Line Integrals
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Transcript of 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
EJEMPLO
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
EJEMPLO
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,…. .
EJEMPLO
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
EJEMPLO
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
EJEMPLO