Line Integrals

RAUL ANDRES TORRES BALLEN

UNIVERSIDAD INDUSTRIAL DE SANTANDER

A curve, C, in three-dimensional space may be represented by parametric equations:

Fig.1

or in vector notation

where (see Fig.1)

LINE INTEGRALS

Then, a vector A[x(t), y(t), z(t)] or a scalar, , is pictured on the domain C, which according to the parametric representation, is referred to the real number interval a≤t≤b

The integral of a vector field A defined on a curve segment C is called Line Integral. The integrand has the representation obtained by expanding the dor product. The ecalar an vector integrals

The following three basic ways are used to evaluate the

line integral: * The parametric equations are used to express the integrand through the parameter t.

* If the curve C is a plane curve and has one of the representations y=f(x) or x=g(y), then the two

integrals that arise are evaluated whit respect to x or y, whichever

is more convenient. * If the integrand is a perfect differential, then it may be

evaluated through knowledge of the end points.

If the equation of a curve C in the plane z=0 is given as y=f(x), then line integral is evaluated by placing y=f(x), dy=f’(x)dx in the integrand to obtain the definite integral

EVALUATION OF LINE INTEGRALS FOR PLANE CURVES

Similarly, is C is given as x=g(y), then dx=g’(y)dy

If C is given in parametric form x=Ø(t), y=ψ(t)

combinations of the above methods may be used in the evaluation. If the integrant A.dr is a perfect differencial,

, then

Similar methods are used for evaluating line integrals along space curve.

PROPERTIES OF LINE INTEGRALS EXPRESSED FOR PLANE CURVE

EJEMPLO

let P, Q, ∂P/ ∂y, ∂Q/ ∂x be single-valued and continuous in a simply connected region R bounded by a simple closed curve C. Then

GREEN’S THEOREM IN THE PLANE

when is used to emphasize that C is closed and that it is described in the positive direction

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The line integral of a vector field A is independent of path if its value is the same regardless of the (allowable) path from initial to terminal point.

For example, the integral of the vector field A=yi+xj is independent of path since

CONDITIONS FOR A LINE INTEGRAL TO BE INDEPEND OF THE PATH

Theorem 1. A necessary and sufficient condition that be independent of path is that

there exists a scalar function such that

Theorem 2. A necessary and sufficient condition that the line integral, be independent of

path is that

Theorem 3. If then the line integral of A

over an allowable closed path is 0, i.e., If C is a plane curve, then Theorem 3 follows

inmediately from Green’s theorem, since in the plane case reduces to

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The surface is thought of as embeddded in a three-dimensional Euclidean space and expressed thriugh a two-parameter vector representation:

SURFACE INTEGRALS

In terms of rectangular cartesian coordinates

In parametric representation

The parameter curves v2=const and v1=const establish a coordinate system on the surface. The key to establishing the surface integral of a function is the differencial elemt of surface area. At the point, P, of surface

In particular, the directions of the coording curves v2=const and v1=const are designated by (Fig.2)

The cross product

Fig.2

is normal to the tangent plane at P, and its manitude is the area of a differential coordinate parallelogram

Definition. The differential element of surface area is

For a function :

If the surface has the cartesian representation z=f(x,y) and the identifications v1=x, v2=y, z=f(v1,v2) are made then

and

Therefore,

if the surface is given in the implicit form F(x,y,z)=0, then the gradient may be employed to obtain another representation.

We again let v1=x, v2=y, z=f(v1,v2). Then

Taking the dot product of both sides of yields

The ambiguity of sing can be eliminated by taking the absolute value.

and the surface integral takes the form

let S be a two-sided surface having projection R on the xy plane as in the adjoining (Fig.3)

Assume that an equation for S is z=f(x,y), where f is single-valued. Divide R into n subregions of area

Fig.3

and erect a vertical column on each of these subregions to intersect S in an area

Form de sum where is

some point of If the limit of this sum as in such a way that each the resulting limit is called the Surface

Integral of Ø(x,y,z) over S and is designated by

Since approximately, where is the angle between the normal line S and the positive z-axis, the limit of the sum can be written

where

Then assuming that z=f(x,y) has continuous

In case the equation for S is given as F(x,y,z)=0, can also be written

In the above we have assumed that S is such that any line parallel to the z-axis intersects S in only one point. In case S is not of this type, we can usually subdivide S into surfaces S1, S2,…., which are of this type. Then the surface integral over S is defined as the sum of the surface integrals over S1, S2,…. .

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Let A be a vector field that is continuously differentiable on a closed-space region, V, bound by a smooth surface,

S (Fig.4) Then

THE DIVERGENCE THEOREM

when n is an outwardly drawn normal. If n is expressed through direction cosines, i.e.,

Fig.4

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Suppose a closed curve, C, bounds a smooth surface portion, S. If the component functions of x=r(v1,v2) have continuous mixed partial derivatives, then for a vector field A with continuous partial derivatives on S (Fig.5)

STOKES’ THEOREM

where representing the angles made by the outward normal n and i,j, and k, respectively.

Fig.5

then the component form of is

if , Stokes’ theorem tell us that

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