Surface Integral - Tích phân mặt

Gii tch hm nhiu binChng: TCH PHN MT

u Th Phit

Ngy 26 thng 4 nm 2014

u Th Phit () Gii tch hm nhiu bin Chng: TCH PHN MTNgy 26 thng 4 nm 2014 1 / 38

Ni dung

1 Mt cong

2 Tch phn mt loi mt

3 Tch phn mt loi hai


Mt cong

Mt cong


Mt cong

nh ngha

Tng t biu din ng cong bng hm vector r(t) vi mt tham s t,ta biu din mt cong l hm vector r(u, v) vi hai tham s u, v

r(u, v) = x(u, v)~i + y(u, v)~j + z(u, v)~k

y l hm vector xc nh trn min D trong mt phng Ouv . x , y , z lhm theo tham s (u, v) trn min D.

1070 | | | | CHAPTER 16 VECTOR CALCULUS

PARAMETRIC SURFACES AND THEIR AREAS

So far we have considered special types of surfaces: cylinders, quadric surfaces, graphs offunctions of two variables, and level surfaces of functions of three variables. Here we usevector functions to describe more general surfaces, called parametric surfaces, and com-pute their areas. Then we take the general surface area formula and see how it applies tospecial surfaces.

PARAMETRIC SURFACES

In much the same way that we describe a space curve by a vector function of a singleparameter , we can describe a surface by a vector function of two parameters and. We suppose that

is a vector-valued function defined on a region in the -plane. So x, y, and , the com-ponent functions of r, are functions of the two variables u and with domain D. The setof all points in such that

and varies throughout , is called a parametric surface and Equations 2 are calledparametric equations of . Each choice of u and gives a point on S; by making all choices, we get all of S. In other words, the surface is traced out by the tip of the posi-tion vector as moves throughout the region . (See Figure 1.)

EXAMPLE 1 Identify and sketch the surface with vector equation

SOLUTION The parametric equations for this surface are

So for any point on the surface, we have

This means that vertical cross-sections parallel to the -plane (that is, with y constant)are all circles with radius 2. Since and no restriction is placed on , the surface is acircular cylinder with radius 2 whose axis is the y-axis. (See Figure 2.) M

vy ! vxz

x 2 ! z2 ! 4 cos2u ! 4 sin2u ! 4

!x, y, z"

z ! 2 sin uy ! vx ! 2 cos u

r!u, v" ! 2 cos u i ! v j ! 2 sin u k

0

z

x y

S

r(u,)0

u

D (u,)r

FIGURE 1A parametric surface

D!u, v"r!u, v"SvS

SD!u, v"

z ! z!u, v"y ! y!u, v"x ! x!u, v"2

! 3!x, y, z"v

zuvD

r!u, v" ! x!u, v" i ! y!u, v" j ! z!u, v" k1

vur!u, v"t

r!t"

16.6

FIGURE 2

0

(0,0,2)

(2,0,0)

x y

z


Mt cong

V d

1) Mt cho bi

r(u, v) = 2 cos u~i + v~j + 2 sin uk

Ta c x = 2 cos u

y = v

z = 2 sin u

x2 + z2 = 4.


PARAMETRIC SURFACES AND THEIR AREAS

So far we have considered special types of surfaces: cylinders, quadric surfaces, graphs offunctions of two variables, and level surfaces of functions of three variables. Here we usevector functions to describe more general surfaces, called parametric surfaces, and com-pute their areas. Then we take the general surface area formula and see how it applies tospecial surfaces.

PARAMETRIC SURFACES

In much the same way that we describe a space curve by a vector function of a singleparameter , we can describe a surface by a vector function of two parameters and. We suppose that

is a vector-valued function defined on a region in the -plane. So x, y, and , the com-ponent functions of r, are functions of the two variables u and with domain D. The setof all points in such that

and varies throughout , is called a parametric surface and Equations 2 are calledparametric equations of . Each choice of u and gives a point on S; by making all choices, we get all of S. In other words, the surface is traced out by the tip of the posi-tion vector as moves throughout the region . (See Figure 1.)

EXAMPLE 1 Identify and sketch the surface with vector equation

SOLUTION The parametric equations for this surface are

So for any point on the surface, we have

This means that vertical cross-sections parallel to the -plane (that is, with y constant)are all circles with radius 2. Since and no restriction is placed on , the surface is acircular cylinder with radius 2 whose axis is the y-axis. (See Figure 2.) M

vy ! vxz

x 2 ! z2 ! 4 cos2u ! 4 sin2u ! 4

!x, y, z"

z ! 2 sin uy ! vx ! 2 cos u

r!u, v" ! 2 cos u i ! v j ! 2 sin u k

0

z

x y

S

r(u,)0

u

D (u,)r

FIGURE 1A parametric surface

D!u, v"r!u, v"SvS

SD!u, v"

z ! z!u, v"y ! y!u, v"x ! x!u, v"2

! 3!x, y, z"v

zuvD

r!u, v" ! x!u, v" i ! y!u, v" j ! z!u, v" k1

vur!u, v"t

r!t"

16.6

FIGURE 2

0

(0,0,2)

(2,0,0)

x y

z


Mt cong

V d

2) Mt cho bi

r(u, v) = (2 + sin v) cos u, (2 + sin v) sin u, u + cos v

vi 0 u 4, 0 v 2

In Example 1 we placed no restrictions on the parameters and and so we obtainedthe entire cylinder. If, for instance, we restrict u and by writing the parameter domain as

then and we get the quarter-cylinder with length 3 illustrated inFigure 3.

If a parametric surface S is given by a vector function , then there are two usefulfamilies of curves that lie on S, one family with u constant and the other with constant.These families correspond to vertical and horizontal lines in the -plane. If we keep constant by putting becomes a vector function of the single parame-ter and defines a curve lying on . (See Figure 4.)

Similarly, if we keep constant by putting given by that lies on . We call these curves grid curves. (In Example 1, for instance, the grid curves obtained by letting u be constant are horizontal lines whereas the grid curves with

constant are circles.) In fact, when a computer graphs a parametric surface, it usuallydepicts the surface by plotting these grid curves, as we see in the following example.

EXAMPLE 2 Use a computer algebra system to graph the surface

Which grid curves have u constant? Which have constant?

SOLUTION We graph the portion of the surface with parameter domain in Figure 5. It has the appearance of a spiral tube. To identify the grid

curves, we write the corresponding parametric equations:

If is constant, then and are constant, so the parametric equations resemblethose of the helix in Example 4 in Section 13.1. So the grid curves with constant arethe spiral curves in Figure 5. We deduce that the grid curves with u constant must be thecurves that look like circles in the figure. Further evidence for this assertion is that if uis kept constant, , then the equation shows that the -values varyfrom to . M

In Examples 1 and 2 we were given a vector equation and asked to graph the corre-sponding parametric surface. In the following examples, however, we are given the morechallenging problem of finding a vector function to represent a given surface. In the rest ofthis chapter we will often need to do exactly that.

u0 ! 1u0 " 1zz ! u0 ! cos vu ! u0

vcos vsin vv

z ! u ! cos vy ! !2 ! sin v" sin ux ! !2 ! sin v" cos u

0 # v # 2$0 # u # 4$,

v

r!u, v" ! #!2 ! sin v" cos u, !2 ! sin v" sin u, u ! cos v $

v

Sr!u, v0 "v ! v0, we get a curve C2v

FIGURE 4

r

0

z

yx

C

C0

D

=(u,)

u=u

u

SC1vu ! u0, then r!u0, v"

uuvv

r!u, v"

x % 0, z % 0, 0 # y # 3,

0 # v # 30 # u # $%2

vvu

SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS | | | | 1071

FIGURE 3

0

(0,3,2)

x y

z

Visual 16.6 shows animated versions of Figures 4 and 5, with moving grid curves,for several parametric surfaces.

TEC

z

yx

u constant

constant

FIGURE 5


Mt cong

V d

3) Tm phng trnh tham s cho mt cu

x2 + y2 + z2 = a2

S dng to cu vi = a = hng s , ta c

x = a sin cos, y = a sin sin, z = a cos

Trong : l gc to bi vector (x , y , z) v trcOz , l gc to bi Ox v vector (x , y , 0).Vy

r(, ) = a sin cos~i + a sin sin~j + a cos~k

vi 0 v 0 2.

EXAMPLE 3 Find a vector function that represents the plane that passes through the point with position vector and that contains two nonparallel vectors a and b.

SOLUTION If P is any point in the plane, we can get from to by moving a certain distance in the direction of and another distance in the direction of . So there arescalars u and such that A . (Figure 6 illustrates how this works, by means of the Parallelogram Law, for the case where and are positive. See also Exercise 40 in Section 12.2.) If r is the position vector of P, then

A ASo the vector equation of the plane can be written as

where u and are real numbers.If we write , , , and ,

then we can write the parametric equations of the plane through the point asfollows:

M

EXAMPLE 4 Find a parametric representation of the sphere

SOLUTION The sphere has a simple representation in spherical coordinates, so letschoose the angles and in spherical coordinates as the parameters (see Section 15.8).Then, putting in the equations for conversion from spherical to rectangular coordi-nates (Equations 15.8.1), we obtain

as the parametric equations of the sphere. The corresponding vector equation is

We have and , so the parameter domain is the rectangle. The grid curves with constant are the circles of constant lati-

tude (including the equator). The grid curves with constant are the meridians (semi-circles), which connect the north and south poles. M

FIGURE 8FIGURE 7

!"D ! !0, #" $ !0, 2#"

0 % ! % 2#0 % " % #

r#", !$ ! a sin " cos ! i & a sin " sin ! j & a cos " k

z ! a cos "y ! a sin " sin !x ! a sin " cos !

' ! a!"

' ! a

x 2 & y 2 & z2 ! a 2

V

z ! z0 & ua3 & vb3y ! y0 & ua2 & vb2x ! x0 & ua1 & vb1

#x0, y0, z0 $b ! %b1, b2, b3 &a ! %a1, a2, a3 &r0 ! %x0, y0, z0 &r ! %x, y, z&

v

r#u, v$ ! r0 & ua & vb

! r0 & ua & vbP0P&r ! OP0

vu! ua & vbP0Pv

baPP0

r0P0


P

uaP

b

a

b

FIGURE 6

N One of the uses of parametric surfaces is incomputer graphics. Figure 7 shows the result oftrying to graph the sphere by solving the equation for and graphing thetop and bottom hemispheres separately. Part ofthe sphere appears to be missing because of therectangular grid system used by the computer.The much better picture in Figure 8 was pro-duced by a computer using the parametric equations found in Example 4.

zx 2 & y 2 & z2 ! 1


Mt cong

V d

3) Tm phng trnh tham s cho mt ellipticparaboloid

z = x2 + 2y2

Ta xem (x , y) nh tham s, khi ta c

x = x , y = y , z = x2 + 2y2

hayr(x , y) = x~i + y~j + (x2 + 2y2)~k

EXAMPLE 3 Find a vector function that represents the plane that passes through the point with position vector and that contains two nonparallel vectors a and b.

SOLUTION If P is any point in the plane, we can get from to by moving a certain distance in the direction of and another distance in the direction of . So there arescalars u and such that A . (Figure 6 illustrates how this works, by means of the Parallelogram Law, for the case where and are positive. See also Exercise 40 in Section 12.2.) If r is the position vector of P, then

A ASo the vector equation of the plane can be written as

where u and are real numbers.If we write , , , and ,

then we can write the parametric equations of the plane through the point asfollows:

M

EXAMPLE 4 Find a parametric representation of the sphere

SOLUTION The sphere has a simple representation in spherical coordinates, so letschoose the angles and in spherical coordinates as the parameters (see Section 15.8).Then, putting in the equations for conversion from spherical to rectangular coordi-nates (Equations 15.8.1), we obtain

as the parametric equations of the sphere. The corresponding vector equation is

We have and , so the parameter domain is the rectangle. The grid curves with constant are the circles of constant lati-

tude (including the equator). The grid curves with constant are the meridians (semi-circles), which connect the north and south poles. M

FIGURE 8FIGURE 7

!"D ! !0, #" $ !0, 2#"

0 % ! % 2#0 % " % #

r#", !$ ! a sin " cos ! i & a sin " sin ! j & a cos " k

z ! a cos "y ! a sin " sin !x ! a sin " cos !

' ! a!"

' ! a

x 2 & y 2 & z2 ! a 2

V

z ! z0 & ua3 & vb3y ! y0 & ua2 & vb2x ! x0 & ua1 & vb1

#x0, y0, z0 $b ! %b1, b2, b3 &a ! %a1, a2, a3 &r0 ! %x0, y0, z0 &r ! %x, y, z&

v

r#u, v$ ! r0 & ua & vb

! r0 & ua & vbP0P&r ! OP0

vu! ua & vbP0Pv

baPP0

r0P0


P

uaP

b

a

b

FIGURE 6

N One of the uses of parametric surfaces is incomputer graphics. Figure 7 shows the result oftrying to graph the sphere by solving the equation for and graphing thetop and bottom hemispheres separately. Part ofthe sphere appears to be missing because of therectangular grid system used by the computer.The much better picture in Figure 8 was pro-duced by a computer using the parametric equations found in Example 4.

zx 2 & y 2 & z2 ! 1


Mt cong

Din tch mt cong

nh ngha

Nu mt cong S cho bi phng trnh

r(u, v) = x(u, v)~i + y(u, v)~j + z(u, v)~k (u, v) D;

Khi din tch mt D cho bi

A(S) =

D|ru rv | dA

vi ru =x

u~i +

y

u~j +

z

u~k , v rv =

x

v~i +

y

v~j +

z

v~k


Mt cong

Din tch mt ca th hm s

Cho mt S xc nh bi phng trnh z = f (x , y) vi (x , y) D v c cco hm ring. Xem x v y l tham s, ta c phng trnh tham s

x = x y = y z = f (x , y)

do ta c

rx =~i +

(f

x

)~k ry =~j +

(f

x

)~k

rx ry =

~i ~j ~k

1 0f

x

0 1f

y

= f

x~i f

y~j + ~k

A(S) =

D

1 +

(z

x

)2+

(z

y

)2dA


Tch phn mt loi mt

Tch phn mt loi mt


Tch phn mt loi mt

Tch phn mt loi mt

Cho mt S bi phng trnh vector

r(u, v) = x(u, v)~i + y(u, v)~j + z(u, v)~k (u, v) D

Gi s min D l hnh ch nht, ta chia thnh cchnh ch nht con Rij vi cc cnh tng ng lu, v . Tng ng, mt S c chia thnh ccmt Sij .Tnh gi tr f (Pij ) ti im bt k trong Sij v lp

tng Riemann

mj=1

ni=1

f (Pij )Sij

SECTION 16.7 SURFACE INTEGRALS | | | | 1081

0

(x, y, z)

(b, 0, 0)

z

x

y

59. Find the area of the part of the sphere thatlies inside the cylinder .

60. (a) Find a parametric representation for the torus obtained by rotating about the -axis the circle in the -plane withcenter and radius . [Hint: Take as parame-ters the angles and shown in the figure.]

; (b) Use the parametric equations found in part (a) to graphthe torus for several values of a and b.

(c) Use the parametric representation from part (a) to find thesurface area of the torus.

!"a # b!b, 0, 0"

xzz

x 2 $ y 2 ! axx 2 $ y 2 $ z2 ! a 2

SURFACE INTEGRALS

The relationship between surface integrals and surface area is much the same as the rela-tionship between line integrals and arc length. Suppose is a function of three variableswhose domain includes a surface . We will define the surface integral of over in sucha way that, in the case where , the value of the surface integral is equal to thesurface area of . We start with parametric surfaces and then deal with the special casewhere is the graph of a function of two variables.

PARAMETRIC SURFACES

Suppose that a surface has a vector equation

We first assume that the parameter domain is a rectangle and we divide it into subrect-angles with dimensions and . Then the surface is divided into correspondingpatches as in Figure 1. We evaluate at a point in each patch, multiply by the area

of the patch, and form the Riemann sum

Then we take the limit as the number of patches increases and define the surface integralof f over the surface S as

Notice the analogy with the definition of a line integral (16.2.2) and also the analogy withthe definition of a double integral (15.1.5).

To evaluate the surface integral in Equation 1 we approximate the patch area by thearea of an approximating parallelogram in the tangent plane. In our discussion of surfacearea in Section 16.6 we made the approximation

%Sij # $ ru & rv $%u %v

%Sij

yyS

f !x, y, z" dS ! lim m, n l'

%m

i!1 %

n

j!1 f !Pij*"%Sij1

%m

i!1 %

n

j!1 f !Pij*"%Sij

%SijPij*fSij

S%v%uRijD

!u, v" ! Dr!u, v" ! x!u, v" i $ y!u, v" j $ z!u, v" k

S

SS

f !x, y, z" ! 1SfS

f

16.7

FIGURE 1

0

u

Rij

u

0

z

yx

P*ijSSij

D

r


Tch phn mt loi mt

Khi gii hn i vi mt th hm s, gi s S cho bi z = z(x , y)

x = x y = y z = z(x , y)

th ta c

|rx ry | =zx~i zy~j + k

=(

z

x

)2+

(z

y

)2+ 1

Khi S

f (x , y , z)dS =

Dxy

f (x , y , z(x , y))

1 +

(z

x

)2+

(z

y

)2dA

Vi Dxy l hnh chiu ca mt S ln mt phng Oxy .


Tch phn mt loi mt

Tng t, nu x = x(y , z), ta chiu mt S xung mt phng Oyz v ccng thc

S

f (x , y , z)dS =

Dyz

f (x(y , z), y , z)

1 +

(x

y

)2+

(x

z

)2dA

Nu y = y(x , z), ta chiu mt S xung mt phng Oxz v c cng thc

S

f (x , y , z)dS =

Dxz

f (x , y(x , z), z)

1 +

(y

x

)2+

(y

z

)2dA

CH . Nu hnh chiu ca S xung mt phng Oxy l 1 ng cong(khi S l mt tr song song Oz) th ta phi chiu S xung cc mtphng to khc.


Tch phn mt loi mt

V d

Tnh

S ydS vi S l mt cho bi z = x + y2, 0 x 1, 0 y 2.

Ta cz

x= 1

z

y= 2y

S

ydS =

D

y

1 +

(z

x

)2+

(z

x

)2dA

=

10

dx

20

y

1 + 1 + 4y2dy

=

10

dx

2

20

y

1 + 2y2dy

=

2 14 2

3(1 + 2y2)3/2

20

=13

2

3

and the density (mass per unit area) at the point is , then the total massof the sheet is

and the center of mass is , where

Moments of inertia can also be defined as before (see Exercise 39).

GRAPHS

Any surface with equation can be regarded as a parametric surface with para-metric equations

and so we have

Thus

and

Therefore, in this case, Formula 2 becomes

Similar formulas apply when it is more convenient to project onto the -plane or -plane. For instance, if is a surface with equation and is its projection on

the -plane, then

EXAMPLE 2 Evaluate , where is the surface , , .(See Figure 2.)

SOLUTION Since!z!y

! 2yand!z!x

! 1

0 " y " 20 " x " 1z ! x # y 2SxxS y dS

yyS

f !x, y, z" dS ! yyD

f (x, h!x, z", z)#$!y!x%2 # $!y!z%2 # 1 dAxz

Dy ! h!x, z"SxzyzS

yyS

f !x, y, z" dS ! yyD

f (x, y, t!x, y")#$ !z!x%2 # $ !z!y%2 # 1 dA4

& rx $ ry & !#$ !z!x%2 # $ !z!y%2 # 1rx $ ry ! %

!t!x

i %!t!y

j # k3

ry ! j # $ !t!y% krx ! i # $ !t!x% kz ! t!x, y"y ! yx ! x

z ! t!x, y"S

z !1m

yyS

z &!x, y, z" dSy !1m

yyS

y &!x, y, z" dSx !1m

yyS

x &!x, y, z" dS

!x, y, z"

m ! yyS

&!x, y, z" dS

&!x, y, z"!x, y, z"S


FIGURE 2

y

x

z


Tch phn mt loi mt

Nu S l mt trn tng phn, tc l hp ca hu hn mt trnS1,S2, . . . ,Sn v cc mt ny ch ct nhau ti bin ca chng, khi tchphn mt ca hm f trn S cho bi

S

f (x , y , z)dS =

S1

f (x , y , z)dS + . . .+

Sn

f (x , y , z)dS


Tch phn mt loi mt

V d

Tnh

S zdS vi S l mt ca vt th gii hn bi tr x2 + y2 = 1,

z = 0 v z = 1 + x .

T hnh, ta thy mt S l hp bi cc mtS1, S2, S3.Vi mt S1 ta s dng to tr vi bnknh r = 1

x = cos, y = sin, z = z

vi 0 2 v0 z 1 + x = 1 + cos

Formula 4 gives

M

If is a piecewise-smooth surface, that is, a finite union of smooth surfaces , that intersect only along their boundaries, then the surface integral of over is defined

by

EXAMPLE 3 Evaluate , where is the surface whose sides are given by thecylinder , whose bottom is the disk in the plane , andwhose top is the part of the plane that lies above .

SOLUTION The surface is shown in Figure 3. (We have changed the usual position of the axes to get a better look at .) For we use and as parameters (see Example 5 in Section 16.6) and write its parametric equations as

where

Therefore

and

Thus the surface integral over is

! 12 [ 32! " 2 sin ! " 14 sin 2!]02#! 3#2

! 12 y2#0

!1 " 2 cos ! " 12 "1 " cos 2!#$ d!

! y2#0

y1"cos !

0 z dz d! ! y2#

0 12 "1 " cos !#2 d!

yyS1

z dS ! yyD

z % r! $ rz %dAS1

% r! $ rz % ! scos 2! " sin2! ! 1

r! $ rz ! % i%sin !0 jcos !0 k01 % ! cos ! i " sin ! j0 & z & 1 " x ! 1 " cos !and0 & ! & 2#

z ! zy ! sin !x ! cos !

z!S1SS

S2z ! 1 " xS3z ! 0x 2 " y 2 & 1S2x 2 " y 2 ! 1

S1SxxS z dSV

yyS

f "x, y, z# dS ! yyS1

f "x, y, z# dS " ' ' ' " yySn

f "x, y, z# dS

SfSnS2, . . . ,S1S

! s2 ( 14 ) 23 "1 " 2y 2 #3&2]02 ! 13s2

3

! y10 dx s2 y2

0 ys1 " 2y 2 dy

! y10 y2

0 ys1 " 1 " 4y 2 dy dx

yyS

y dS ! yyD

y'1 " ( (z(x)2 " ( (z(y)2 dA


FIGURE 3

0

S(+=1)

S

S (z=1+x)

x

z

y


Tch phn mt loi mt

r rz =

~i ~j ~k

sin cos 00 0 1

= cos~i + sin~j |r rz | = 1Tch phn trn mt S1

S1

zdS =

D

z |r rz |dA

=

20

d

1+cos0

zdz =

20

1

2(1 + cos)2d

=1

2

20

[1 + 2 cos+

1

2(1 + cos 2)

]d

=1

2

[3

2

]20

=3

2


Tch phn mt loi mt

Trn mt S2, z = 0 nnS2

zdS =S2

0dS = 0.

Chiu mt S3 xung mt phng Oxy , ta c hnh trn n v D.S3

zdS =

D

(1 + x)

1 +

(z

x

)2+

(z

y

)2dA

=

20

10

(1 + cos)

1 + 1 + 0rdrd

=

2

20

10

(r + r2 cos)drd

=

22

10

rdr =

2

Vy SzdS =

S1

zdS +

S2

zdS +

S3

zdS =

(3

2+

2

)


Tch phn mt loi mt

Bi tp

1 Tnh tch phn

xyzdS trong S l phn mt phng

x + y + z = 1 nm trong gc 1/8 th nht.

2 Tnh tch phnS

1

(1 + x + y)2dS trong S l mt xung quanh ca

t din x + y + z = 1, x = 0, y = 0, z = 0.

3 Tnh tch phnS

|xyz |dS trong S l phn mt z = x2 + y2 nm

gia 2 mt phng z = 0, z = 1.

4 Tnh tch phnS

x

x2 + y2dS trong S l mt cu

x2 + y2 + z2 = R2 trong gc 1/8 th nht.

5 Tnh tch phnS

(xy + yz + zx)dS trong S l phn ca mt nn

z =

x2 + y2 b ct bi mt tr x2 + y2 = 2ax


Tch phn mt loi hai

Tch phn mt loi hai


Tch phn mt loi hai

nh ngha

Cho trng vectorF = (P(x , y , z),Q(x , y , z),R(x , y , z)) xcnh trn mt nh hng S .Gi ~n l vector php tuyn n v ca mt S .Khi tch phn mt loi hai cho bi

S

F dS =S

F ~ndA

SURFACE INTEGRALS OF VECTOR FIELDS

Suppose that is an oriented surface with unit normal vector , and imagine a fluid withdensity and velocity field flowing through . (Think of as an imagi-nary surface that doesnt impede the fluid flow, like a fishing net across a stream.) Then therate of flow (mass per unit time) per unit area is . If we divide into small patches ,as in Figure 10 (compare with Figure 1), then is nearly planar and so we can approxi-mate the mass of fluid crossing in the direction of the normal per unit time by thequantity

where , , and are evaluated at some point on . (Recall that the component of the vec-tor in the direction of the unit vector is .) By summing these quantities and tak-ing the limit we get, according to Definition 1, the surface integral of the function over :

and this is interpreted physically as the rate of flow through .If we write , then is also a vector field on and the integral in Equation 7

becomes

A surface integral of this form occurs frequently in physics, even when is not , and iscalled the surface integral (or flux integral ) of over .

DEFINITION If is a continuous vector field defined on an oriented surface with unit normal vector , then the surface integral of over S is

This integral is also called the flux of across .

In words, Definition 8 says that the surface integral of a vector field over is equal tothe surface integral of its normal component over (as previously defined).

If is given by a vector function , then is given by Equation 6, and from Def-inition 8 and Equation 2 we have

where is the parameter domain. Thus we have

yyS

F ! dS ! yyD

F ! !ru ! rv " dA9

D

! yyD

#F!r!u, v"" ! ru ! rv$ ru ! rv $ %$ ru ! rv $ dA yyS

F ! dS ! yyS

F !ru ! rv

$ ru ! rv $ dS

nr!u, v"SS

S

SF

yyS

F ! dS ! yyS

F ! n dS

FnSF8

SF"vF

yyS

F ! n dS

! 3FF ! "vS

yyS

"v ! n dS ! yyS

"!x, y, z"v!x, y, z" ! n!x, y, z" dS7

S"v ! n

"v ! nn"vSijnv"

!"v ! n"A!Sij"

nSijSij

SijS"v

SSv!x, y, z""!x, y, z"nS


0

y

z

x

nF=v

SSij

FIGURE 10

N Compare Equation 9 to the similar expres-sion for evaluating line integrals of vector fieldsin Definition 16.2.13:

yC F ! dr ! yb

a F!r!t"" ! r#!t" dt


Tch phn mt loi hai

tnh tch phn mt loi hai, ta c th a v tch phn mt loi mt.Tuy nhin trong mt s trng hp vector php tuyn n v phc tp, do ta tnh theo cch sau

I =

S

Pdydz + Qdzdx + Rdxdy

=

S

Pdydz +

S

Qdzdx +

S

Rdxdy

= I1 + I2 + I3


Tch phn mt loi hai

Tnh I3

Gi Dxy l hnh chiu ca S xung mt phng Oxy . Khi S

R(x , y , z)dxdy = Dxy

R(x , y , z(x , y))dxdy

Trong

1 Du "+" nu php vector to vi chiu dng ca Oz mt gc nhn.

2 Du "-" nu php vector to vi chiu dng ca Oz mt gc t.

3 Nu php vector vung gc vi tia Oz , tch phn I3 = 0.

Ch . Nu hnh chiu Dxy l mt ng cong th tch phn I3 = 0.


Tch phn mt loi hai

V d

Tnh I =S

zdxdy vi S l pha ngoi mt cu x2 + y2 + z2 = R2.

Gi S+ v S l hai na mt cu tng ng vi z 0 v z 0.Trn S+ ta c z =

R2 x2 y2

I1 =

S+

zdxdz =

D

R2 z2 y2dxdy

vi D : x2 + y2 R2.Trn S ta c z =

R2 x2 y2

I1 =

S

zdxdz = D

R2 z2 y2dxdy

Do

I = 2

D

R2 x2 y2dxdy = 2

20

d

R0

r

R2 r2dr = 43R3


Tch phn mt loi hai

V d

TnhS

x2dydz + y2dxdz + z2dxdy vi S l pha ngoi na mt cu

x2 + y2 + z2 = R2, z 0.

Ta c

I3 =

S

z2dxdy =

x2+y2R2

(R2 x2 y2)dxdy

=

20

d

R0

(R2 r2)rdr = R4

2


Tch phn mt loi hai

Ta c

I2 =

S

y2dzdx =

S1

y2dzdx +

S2

y2dzdx

vi S1 tng ng vi y 0 v S2 tng ng viy 0. (Tch phn trn na hnh trn v ngc du)

S1

y2dzdx = S2

y2dzdx I2 = 0

S1

y2dzdx =

0

R0

(R2 r2)rdr = R4

4

Tng t I1 = 0.


Tch phn mt loi hai

Bi tp

1 TnhS

zdxdy trong S l mt xung quanh, pha ngoi ca mt

gii hn bi z = x2 + y2, z = 0, z = 12 Tnh

S

yzdydzxzdzdx + xydxdy trong S l mt ngoi ca t din

cho bi x + y + z a, x 0, z 0, z 0.3 Tnh

S

y2zdydzxzdydz + x2ydxdy trong S l mt ngoi ca vt

th gii hn bi z = x2 + y2, x2 + y2 = 1 nm trong gc 1/8 thnht.

4 Tnh tch phnS

x2zdydz + ydzdx + 2dxdy trong S mt pha

ngoi ca mt 4x2 + y2 + 4z2 = 4 nm trong gc x 0, y 0,z 0.

5 Tnh tch phnS

zdxdy trong S mt pha ngoi ca Ellipsoid

x2

a2+ y

2

b2+ z

2

c2= 1.


Tch phn mt loi hai

Cng thc Gauss - Ostrogratxki

Cho S l mt mt kn vi chiu hng ra ngoi, V v vt th bao quanhbi S . Nu cc hm P(x , y , z), Q(x , y , z), R(x , y , z) v cc o hm ringca n lin tc trn min V th

S

Pdydz + Qdzdx + Rdxdy =

V

(P

x+Q

y+R

z

)dxdydz

Nh vy, ta chuyn tch phn mt loi hai thnh mt tch phn bi ba.


Tch phn mt loi hai

V d.

V d 1.Tnh

S

(y z)dydz + (z y)dzdx + (z x)dxdy vi S l mt ngoi

hnh lp phng 1 x 1, 1 y 1, 1 z 1.

V d 2.Tnh tch phn

S

ydydz +xydzdx zdxdy vi S l mt bin pha trong

ca min gii hn bi x2 + y2leq4, 0 z x2 + y2.


Tch phn mt loi hai

nh l Stokes

Cho F = P~i + Q~j + R~k l mt trng vector trong R3. P,Q,R v cc ohm ring tn ti. Khi ta nh ngha curl ca F l mt trng vectortrn R3

curlF =

(R

y Qz

)~i +

(P

z Rx

)~j +

(Q

x Py

)~k

d nh, ta c

curlF =

~i ~j ~k

x

y

zP Q R

=

(R

y Qz

)~i +

(P

z Rx

)~j +

(Q

x Py

)~k


Tch phn mt loi hai

nh l Stokes c xem nh nh l Green trongkhng gian nhiu chiu.Trong nh l Green, ta ch ra mi lin h gi tchphn kp trn min phng D v tch phn ngtrn ng bin.nh l Stokes cho ta mi lin h gia tch phnmt v tch phn ng trn ng bin ca mtcong S .

an oriented surface with unit normal vector . The orientation of induces the positiveorientation of the boundary curve C shown in the figure. This means that if you walk inthe positive direction around with your head pointing in the direction of , then the sur-face will always be on your left.

STOKES THEOREM Let be an oriented piecewise-smooth surface that is boundedby a simple, closed, piecewise-smooth boundary curve with positive orientation.Let be a vector field whose components have continuous partial derivatives on an open region in that contains . Then

Since

Stokes Theorem says that the line integral around the boundary curve of of the tangen-tial component of is equal to the surface integral of the normal component of the curl of .

The positively oriented boundary curve of the oriented surface is often written as , so Stokes Theorem can be expressed as

There is an analogy among Stokes Theorem, Greens Theorem, and the Fundamental Theorem of Calculus. As before, there is an integral involving derivatives on the left sideof Equation 1 (recall that is a sort of derivative of ) and the right side involves thevalues of only on the boundary of .

In fact, in the special case where the surface is flat and lies in the -plane withupward orientation, the unit normal is , the surface integral becomes a double integral,and Stokes Theorem becomes

This is precisely the vector form of Greens Theorem given in Equation 16.5.12. Thus wesee that Greens Theorem is really a special case of Stokes Theorem.

Although Stokes Theorem is too difficult for us to prove in its full generality, we cangive a proof when is a graph and , , and are well behaved.

PROOF OF A SPECIAL CASE OF STOKES THEOREM We assume that the equation of is , where has continuous second-order partial derivatives and is a simple

plane region whose boundary curve corresponds to . If the orientation of isupward, then the positive orientation of corresponds to the positive orientation of .(See Figure 2.) We are also given that , where the partial deriva-tives of , , and are continuous.RQP

F ! P i ! Q j ! R kC1C

SCC1Dt!x, y" ! D

z ! t!x, y",S

CSFS

yC F " dr ! yy

S

curl F " dS ! yyS

!curl F" " k dA

kxyS

SFFcurl F

yyS

curl F " dS ! y"S

F " dr1

"SS

FF

S

yyS

curl F " dS ! yyS

curl F " n dSandyC F " dr ! y

C F " T ds

yC F " dr ! yy

S

curl F " dS

S! 3F

CS

nC

Sn

SECTION 16.8 STOKES THEOREM | | | | 1093

S

y

z

x

C0

n

n

FIGURE 1

N Stokes Theorem is named after the Irishmathematical physicist Sir George Stokes(18191903). Stokes was a professor at Cam-bridge University (in fact he held the same position as Newton, Lucasian Professor of Mathematics) and was especially noted for hisstudies of fluid flow and light. What we callStokes Theorem was actually discovered by the Scottish physicist Sir William Thomson(18241907, known as Lord Kelvin). Stokeslearned of this theorem in a letter from Thomsonin 1850 and asked students to prove it on anexamination at Cambridge University in 1854.We dont know if any of those students was able to do so.

FIGURE 2

0

D

CS

z=g(x,y)

C

n

y

z

x

Gi ~n l vector php tuyn n v trn mt S . Chiu dng ca ngcong bin C l chiu gp ca bn tay phi khi ngn tay ci cng chiuvi ~n.


Tch phn mt loi hai

nh l.

nh l

Cho S l mt cong nh hng trn tng mng vi bin l mt ngcong n, ng, trn tng khc C theo chiu dng. Cho F l trngvector vi cc thnh phn c cc o hm ring lin tc trn min trongR3 cha S . Khi

C

F dr =S

curlF dS =

~i ~j ~k

x

y

zP Q R

dS=

(R

y Qz

)dydz +

(P

z Rx

)dzdx +

(Q

x Py

)dxdy


Tch phn mt loi hai

V d

Tnh tch phnC

x2y3dx +dy +zdz vi C l giao tuyn ca x2 +y2 = 4

vi mt z = 1 theo chiu ngc chiu kim ng h theo hng dngca trc Oz

Ta c P = x2y3,Q = 1,R = z . Theo cng thcStokes

C

x2y3dx + 1dy + zdz

=

S

0dydz + 0dxdz 3x2y2dxdy

=

20

203r4 cos2 sin2 rdr = 8


Tch phn mt loi hai

Bi tp 1.

1 TnhS

zdxdy trong S l mt xung quanh, pha ngoi ca mt

gii hn bi z = x2 + y2, z = 0, z = 12 Tnh

S

yzdydz + xzdzdx + xydxdy trong S l mt ngoi ca t

din cho bi x + y + z a, x 0, z 0, z 0.3 Tnh

S

y2zdzdx + xzdydz + x2ydxdy trong S l mt ngoi ca

vt th gii hn bi z = x2 + y2, x2 + y2 = 1 nm trong gc 1/8 thnht.

4 Tnh tch phnS

x2zdydz + ydzdx + 2dxdy trong S mt pha

ngoi ca mt 4x2 + y2 + 4z2 = 4 nm trong gc x 0, y 0,z 0.

5 Tnh tch phnS

zdxdy trong S mt pha ngoi ca Ellipsoid

x2

a2+ y

2

b2+ z

2

c2= 1.


Tch phn mt loi hai

Bi tp 2.

1 Tnh

S z2dydz + xdzdx zdxdy vi S l mt xung quanh hng

bn ngoi ca vt th gii hn bi z = 4 y2, z = 0, x = 1, x = 0.2 Tnh

S(y + z)dydz + (x z)dzdx + (z + 1)dxdy vi S l mt pha

trn ca na mt cu x2 + y2 + z2 = R2, z 0.3 Tnh

C

y2dx + z2dy + x2dz trong C l chu tuyn ca tam gic

ABD vi cc nh A(2, 0, 0), B(0, 0, 2), D(0, 0, 2) ly theo chiungc chiu kim ng h nhn t chiu dng trc Oz .


Mt cong Tch phn mt loai mt Tch phn mt loai hai

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Surface Integral - Tích phân mặt