Line (path) integrals
Example. A particle moves under the influence of a force
~F = 6x2 y~i + 10xy2~j along a path C described in parametric
form by g(t) = (t, t3 ) as t goes from 0 to 1. Find the work done
by the force.
R
Solution. Compute C ~F · d~x, where d~x is the vector (dx, dy).
R
R1 5
R
2
2
7
2
~ x=
C F · d~
C 6x y dx + 10xy dy = 0 6t dt + 10t × 3t dt
= 1 + 3 = 4.
R
Example. Compute C (x + y) dx + (x − y) dy, where C is the
triangle with vertices at (0, 0), (1, 0), and (0, 1) (oriented
counterclockwise).
Math 311-102
Harold P. Boas
boas@tamu.edu
Math 311-102
June 24, 2005: slide #1
Math 311-102
Green’s theorem (in the plane)
Z
closed
curve
P(x, y) dx + Q(x, y) dy =
ZZ
region
inside
µ
∂Q ∂P
−
∂x
∂y
June 24, 2005: slide #2
Stokes’s theorem (in the plane)
¶
Reinterpretation of
dx dy
³
∂Q
∂x
−
∂P
∂y
´
∂
∂
+ ~j ∂y
, a vector
. Write ∇ = ~i ∂x
where the curve is traversed in the direction that leaves the
region on the left-hand side.
~
~
differential operator.³Write ~F =
´ iP + jQ, a vector force.
∂Q
∂P
~
~
−
Then (∇ × F) · k =
.
Example. For the preceding example
³ with the triangle,
´
RR
R
∂
∂
C (x + y) dx + (x − y) dy =
inside ∂x (x − y) − ∂y (x + y) dx dy
RR
= inside (1 − 1) dx dy = 0.
Reinterpretation of ~F · d~x. If our path is described in
parametric form as g(t) = (g1 (t), g2 (t)), then d~x = ~i dx + ~j dy
³ 0
´
g1 (t) ~
g0 (t)
def
i + 20 ~j |g0 (t)| dt = ~t ds,
= (g10 (t)~i + g20 (t)~j) dt =
0
∂x
|g (t)|
|g (t)|
Green’s
theoremZrewritten:
Z
Z
~F · ~t ds =
(∇ × ~F) · ~k dx dy.
Solution. You could parametrize the path via x = 2 cos(θ),
y = 2 sin(θ) and evaluate
RR the line integral. Using Green’s
theorem instead gives
inside (−1 − 1) dx dy
closed
curve
of circle
region
inside
The path integral is the circulation of ~F around the curve.
The quantity ∇ × ~F is the curl of ~F.
= −2 × (area of circle) = −8π.
June 24, 2005: slide #3
∂y
where ~t is the unit tangent vector to the curve, and ds is the
arc-length element.
Example. Let C be the circle centeredR at (0, 0) with radius 2,
oriented counterclockwise. Compute C y dx − x dy.
Math 311-102
Solution. Choose parametrizations for each leg of the triangle.
For the hypotenuse, you could use the parametrization y = t,
x = 1 − t. The integral becomes the sum
R1
R0
R1
1
1
0 x dx + 0 1 × (−dt) + (1 − 2t) dt + 1 −y dy = 2 − 1 + 2 = 0.
Math 311-102
June 24, 2005: slide #4
Divergence theorem (in the plane)
If the curve is g(t) = (g1 (t), g2 (t)), then the unit normal vector
g0 (t)
g0 (t)
is ~n = 20 ~i − 10 ~j. If we write ~F = F1~i + F2~j, then ~F · ~n ds =
|g (t)|
|g (t)|
(F1 g20 (t) − F2 g10 (t)) dt = F1 dy − F2 dx.
Z
~F · ~n ds =
Green’s theorem implies that
closed
curve
µ
¶
ZZ
ZZ
∂F1
∂F2
∇ · ~F dx dy.
+
dx dy =
region
region
∂x
∂y
inside
inside
The path integral is the flux of ~F across the curve.
The quantity ∇ · ~F is the divergence of ~F.
Math 311-102
June 24, 2005: slide #5