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Chapter 10 Vector Integral Calculus Integral Theorems

Page history last edited by CAI, Sheng 12 years, 2 months ago

Chapter 10 Vector Integral Calculus. Integral Theorems

 


1 Line Inegrals
The concept of a line integral is a simple and natural generalization of a definite integral.

                                                                                         Formula

known from calculus. In a line integral we shall integrate a given function, also called the integrand, along a curve C in space. Hence curve integral would be a better name, but line integral is standard

 

We represent the curve C by a parametric representation i

                             Formula 

The curve C is called the path of integration, A: Formula its initial point, and B: Formula its terminal point. C is now oriented. For figure b, the curve C is called a closed path.

 

C is called a smooth curve if it has at each point a unique tangent whose direction varies continuosly as we move along C.

 

Formula in (2) is differentiable and the derivative Formula is continous and different from the zero vector at every point of C.'

 

 

1.1 Definition and Evaluation of Line Integrals


A line integral of a vector function F(r) over a curve C: Formula [as in (2)] is defined by

                                                               Formula


(See also chapter 9 for dot produnct). In terms of components, with Formula and Formula, formula 3 becomes

 

                                                              Formula 

 

If the path of integration of C in (3) is a closed curve, then instead of Formula, we also write Formula

 

Note: The integrand in (3) is a scalar because we take dot product.

 

Example 1 Find the value of the line integral (3) when Formula and C is the circular arc as shown in the following figure form A to B.

 

Solution We may represent C by Formula whereFormula. Then Formula, Formula and,

                              Formula

 

By differentiation,, Formula, so that by (3), we have,

 

                                        Formula

 

Exercise 1 Find the value of (3) when Formula and C is the helix.

                              Formula

 

(The evaluation of the line integrals in space is practically the same as it is in the plane.)

 

 

 

 

 

 

 

 

 

 

 


Simple general properties of the line integral (3) follow directly form corresponding properties of the definition integral in calculus, namely,

                                                    Formula

 

1.2 Motivation of the Line Integral (3): Work Done by the Force

 

The work W done by a constance force F in the displacement along a straight segment d is Formula. This suggests that we define the work W done by a variable force F in the displacement along a curve C:  r(t) as the limit of sums of tworks done in displacements along small chords of C.

 

Example 2 In example 1, if F is a force, the work done by F in the displacement along the quarter-circle will be 0.4521, measured in newton-meters.

 

Exercise 2 - (PS 10.1 q1) Calculate Formula. If F is a force, this gives the work done in the displacement along C.

 

(Hint: As in example 1, we should represent F in parametric form first.)

 


Formula, C the parabola Formula from A(0,0) to B (2, 20).

 

 

 

 

 

1.3 Other Forms of Line Integals

 

The line integrals

                                                          Formula

are special cases of (3) when Formularespectively.

 

Furthermore, without taking a dot product as in (3), we can obtain a line integral whose value is a vector rather than a scalar, namely,

                     Formula

 

If we take Formula, Formula. Then

                                                                           Formula

with C as in (2).

 

Example 3 Integrate Formula along the helix in Exercise 2.

 

Solution Formula inegrated with respect to t form 0 togives

 

               Formula

 

Exercise 3 -PS10.1 q4 Integrate Formula

 

 

 

 

1.4 Path Dependence

 

Theorem 2 - Path Dependence The line integral (3) generally depends not only on F and on the endpoints A and B of the path, but also on the path itself along which the integral is taken.

 

Example 4 - Proof of theorem 2 Almost any example will show this. Take, for instance, the straight segment

 

Formula and the parabola Formula with Formula and integrate Formula

Then

                                                                 Formula

 

so that the integration gives Formulaand Formula respectively.

 

 

 

 

 

 

 

 

 

2 Path Independence of Line Integrals

 

In this section we consider line integrals

                        Formula

 

as before, and we shall now find conditions under which (8) is path independent in a domain D in space.

 

By definition this means that for ever pair of endpoints A, B in D, the integral (8) has the same value for all paths in D that begin at A and end at B.

We shall follow up three ideas that will give path independence of (8) in a domain D if and only if:

 

  • (Theorem 1) Formula
  • (Theorem 2) Integration around closed curves C in D always gives 0.
  • (Theorem 3) curl F = 0 (privided D is simply connected)

 

 

2.1 Theorem 1

 

A line integral (8) with continuous Formula in a domain D in space is path independent in D if and only if Formula is the gradient of some function f in D.

 

                                  Formula

 

Proof (a) We assume that (9) holds for some function f in D and show that this implies path independence.

 

Let C be any path in D from any point A to any point B in D, given by Formula ,where Formula . Then from (9), the chain rule and (3) we have

 

                      Formula

 

The last formula in part (a) of the proof,

 

                          Formula

is the analog of the usual formula for definite integrals in calculus,

 

                                           Formula

 

Formula (10) should be applied when a line integral is independent of the path.

 

Example 6 Show that the integral Formula is path independent in any domain in space and

find its value in the integration from A: Formula to B: Formula.

 

Solution Formula= grad Formula , where Formula because

 

                                                                   Formula

 

Hence the integral is independent of path according to theorem 1 and (10) gives Formula

 

If you want to check this, use the most convenient path C:

 


Formula, on which Formula

 

so that Formula, and integration from 0 to 2 gives Formula.

 

Exercise 6 Evaluate the integral Formula from A:Formula to B: Formula by showing that Formula has a potential (that is to find Formula ) and applying (10).

 

(Hint: let Formula,Formula and Formula)

 

2.2 Theorem 2 - Path Independence and Integration Around Closed Curves

 

The integral (10) is path independent in a domain D if and only if its value around every closed path in D is zero.

 

The idea is that two paths with common end-points make up a single closed curve.

 

Proof

  • If we have path independence, then integration from A to B along Formula in the figure above gives the same value.

 

  • Now Formulaand Formula together make up a closed curve C and if we integratefrom A along Formula to B as before, but then in the opposite sense along Formula back to A, the sum of the two integrals is zero, but this is the integral around the crosed curve C.
  • Conversely, assume that the integral around any closed path C in D is zero.

 

  • Given any points A and B and any two curves Formula and Formula from A to B in D, we see that Formula with the orientation reversed and Formula together form a closed path C.

 

  • By assumpltion, the integral over C is zero.

 

  • Hence the integrals overFormulaand Formula, both taken from A to B, must be equal.

 

2.2.1 Work Conservative and Noconservative Physical System

 

  • From the last section, we know that the integral (8) gives the work done by a forceFormulain the displacement of a body along curve C.

 

  • Then Theorem 2 states tahat work is path independent in D if and only if its value is zero for diplacement around every closed path in D.

 

  • Furthermore, theorem 1 tells us that this happens if and only if Formula is the fradient of a potential in D.

 

  • In this case, Formula and the vector field defined by Formula are called conservative in D because in this case mechanical energy is conserbved;

 

  • that is, now work is done in the displacement from a point A and back to A.

 

2.3 Theorem 3* - Path Independence and Exactness of Differential Forms

 

Theorem 1 relates path independence of the line integral (1) to the gradient and Theorem 2 to integration around closed curves. A third idea relates path independence to the exactness of the differential form

 

                                            Formula

under the integtral sign in (8). This form (11) is called exact in a domain D in space if it is differential

 

                                              Formula

of a differentiable function f(x,y,z) evertwhere in D, that is, if we have

                                                                                                    Formula

Comparing these two formulae, we see that the form (4) is exact if and only if there is differentiable funcion f(x,y,z) in D such that everywhere in D,

                                 Formula

 

Hence theorem 1 implies theorem 3.

 

The integral (8) is path independent in a domian D in space if and only if the differential from (11) has continous coefficient functions Formula, Formula, Formula and is exact in D.

 

A domain D is called simply connected if every clsoed curve in D can be continuously strucnk to any point in D without leaving D.

 

2.4 Theorem 3 - Criterion for Exactness and Path Independence

 

Let Formula, Formula, Formula in the line integral (8),

                                                      Formula

 

be continuos and have continuos first partial derivatives in a domain D in space. Then:

 

  • If the differential form (11) is exact in D - and thus (8) is path independent by theorem 3* - , then in D,

                                                                          Formula

in components (6') , refer to chapter 9

                                                                             Formula

  • If (12) holds in D and D is simply connected, then (11) is exact in D - and thus (8) is path independent by theorem 3*.

 

Proof

  • If (11) is exact in D, then Formula in D by theorem 3*, and, futhermore, curl Formula by theorem of Grad, Div, Curl in chapter 9.

 

Linde Integral in the Plane.

For Formula the curl has only one component (z-component), so that (6') can be reduced to the single rotation

                                                                                    Formula

 

Example 6 - Exactness and Independence if Path. Determination of a Potential Using (6'), show that the differential form under the integral sign of

 

                               Formula

 

is exact, so taht we have independence of path in any domain, and find the value of I from A: Formula to B: Formula.

 

Solution Eactness follows from (6'), which gives

                                             Formula

 

To find Formula we integrateFormula (which is ``long'' ,so that we save work) and then diferentiate to compare with Formula and Formula.

 

                                 Formula

 

Formulaimplies Formula and we can take Formula, so that Formula in the first line. This gives, by (10)

 

                         Formula

 

The assumption in Theorem 3 that D is simply connected is essential and cannot be omitted.

 

Exercise 7 PS10.2 q5 Show that the form under the integral sign is exact in the space.

 

                                                         Formula

 

 

 

3 Double Integrals


  • In a definite integral (1), we integrate a function f(x) iver an interval (a segment) of the x-axis.

 

  • In a double integral we integrate a function f(x,y), called the integrand, over a closed bounded region R in the x-y plane, whose boundary curve has a unique tangent at each point, but may have finitely many cusps.

 

  • This double integral can be considered as a limit of the following sum Formula whereFormulais the area of theFormulath rectangle as shown in the figure below.

 

 

The Double Integral of f(x,y) over the region R is denoted by

                                                       Formula

 

 

3.1 Properties of the double integrals

 

Double integrals have properties quite similar to those of definite integrals.

                                   Formula

 

Furthermore if R is simply connected, then there exists at least one point Formula in R such that we have

                                                     Formula

 

where A is the area of R. This is called the mean value theorem for double integrals.

 

3.2 Evaluation of Double Integrals by Two Successive Integrations

 

Double integrals over a region R may be evaluated by two successive integrations. We may integrate first over y and then over x. Then the formula is

                                          Formula

 

  • HereFormula and Formula represent the boundary curve of ZR and,

 

  • keeping x constant, we integrate Formula over Formula fromFormula to Formula The result will be a funciton of x.

 

  • Then we integrate the result from Formula to Formula.

 

Similarly, for integrating first over x and then over y the formula is

                                     Formula

 

 

  • The boundary cuve of R is now represented by Formula and Formula

 

  • Treating y as constant, we first integrate Formulaover x from Formulato Formulaand then

 

  • we integrate the resulting function of y from Formula to Formula.

 

 

3.3 Application of Double Integrals

 

Double Integrals have various phyiscal and geometric application. For instance, the area A of a reguin R in the xy-plane is given by the double integral.

                                                                     Formula

 

The volume V beneath the surface z=f(x,y) (>0) and above a region R in xy-plane is

                                                           Formula

 

because the termFormula in Formula at the beginning of this section represents the volume of a rectangular box with base of areaFormulaand altitude Formula

 

As another application, let f(x,y) be the density (mass per unit area) of a distribution of mass in the xy-plane. Then the total mass M in R is

                                                                 Formula

the center of gravity of the mass in R has the coordinates Formula, Formula, where

                              Formula

 

the moments of inertia Formula andFormulaof the mass in R about the x- and y- axes, respectively, are

                                         Formula

 

 

3.4 Change of Variables in Double Integrals. Jacobian

 

                                                 Formula

  • Here we assume that Formula is continous and

 

  • Formula has a continuous derivative in some intervalFormula such that

 

  • Formula, Formula, or [ Formula, Formula] and Formula varies between a and b when u varies between Formulaand Formula .

 

The formula for a change of variables in double integrals form Formula to Formula is

                           Formula

that is, the integrand is expressed in terms of u and v, and dx dy is replaced by du dv times the absolute value of the Jacobian.(where Formularepresents tohe corresponding region of R on the u-v plane.

                                     Formula

 

Example 7 -Change if Variables in Double Integral Evaluate the following doyble integral over the squre R in the following figure.

                                                              Formula

 

Solution The shape of R suggests the transformation Formula, Formula Then Formula, Formula, the Jacobia is

                                                Formula

 

R corresponds to the squareFormula, Formula. Therefore,

                                          Formula

 

If polar coordinates Formula andFormulaare to be used, we can set Formula, Formula .Then

                                                  Formula

and

                                Formula

 

Exercise 8 Let Formulabe the mass density in the region as shown in the figure. Find the total mass, the center of gravity, and the moment of inertia Formula, Formula.

 

 

 

4 Green's Theorem in the plane

 

Double iintegrals over a plane region may be transformed into line integrals over the boundary of the region and conversely. This is of practical interest because it may simplify the evaluation of an integral.

 

4.1 Theorem 4 - Green's Theorem in the Plane (Transformation between Double Integrals and Line Integrals

 

Let R be a closed bounded region in the xy-plane whose boundary C consists of finitely many smooth curves. Let Formulaand Formulabe functions taht are continuous and have continuous partial derivativesFormulaandFormula everywhere in some domain containing R. Then

                                Formula

 

Here we integrate along the entire boundary C of R in such a sense that R is on the left as we advance in the direction of integration.

 

Region R whose boundary C consists of two parts:Formula is traversed counterclockwise, while Formulais traversed clockwise in such a way that R is on the left for both curves.

 

Setting Formula and using the curl theorem in chapter 9, we have

                                                Formula

 

Example 9 -Verification of Green's Theorem in the plane Green's theorem in the plane will be quite important in our further work. Before proving it, let us get used to it by verifying it for Formula and C the circle Formula

 

Solution In (22) on the left we get

                    Formula

 

since the circular disk R area isFormula.

 

Now, we have to show that the line integral in (22) on the right gives the same value. We must orient C counterclockwise, say,

 

FormulaThen Formula,and on C

                          Formula

 

Hence the line integral in (22) becomes, verifying Green's theorem,

              Formula

 

 

4.2 Some Applications of Green's theorem

 

Example 10 - Area of a Plane Region as a Line Integral Over the Boundary In (22), we first chooseFormulaand then Formula.This gives

                               Formula

respectively. The double integral is the area A of R. By addition we have

 

                                                          Formula

 

where we integrate as indicated in Green's theorem. This interesting formula expresses the area of R in therms of line integral over the boundary. It is used, for instance, in the theory of certain planimeters.

 

For an ellipse Formula or Formula we get Formula, Formula. Thus from (25) we obtain the familiar formula for the area of region bounded by an ellipse.

 

                           Formula

 

Example 11 - Area if Plane Region in Polar Coordinates Let Formula and Formula be polar coordinates defined by Formula,Formula. Then

                                             Formula

and (25) becomes a formula that is well known from calculus, namely,

                                                              Formula

As an application of (26), we consider the cardioid Formula,where Formula We find

                                                       Formula

 

 

 

5 Surface for surface Integral

 

5.1 Representation of surfaces

Represebtatuibs if a syrface S in xyz-space are

                                                         Formula

For example, Formula represents a hemisphere of radius a and center 0.

 

Now, for curve C in line integrals, it was more practical and gave greater flexibility yo use a parametric representation Formula, where Formula This is a mapping of the interval Formula,located on the t-axis, onto the curve C. It maps every t in that interval onto the point of C with position vector Formula

 

Similarly, for surfaces S in surface integrals, it will often be more practical to use a parametric representation. Surfaces are two dimensional. Hence we need two parameters, which we call u and v. Thus a Formula of a surface S in space is of the form

            Formula

 

Example 12 -Parametric Representation of a Cylinder The circle cylinder Formula,Formula, has radius a, height 2 and the z-axis as axis. A parametric representation is

                              Formula

 

  • The components of Formula and Formula

 

  • The parametrics u,v vary in the rectangle R: Formula in the uv-plane.

 

  • The curves u=const are vertical straight lines.

 

  • The curves v=const are parallel circles. The point P in the following figure corresponds to Formula

 

Example 13 - Parametric Representation of a Sphere A sphere Formula can be represented in the form

                        Formula

where the parameters u, v vasry in the rectangle R in the uv-plane and give by the inequalities, Formula, Formula. The components of Formula are

                                           Formula

 

The curves u=const and v - const are the ``meridians'' and parallel on S. This representation is used in geography for measuring the latitude and longitudinal of points on the globe.

 

Another parametric representation of the sphere also used in mathematic is

                         Formula

whereFormula

 

 

 

Example 14- Parametric Representation of a Cone A circular coneFormula,Formula can be represented by

                                Formula

 

in componentsFormula. THe parameters vary in the rectangle R: Formula. Check that Formula, as it should be.

 

5.2 Tangent Plane and Surface Normal

  • Recall from chapter 9 that the tangent vectors of all the curves on a surface S through a point P oof S from a plane, called the tangent plane of S at P.

 

  • Excepteions are points where S has an edge or a cusp (like a cone),so that S cannot have a tangent plane at such a point.

 

  • Furthermore, a vector perpendicular to the tangent plane is called a normal vector of S at P.

 

  • Now, since S can be given by Formula in the parametric representation, the new idea is that we get a curve C on S by taking a pair of differentiable functions

                                                               Formula

 

  • whose derivatives u;, v; are continuous. Then C has the position vector Formula. By differentation and the use of the chain rule, we ontain a tangent vector of C on S

                                                           Formula

 

  • Hence, the partial derivatives Formula at P are tangential to S at P. We assume that they are linearly independent, which geometrically means that the curves u=const and v=const on S intersect at P at a nonzero angle. Then Formula span the tangent plane of S at P.

 

  • Hence their cross product gives a normal vector N of S at P

                                                         Formula

 

  • The corresponding unit normal vector n of S at P is

                                        Formula 

 

Also, if S is represented by g(x,y,z)=0, then the unit normal vector will be

                                                  Formula

 

A surface S is called a smooth surface if its surface normal depends continuously on the ponet of S.

 

Example 15 - Unit Normal Vector of a Sphere From (30) we find that the sphere Formula has the unit normal vector

                                             Formula

 

 

6 Surface Integral

  • For a given vector function F we can now define the surface integral over S by

                    Formula

  • Here Formula and Formulais the area of the parallelogram with sides Formula

 

  • By definition of cross product. Hence

                                          Formula 

  • And we see that dA=Formula is the element of area of S.

 

  • AlsoFormula is the normal component of Formula.This integral arises naturally in flow problem where it gives the flux(mass of fluid acrossing S per unit time) across S.

 

  • We can express (30) in components, using Formula and Formula

 

  • Here,Formula are the angles betweenFormulaand the coordinate are indeed, for the angel between Formula and Formula, as Formula and so on. Therefore,

                    Formula 

 

  • In (33) we can writeFormulaThen (33) becomes the following integral for the flux:

                      Formula

 

Example 16 - Surface Integral Evaluate (30) when Formula and S is the portion of the plane x+y+z=1 in the first octant as shown.

 

Solution Writing x=u and y=v, we have z=1-x-y=1-u-v.

 

Hence we can represent theplane x+y+z=1 in the form Formula. We obtain the first-octant portion S of this plane by restricting x=u and y=v to the projection R of S in the xy-plane.

 

R is the triangle bounded by the two coordinate axes and the straight line x+y=1, obtained from x+y+z=1 by setting z=0.

 

Thus, Formula

 

By differentiation

 

                         Formula 

 

HenceFormula. By (30),

                    Formula

 

Example 16 -Surface Integral of a cylinder (PS10.6 q4) Given Formula, and S: Formula

Find the surface integral.

 

Solution This is a circular cylinder of radius 3 and height 2 in the first octant with the z-axis as axis.

 

A parametric representation is

                         Formula

 

From this we have

 

                                                  Formula

                                             Formula

 

Integration over u from 0 to Formulagives Formula. Integration of this over v from 0 to 2 gives the answer:

 

                                   Formula

 

Exercise 17 - Surface Integral of a sphere (PS10.6 q 9) Given Formula,Formula

Find the surface integral.

 

 

 

 

 

 

7 Triple intergals.
     Divergence Theorem of gauss

 

7.1 Triple intergal

A triple integral is an integral of a function Formula taken over a closed bounded region T in space.It is denoted by

               Formula

 

Triple integerals can be evaluated by three successive integrations.

 

7.2 Divergence Theorem of gauss
               (Transformation between triple and surface integrals)

 

Let T be a closed bounded region in space whose boundary is a piecewise smooth orientable surface S. Let Formula be a vector function that is continuous and has continuous first partial derivatives in some domain containing T. Then

                                             Formula

 

In components of Formula and of the outer unit normal vector Formula of S, (35) becomes

                         Formula

 

Example 18 - Evalution of a surface integral by the Divergence Theorem Evaluate

                                   Formula

where S is the closed surface in (36) consisting of the cylinder Formula and the circular disks Formula and Formula

 

Solution.Formula, Formula, Formula Hence div Formula. The from of the surface suggests that we introduce polar coordinates Formula defined by Formula, Formula(thus cylindrical coordinates Formula). Then the volume element is Formula, and we obtain

 

                         Formula

 

Example 19 - Verification of the Divergence Theorem Evaluate Formula over the sphere Formula (a) by (35), (b) directly.

Solution.

  1. div Formula 

     Formula the answer is Formula.

 

     2. Wirte S in parameter form of Formula and using Formula,

                                   Formula

 

Then

                    Formula

 

Now on S we have Formula, Formula, so that Formula become on S

                                        Formula

 

and

               Formula

 

On S we have to integrate over u form 0 to Formula. This gives

                                                  Formula

 

Then integrating over v form Formula to Formula, we get

                                             Formula

 

 

 

7.3 Invariance of the Divergence

 

The divergence of a vector function Formula with continuous first partial derivatives in a region Formula is independent of the particular choice of Cartesian coordinates. For any point Formula in Formula it is given by

 

                              Formula

 

 

8 Stokes's Theorem
(Transformation between surface an line integrals)

 

Let S be a piecewise smooth oriented surface in space and let the boundary of S be a piecewise smooth simple close curve C. Let Formula be a countions vector function that has continuous first partial derivatices in a domain in space containing S. Then

                                   Formula

 

Here Formula is a unit normal vector of S and, depending on Formula, the integration around C is taken in the sense shown in (38). Furthermore, Formula is the unit tangent vector and s the arc length of C.

 

In components, (38) becomes

Formula

 

Here, Formula and R is the region with boundary curve Formula in the Formula-plane corresponding to S represented by Formula.

 

Example 20 - Verifying Stokes's Theorem for Formula and S the paraboloid

                    Formula

 

Solution. The curve C, oriented as in (), is the circle Formula. Its unit tangent vector is Formula.The function Formula on C is Formula.Hence

     Formula

 

 

We now consider the surface integral. We have Formula, so that in (38) we obtain

          Formula

 

A normal vector of S is Formula.Hece Formula Now FormulaUsing polar coordinates Formula defined by Formula Formula and denoting the projection of S into xy-plane by R, we thus obtain

 

     Formula

 

Note. Green's Theorem in the plane as a special case of Stokes's Theorem

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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