Post on 24-Sep-2015
description
Line Integrals of Scalar Fields
Math 55 - Elementary Analysis III
Institute of MathematicsUniversity of the Philippines
Diliman
Math 55 Line Integrals 1/ 16
Curtain Area Problem
Problem
Let C be a smooth curve defined by a vector function~R(t) = x(t) + y(t), with t [a, b] and suppose f(x, y) iscontinuous on C. Find the area of the curtain whose base is aportion of C and whose height at a point (x, y) on C is f(x, y).
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Solving the Curtain Area Problem
1 Partition [a, b] into n sub-intervals[ti1, ti] such that the correspondingarcs in C has equal lengths s.
2 Choose ti [ti1, ti]. This correspondsto a point P i (x
i , yi ) in each subarc.
3 Construct cylinders with each subarc as base andheight f(xi , y
i ).
4 Each cylinder will have area A = f(xi , yi )s and the
area of the curtain is approximately
A ni=1
A =
ni=1
f(xi , yi )s.
5 As n, the error vanishes and the area A of the curtain is
A = limn
ni=1
f(xi , yi )s.
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Line Integrals with respect to Arclength
Definition
Let f(x, y) be continuous on a smooth plane curve C defined bya vector function ~R(t) = x(t), y(t), t [a, b]. The lineintegral of f along C (with respect to the arclengthparameter) is
Cf(x, y) ds = lim
n
ni=1
f(xi , yi )s
provided this limit exists.
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Line Integrals with respect to Arclength
Remarks.
1 The line integral is independent of the parametrization ofC.
2 If C denotes the curve C traced in the opposite direction,then
Cf(x, y) ds =
C
f(x, y) ds
3 If C is piecewise smooth, i.e, C is the union of finite number ofsmooth curves C1, C2, . . . , Cn, thenC
f(x, y) ds =
C1
f(x, y) ds+
C2
f(x, y) ds+ . . .+
Cn
f(x, y) ds
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Evaluating Line Integrals
Recall thatds
dt= ~R(t). Hence,
Definition Cf(x, y) ds =
baf(x(t), y(t)) ~R(t) dt
If ~R(t) = x(t), y(t), then ~R(t) =
[x(t)]2 + [y(t)]2 andtherefore,
DefinitionCf(x, y) ds =
baf(x(t), y(t))
[x(t)]2 + [y(t)]2 dt
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Some Examples
Example
Evaluate
C
2xy ds where C is the part unit circle in the first
quadrant.
Solution. C can be parametrized by ~R(t) = cos t, sin t,0 t pi2 . Therefore,
C2xy ds =
pi2
02 cos t sin t
( sin t)2 + (cos t)2 dt
=
pi2
02 cos t sin t dt
= sin2 t
pi20
= 1
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Some Examples
Example
Evaluate
Cx ds where C is the part of the parabola y = x2
from (0, 0) to (1, 1), followed by the line segments from (1, 1) to(0, 2) and from (0, 2) to (0, 0).
Solution. C is a piecewise smooth curve which can be wittenas the union of the curves C1, C2 and C3. Therefore,
Cx ds =
C1
x ds +
C2
x ds +
C3
x ds
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Some Examples
Solution(contd).C1 is the part of the parabola y = x
2 from (0, 0) to (1, 1).
The parametrization is ~R1(t) =t, t2
, 0 t 1. Hence,
C1
x ds =
10
t
1 + 4t2 dt =5
5 112
C2 is the line segment from (1, 1) to (0, 2). Aparametrization will be~R2(t) = (1 t) 1, 1+ t 0, 2 = 1 t, 1 + t, 0 t 1.Hence,
C2
x ds =
10
(1 t)1 + 1 dt =
2
2
C3 is the line segment from (0, 2) to (0, 0). Using the
parametrization ~R3(t) = 0, t where 0 t 2, we haveC3
x ds =
20
0
0 + 1 dt = 0
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Some Examples
Solution(contd). Finally, we haveCx ds =
C1
x ds +
C2
x ds +
C3
x ds
=5
5 112
+
2
2+ 0
=5
5 + 6
2 112
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Line Integrals with respect to x and y
Definition
Let f(x, y) be continuous on a smooth curve C given by~R(t) = x(t), y(t), a t b.
1
Cf(x, y) dx = lim
n
ni=1
f(xi , yi )x
2
Cf(x, y) dy = lim
n
ni=1
f(xi , yi )y
provided these limits exist.
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Evaluating Line Integrals
Equivalently, we have the following definition
Definition
1
Cf(x, y) dx =
baf(x(t), y(t))x(t) dt
2
Cf(x, y) dy =
baf(x(t), y(t)) y(t) dt
Oftentimes, the line integrals with respect to x and y occurtogether. In this case, we have the following
DefinitionCP (x, y) dx +
CQ(x, y) dy =
CP (x, y) dx + Q(x, y) dy
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Evaluating Line Integrals
Remarks.1 If C denotes the curve C traced in the opposite direction,
then C
f(x, y) dx = C
f(x, y) dxC
f(x, y) dy = C
f(x, y) dy
2 If C is piecewise smooth such that C = C1 C2 . . . Cnwhere each Ci is smooth for i = 1, 2, . . . , n, thenC
f(x, y) dx =
C1
f(x, y) dx+
C2
f(x, y) dx+ . . .+
Cn
f(x, y) dxC
f(x, y) dy =
C1
f(x, y) dy +
C2
f(x, y) dy + . . .+
Cn
f(x, y) dy
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Line Integrals with respect to x and y
Example
Let C be the curve given by ~R(t) =t2, 2 t, t [0, 1].
Evaluate
Cxy dx + (2x y) dy
Solution.C
xy dx + (2x y) dy = 10
t2(2 t)(2t) dt + [2t2 (2 t)](1) dt
=
10
2t4 + 4t3 2t2 t + 2 dt
= 25t5 + t4 2
3t3 1
2t2 + 2t
10
= 25
+ 1 23 1
2+ 2
=43
30
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Exercises
Evaluate the following line integrals over the given curves C.
1
Cy3 ds, C is given by ~R(t) =
t3, t
, t [0, 2].
2
Cxy4 ds, C is the right half of the circle x2 + y2 = 16.
3
Cxey dx, C is the arc of the curve x = ey from (1, 0) to
(e, 1).
4
C
sinx dx + cos y dy, where C is the top half of the circle
x2 + y2 = 1 from (1, 0) to (1, 0) and the line segment from(1, 0) to (1, 0).
5
Cxyz2 ds, C is the line segment from (1, 5, 0) to (1, 6, 4).
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References
1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008
2 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/
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