ( x, y ) = Q ( f ( t ) , g ( t ) ) = V ( t )
Work done by a vector field F as its point of application
moves along C from A to B.
Calc II
C → line
W = F⋅d
Now
C → curve
r the parametrization of C as a vector
W = ∫ F ⋅ dr
C
Section 9.8 #14
Evaluate ∫ ydx + xdy
C
on C : line segments from ( 0, 0 ) to (1, 0 )
C2
and from (1, 0 ) to (1,1)
∫ ydx + xdy = ∫ ydx + xdy + ∫ ydx + xdy
C
C1
C1 :
∫
C1
C2
x=t
y=0
0 ≤ t ≤1
dx = dt dy = 0
1
ydx + xdy = ∫ ( 0 + 0 ) dt = 0
0
C1
C2 :
x =1
y=t
0 ≤ t ≤1
dx = 0 dy = dt
1
∫
ydx + xdy = ∫ ( 0 + 1) dt = 1
C2
0
∫ ydx + xdy = 1
C
2
Section 9.8 #20
Evaluate
∫ ( x + y ) dx − 2 xydy on the given closed curve C.
) dx − 2 xydy = ∫ ( x + y ) dx − 2 xydy + ∫ ( x + y ) dx − 2 xydy
2
2
C
∫(x
2
+ y2
2
C
2
2
C1
C2
y = t2
0 ≤ t ≤1
dy = 2tdt
x=t
C1 :
dx = dt
∫ (x
2
1
1
0
0
1
t 3 3t 5
1 3 5 − 9 −4
−
= − =
=
3 5 0 3 5
15
15
+ y 2 ) dx − 2 xydy = ∫ ( t 2 + t 4 − 4t 4 ) dt = ∫ ( t 2 − 3t 4 ) dt =
2
C1
x=t
C2 :
y= t
t starts at 1 and ends at 0
1
dx = dt dy =
dt
2 t
∫ (x
2
C2
∫(x
0
1
1
0
2
+ y 2 ) dx − 2 xydy = ∫ ( t 2 + t − t ) dt = − ∫ t dt = −
2
+ y 2 ) dx − 2 xydy =
C
1
t3
1
= −
30
3
−4 1 −4 − 5 −9 −3
− =
=
=
15 3
5
15
15
Section 9.8 #32
Find the work done by the force F ( x, y ) = 2 xyi + 4 y 2 j acting
along the piecewise smooth curve consisting of line segments
from ( −2, 2 ) to ( 0, 0 ) and from ( 0, 0 ) to ( 2,3) .
C2
C1
Work = ∫ F ⋅ dr = ∫ F ⋅ dr + ∫ F ⋅ dr
C
C1 :
C1
C2
y = −t
dx = dt
dy = −dt
−2≤t ≤0
r = ti − tj ⇒ dr = 1, −1 dt
F on C1 : −2t 2 , 4t 2
0
∫ F ⋅ dr = ∫ ( −2t
2
− 4t
2
0
= ∫ −6t 2 dt = −2 t 3
0
−2
C2 :
F on C2 : 3t 2 ,9t 2
) dt

−2
C1
3
y= t
2
0≤t ≤2
3
dx = dt dy = dt
2
3
3
r = ti + tj ⇒ dr = 1, dt
2
2
x=t
x =t
= −2 ( 0 − ( −8 ) ) = −16
∫ F ⋅ dr = ∫  3t
C2
C2
2
2
+
27 2 
33
t  dt = ∫ t 2 dt
2 
2
0
11 2 11
= t 3 = ( 8 − 0 ) = 44
2 0 2
−2
Work = ∫ F ⋅ dr = −16 + 44 = 28
C
3