Section 17.2 Line Integrals. Let C be a smooth plane curve given by x = x(t), y = y(t), a ≤ t ≤...

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Transcript of Section 17.2 Line Integrals. Let C be a smooth plane curve given by x = x(t), y = y(t), a ≤ t ≤...

Section 17.2Line Integrals

LINE INTEGRALSLet C be a smooth plane curve given byx = x(t), y = y(t), a t b.We divide the parameter interval [a, b] into n subintervals [ti 1, ti]of equal width, and we let xi = x(ti) and yi = y(ti). Then the corresponding points Pi(xi, yi) divide C into n subarcs with lengths si. Let be a point on the subarc Ci. If f is defined on a smooth curve C, then the line integral of f along C
if the limit exists.

EVALUATING LINE INTEGRALSRecall from Section 11.2 that the arc length of C is
If f is a continuous function, the limit on the previous slide always exists. The following formula can be used to evaluate the line integral.

INTERPRETATION OF THELINE INTEGRALIf z = f (x, y) 0 and C is a curve in the plane, then line integral
gives the area of the curved curtain below the surface and above C. See Figure 2 on page 1099.

EXAMPLEEvaluate the following line integral where C is the line segment joining (1, 2) to (4, 7)

PIECEWISESMOOTH CURVES AND LINE INTEGRALSIf C is a piecewisesmooth curve then C can be written as a finite union of smooth curves; that is,C = C1 U C2 . . . U CnThe line integral of f along C is defined as the sum of the line integrals of f along each of the smooth pieces of C; that is,

EXAMPLEEvaluate where C is the piecewise smooth curve formed by the boundary region bounded by y = x and y = x2.

AN INTERPRETATION OF THE LINE INTEGRALSuppose that (x, y) represents the density of a thin wire that is shaped like the plane curve C. The mass of the wire is given by
The center of mass of the wire is given by

EXAMPLEA thin wire is bent in the shape of the semicirclex = cos t, y = sin t, 0 t If the density of the wire at a point is proportional to its distance from the xaxis, find the mass and center of mass of the wire.

LINE INTEGRALS WITH RESPECT TO x AND yTwo other line integrals can be obtained by replacing si by either xi = xi xi 1 or yi = yi yi 1. They are called the line integrals of f along C with respect to x and y.

DISTINGUISHING FROM THE ORIGINAL LINE INTEGRALTo distinguish the line integral with respect to x and y from the original line integral C f (x, y) ds, we call C f (x, y) ds the line integral with respect to arc length.

EVALUATING LINE INTEGRALS WITH RESPECT TO x AND y

A SPECIAL NOTATIONThe line integrals with respect to x and y frequently occur together. We write this as follows.

ORIENTATION AND LINE INTEGRALSRecall that a given parametrization x = x(t), y= y(t), a t b, determines an orientation of a curve C. If we let C denote the curve consisting of the same points as C but with opposite orientation, then we have:NOTE: The line integral with respect to arc length DOES NOT change sign.

LINE INTEGRALS IN SPACESuppose that C is a smooth space curve given byx = x(t), y = y(t), z = z(t), a t b.Suppose that f is function of three variables that is continuous on some region containing C, then the line integral of f along C is defined in a similar manner as for plane curves:

EVALUATING LINE INTEGRALS IN SPACE

VECTOR NOTATION FOR LINE INTEGRALSIf r(t) is the vector form of either a plane curve or a space curve, then the formula for evaluating a line integral with respect to arc length can be written compactly as

WORK AND LINE INTEGRALSSuppose that F = P i + Q j + R k is a continuous force field in three dimensions, such as a gravitational field. To compute the work done by this force in moving a particle along the smooth curve C, we divide C into subarcs Pi1Pi with lengths si by dividing the parameter interval [a, b] into subintervals of equal width. Choose a point on the ith subarc corresponding to the parameter . If si is small, then as the particle moves from Pi1 to Pi along the curve, it proceeds approximately in the direction of , the unit tangent vector at . The work done by the force F in moving to particle from Pi1 to Pi is approximately
The total work done in moving the particle along C is approximately

WORK (CONCLUDED)Based on the derivation on the previous slide, we define the work W done by the force field F in moving a particle along C as the limit of the Riemann sums, namely,

EVALUATING A LINE INTEGRAL FOR WORKIf the curve C is given by the vector equationr(t) = x(t)i + y(t)j + z(t)kthen T(t) = r(t)/r(t). So, the line integral for work can be rewritten as

EXAMPLEFind the work done by the force field
on a particle as it moves along the helix given byr(t) = cos ti + sin tj + tkfrom the point (1, 0, 0) to (1, 0, 3)

LINE INTEGRAL OF A VECTOR FIELDDefinition: Let F be a continuous vector field defined on a smooth curve C given by a vector function r(t), a t b. Then the line integral of F along C is
where F = Pi + Qj + Rk.

NOTE 1Even though C F dr = C F T ds and integrals with respect to arc length are unchanged when orientation is reversed, it is still true that
because the unit tangent vector T is replaced by its negative when C is replaced by C.

NOTE 2where F = Pi + Qj + Rk