Bài Giảng GIẢI TÍCH II

8/11/2019 Bi Ging GII TCH II

1/115

TRNG I HC BCH KHOA H NIVIN TON NG DNG & TIN HC

BI XUN DIU

Bi Ging

GII TCHII(lu hnh ni b)

CC NG DNG CA PHP TNH VI PHN, TCH PHN BI, TCH PHNPH THUC THAM S, TCH PHN NG, TCH PHN MT, L THUYT

TRNG

Tm tt l thuyt, Cc v d, Bi tp v li gii

H Ni- 2009


2/115

MC LC

Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Chng 1 .Cc ng dng ca php tnh vi phn trong hnh hc . . . . . . . 5

1 Cc ng dng ca php tnh vi phn trong hnh hc phng . . . . . . . . . . 5

1.1 Phng trnh tip tuyn v php tuyn ca ng cong ti mt im. 51.2 cong ca ng cong. . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Hnh bao ca h ng cong ph thuc mt tham s . . . . . . . . . . 7

2 Cc ng dng ca php tnh vi phn trong hnh hc khng gian . . . . . . . 102.1 Hm vct . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 Phng trnh tip tuyn v php din ca ng cong cho di dng tham s2.3 Phng trnh php tuyn v tip din ca mt cong.. . . . . . . . . . 112.4 Phng trnh tip tuyn v php din ca ng cong cho di dng giao c

Chng 2 .Tch phn bi . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1 Tch phn kp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.2 Tnh tch phn kp trong h to Descartes. . . . . . . . . . . . . . 161.3 Php i bin s trong tch phn kp . . . . . . . . . . . . . . . . . . . 24

2 Tch phn bi ba . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.1 nh ngha v tnh cht . . . . . . . . . . . . . . . . . . . . . . . . . . 352.2 Tnh tch phn bi ba trong h to Descartes . . . . . . . . . . . . 352.3 Phng php i bin s trong tch phn bi ba. . . . . . . . . . . . . 38

3 Cc ng dng ca tch phn bi . . . . . . . . . . . . . . . . . . . . . . . . . . 503.1 Tnh din tch hnh phng . . . . . . . . . . . . . . . . . . . . . . . . . 503.2 Tnh th tch vt th . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.3 Tnh din tch mt cong . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Chng 3 .Tch phn ph thuc tham s. . . . . . . . . . . . . . . . . . . 63

1 Tch phn xc nh ph thuc tham s. . . . . . . . . . . . . . . . . . . . . . 631.1 Gii thiu. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

1


3/115

2 MC LC

1.2 Cc tnh cht ca tch phn xc nh ph thuc tham s. . . . . . . . 631.3 Cc tnh cht ca tch phn ph thuc tham s vi cn bin i. . . . 66

2 Tch phn suy rng ph thuc tham s. . . . . . . . . . . . . . . . . . . . . . 672.1 Cc tnh cht ca tch phn suy rng ph thuc tham s. . . . . . . . 672.2 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3 Tch phn Euler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.1 Hm Gamma. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.2 Hm Beta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.3 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Chng 4 .Tch phn ng. . . . . . . . . . . . . . . . . . . . . . . . . 79

1 Tch phn ng loi I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 791.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 791.2 Cc cng thc tnh tch phn ng loi I . . . . . . . . . . . . . . . . 80

1.3 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 802 Tch phn ng loi II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 822.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 822.2 Cc cng thc tnh tch phn ng loi II. . . . . . . . . . . . . . . . 822.3 Cng thc Green. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 852.4 ng dng ca tch phn ng loi II . . . . . . . . . . . . . . . . . . 912.5 iu kin tch phn ng khng ph thuc ng ly tch phn. 92

Chng 5 .Tch phn mt. . . . . . . . . . . . . . . . . . . . . . . . . . 95

1 Tch phn mt loi I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 951.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 951.2 Cc cng thc tnh tch phn mt loi I . . . . . . . . . . . . . . . . . 951.3 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

2 Tch phn mt loi II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 982.1 nh hng mt cong . . . . . . . . . . . . . . . . . . . . . . . . . . . . 982.2 nh ngha tch phn mt loi II . . . . . . . . . . . . . . . . . . . . . 982.3 Cc cng thc tnh tch phn mt loi II . . . . . . . . . . . . . . . . . 982.4 Cng thc Ostrogradsky, Stokes . . . . . . . . . . . . . . . . . . . . . 102

2.5 Cng thc lin h gia tch phn mt loi I v loi II . . . . . . . . . 105Chng 6 .L thuyt trng. . . . . . . . . . . . . . . . . . . . . . . . . 107

1 Trng v hng. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1071.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1071.2 o hm theo hng . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1071.3 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1081.4 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

2


4/115

MC LC 3

2 Trng vct . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1112.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1112.2 Thng lng, dive, trng ng . . . . . . . . . . . . . . . . . . . . . . . 1112.3 Hon lu, vct xoy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1112.4 Trng th - hm th v . . . . . . . . . . . . . . . . . . . . . . . . . . 112

2.5 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

3


5/115

4 MC LC

4


6/115

CHNG

1CC NG DNG CA PHP TNH VI PHNTRONG HNH HC

1. CC NG DNG CA PHP TNH VI PHN TRONGHNH HC PHNG

1.1 Phng trnh tip tuyn v php tuyn ca ngcong ti mt im.

1. im chnh quy.

Cho ng cong (L) xc nh bi phng trnh f( x,y) = 0. im M(x0,y0)c gi l im chnh quy ca ng cong (L)nu tn ti cc o hm ringf

x(M) ,f

y(M)khng ng thi bng 0.

Cho ng cong (L) xc nh bi phng trnh tham s

x= x(t)

y= y (t). im

M(x(t0) ,y (t0))c gi l im chnh quy ca ng cong( L)nu tn ti cc

o hmx (t0) ,y(t0)khng ng thi bng 0. Mt im khng phi l im chnh quy c gi l im k d.

2. Cc cng thc.

Phng trnh tip tuyn v php tuyn ca ng cong xc nh bi phngtrnh ti im chnh quy:

5


7/115

6 Chng 1. Cc ng dng ca php tnh vi phn trong hnh hc

Tip tuyn(d): f

x(M) . (x x0)+ f

y(M) . (y y0)= 0.

Php tuyn

d

:

x x0fx(M)

= y y0fy(M)

.

Ch :Trng hp c bit, ng cong cho bi phng trnh y = f(x)th phng trnh tip tuyn ca ng cong ti im M(x0,y0)chnh quy ly y0 = f(x0)(x x0). y l cng thc m hc sinh bit trong chngtrnh ph thng.

Phng trnh tip tuyn v php tuyn ca ng cong(L)xc nh bi phng

trnh tham s

x= x (t)

y= y (t)ti im M(x(t0) ,y (t0))chnh quy:

Tip tuyn(d) : x x(t0)

x(t0) = y y (t0)

y(t0) .

Php tuyn d

: x(t0) . (x x(t0))+y(t0) . (y y (t0))= 0.

1.2 cong ca ng cong.1. nh ngha.

2. Cc cng thc tnh cong ca ng cong ti mt im. Nu ng cong cho bi phng trnh y = f(x)th:

C (M)= |y|

(1+y2)3/2

Nu ng cong cho bi phng trnh tham s

x= x (t)

y= y (t)th:

C (M)=

x yx y(x2 +y2)3/2

Nu ng cong cho bi phng trnh trong to ccr = r()th:

C (M)=

r2 +2r2 rr(r2 +r2)3/2

6


8/115

1. Cc ng dng ca php tnh vi phn trong hnh hc phng 7

1.3 Hnh bao ca h ng cong ph thuc mt thams

1. nh ngha: Cho h ng cong(L)ph thuc vo mt hay nhiu tham s. Nu ming cong trong h

(L)u tip xc vi ng cong

(E)ti mt im no trn

Ev ngc li, ti mi im thuc (E)u tn ti mt ng cong ca h (L)tip xcvi(E)ti im th(E)c gi l hnh bao ca h ng cong (L).

2. Quy tc tm hnh bao ca h ng cong ph thuc mt tham s.

nh l 1.1. Cho h ng cong F(x,y, c) = 0 ph thuc mt tham s c. Nu hng cong trn khng c im k d th hnh bao ca n c xc nh bng cch

khc t h phng trnh

F(x,y, c)= 0

Fc(x,y, c)= 0(1)

3. Nu h ng cong cho c im k d th h phng trnh (1) bao gm hnh bao(E)v qu tch cc im k d thuc h cc ng cong cho.

Bi tp 1.1. Vit phng trnh tip tuyn v php tuyn vi ng cong:

a) y= x3 +2x2 4x 3ti(2, 5).

Li gii. Phng trnh tip tuyny= 5

Phng trnh php tuynx =2

b) y= e1x2ti giao im ca ng cong vi ng thngy = 1.

Li gii. TiM1(1, 1),Phng trnh tip tuyn2x y+3= 0Phng trnh php tuyn x +2y 1= 0

TiM2(1, 1),

Phng trnh tip tuyn2x+y 3= 0Phng trnh php tuynx

2y+1= 0

c.

x= 1+t

t3

y= 32t3

+ 12tti A(2, 2).

Li gii. Phng trnh tip tuyny = x.

Phng trnh php tuyn x+y 4= 0.

7


9/115


d. x23 +y

23 =a

23 ti M(8, 1).

Li gii. Phng trnh tip tuyn x +2y 10= 0. Phng trnh php tuyn2x y 15= 0.

Bi tp 1.2. Tnh cong ca:a. y=x3 ti im c honh x = 12 .

Li gii.

C(M)= |y|

(1+y2)3/2 =... =

192

125

b. x= a (t sin t)y= a (t cos t)

(a > 0)ti im bt k.

Li gii.

C (M)=

x yx y

(x2 +y2)3/2 =... =

1

2a

2

11 cosx

c. x23 +y

23 =a

23 ti im bt k(a > 0).

Li gii. Phng trnh tham s: x= a cos3 ty= a sin3 t

, nn

C (M)=

x yx y

(x2 +y2)3/2 =... =

1

3a |sin t cos t|

d. r= aeb, (a, b > 0)

Li gii.

C (M)=

r2 +2r2 rr(r2 +r2)3/2

= 1

aeb

1+b2

Bi tp 1.3. Tm hnh bao ca h ng cong sau:

a. y= xc +c2

8


10/115

1. Cc ng dng ca php tnh vi phn trong hnh hc phng 9

b. cx2 +c2y= 1

c. y= c2 (x c)2

Li gii. a. tF(x,y, c):=y xc c2 =0.iu kin:c

=0.

Xt h phng trnh: Fx(x,y, c)= 0

Fy(x,y, c)= 0 Fx(x,y, c)= 0

1= 0, h phng trnh v

nghim nn h ng cong khng c im k d. Ta c F(x,y, c)= 0

Fc(x,y, c)= 0

y xc c2 =02c+ x

c2 =0

x= 2c3

y= 3c2

nn

x2

2 y33 = 0.Do iu kin c= 0nn x,y= 0. Vy ta c hnh bao ca hng cong l ng

x2

2

y3

3=0tr imO (0, 0).

b. t F(x,y, c) := cx2 +c2y 1 = 0.Nu c = 0th khng tho mn phng trnh cho nn iu kin: c=0.Xt h phng trnh:

Fx(x,y, c)= 0Fy(x,y, c)= 0

2cx = 0

c2 =0 x = c = 0, nhng im k

d khng thuc h ng cong cho nn h ng cong cho khng c im kd. Ta c

F(x,y, c)= 0

Fc(x,y, c)= 0

cx2 +c2y= 1

x2 +2cx = 0

x= 2cy= 1

c2

Do x,y=0v ta c hnh bao ca h ng cong l ng y = x44 tr imO(0, 0).c. tF(x,y, c):=c2 (x c)2 y= 0.

Xt h phng trnh:

Fx(x,y, c)= 0Fy(x,y, c)= 0

Fx =01= 0 , h phng trnh v nghim

nn h ng cong cho khng c im k d.Ta c

F(x,y, c)= 0

Fc(x,y, c)= 0

c2 (x c)2 y= 0 (1)2c (x c) 2c2 (x c)= 0 (2)

(2) c= 0c= x

c= x2

, th vo(1)ta cy = 0,y= x4

16 .

Vy hnh bao ca h ng cong ly = 0,y= x4

16 .

9


11/115


2. CC NG DNG CA PHP TNH VI PHN TRONGHNH HC KHNG GIAN

2.1 Hm vctGi s Il mt khong trong R.

nh x I Rn

tr(t) Rn c gi l hm vct ca bin s txc nh trn R. Nu

n = 3, ta vitr(t)= x (t) .

i +y (t) .

j +z (t) .

k. t M(x (t) ,y (t) ,z (t)), qu tch

Mkhi tbin thin trong Ic gi l tc ca hm vctr(t).

Gii hn: Ngi ta ni hm vct c gii hn la khit t0nu limtt0

r(t) a

=

0, k hiu lim

tt0

r(t)=

a.

Lin tc: Hm vct r(t)xc nh trn Ic gi l lin tc ti t0Inu limtt0

r(t)=

r(t0). (tung ng vi tnh lin tc ca cc thnh phn tng ng x(t) ,y (t) ,z (t))

o hm: Gii hn, nu c, ca t slimh0

rh = limh0

r (t0+h)r (t0)

h c gi l o hm

ca hm vctr(t)tit0, k hiu r (t0)hay d

r (t0)dt , khi ta ni hm vct

r(t)kh

vi tit0.Nhn xt rng nux(t) ,y (t) ,z (t)kh vi tit0th

r(t)cng kh vi tit0v r(t0)=

x(t0) .i +y(t0) .j +z(t0) .k.

2.2 Phng trnh tip tuyn v php din ca ngcong cho di dng tham s

Cho ng cong

x= x(t)

y= y(t)

z= z(t)

vM(x0,y0,z0)l mt im chnh quy.

Phng trnh tip tuyn ti M

(d) : x x(t0)

x(t0) =

y y (t0)y(t0)

= z z (t0)

z(t0) .

Phng trnh php din ti M.

(P): x(t0) . (x x(t0))+y(t0) . (y y (t0))+z(t0) . (z z (t0))= 0.

10


12/115

2. Cc ng dng ca php tnh vi phn trong hnh hc khng gian 11

2.3 Phng trnh php tuyn v tip din ca mtcong.

Cho mt cong Sxc nh bi phng trnh f(x,y,z) = 0v M(x0,y0,z0)l mt imchnh quy ca

S.

Phng trnh php tuyn tiM

(d): x x0fx(M)

= y y0fy(M)

= z z0fz(M)

.

Phng trnh tip din ti M

(P): fx(M) . (x x0)+ fy(M) . (y y0)+ fz(M) . (z z0)= 0.

c bit, nu mt cong cho bi phng trnh z = z (x,y)th phng trnh tip din ti Ml(P) : z z0 =z x(M) . (x x0)+zy(M) . (y y0).

2.4 Phng trnh tip tuyn v php din ca ngcong cho di dng giao ca hai mt cong

Cho ng cong xc nh bi giao ca hai mt cong nh sau

f( x,y,z)= 0

g (x,y,z)= 0.

tnf = fx(M) ,fy(M) ,fz(M), l vct php tuyn ca mt phng tip din ca mtcong f( x,y,z)= 0ti M.tng =

gx(M) ,gy(M) ,gz(M)

, l vct php tuyn ca mt phng tip din ca mt

congg (x,y,z)= 0ti M.Khi nfngl vct ch phng ca tip tuyn ca ng cong cho tiM. Vy phngtrnh tip tuyn l:

PTTQ :

fx(M) . (x x0)+ fy(M) . (y y0)+ fz(M) . (z z0)= 0.gx(M) . (x

x0)+g

y(M) . (y

y0)+g

z(M) . (z

z0)= 0.

PTCT : xx0 f

y(M) f

z(M)

gy(M) gz(M)

= yy0 f

z(M) f

x(M)

gz(M) gx(M)

= zz0 f

x(M) f

y(M)

gx(M) gy(M)

Bi tp 1.4. Gi sp (t) , q (t) , (t)l cc hm vct kh vi. Chng minh rng:

a. ddtp (t)+ q (t)= dp(t)dt + dq (t)dt

11


13/115


b. ddt (t) p (t)= (t) dp(t)dt +(t) p (t)

c. ddtp (t) q (t)=p (t) dq (t)dt + dp(t)dt q (t)

d. ddt

p (t) q (t)

=p (t) dq (t)dt + d

p(t)dt q (t)

Li gii. a. Gi sp (t)=(p1(t) ,p2(t) ,p3(t)) , q (t)=(q1(t) , q2(t) , q3(t)), khi :d

dt

p (t)+ q (t)= ddt

(p1(t)+q1(t) ,p2(t)+q2(t) ,p3(t)+q3(t))

=p1(t)+q

1(t) ,p

2(t)+q

2(t) ,p

3(t)+q

3(t)

=p1(t) ,p

2(t) ,p

3(t)

+

q1(t) , q2(t) , q

3(t)

=

dp (t)dt

+dq (t)

dt

b.

d

dt

(t) p (t)

=

[ (t)p1(t)], [ (t)p2(t)], [ (t)p3(t)]

=(t)p1(t)+ (t)p1(t) ,

(t)p2(t)+ (t)p2(t) , (t)p3(t)+ (t)p3(t)

=(t)p1(t) , (t)p2(t) , (t)p3(t)

+ (t)p1(t) , (t)p

2(t) , (t)p

3(t)

= (t)

dp (t)dt

+(t) p (t)

c. Chng minh tng t nh cu b, s dng cng thc o hm ca hm hp.

d.

d

dt

p (t) q (t)=

d

dt

p2(t) p3(t)q2(t) q3(t) , p3(t) p1(t)q3(t) q1(t)

, p1(t) p2(t)q1(t) q2(t)

=...

= p2(t) p

3(t)

q2(t) q3(t) , p3(t) p

1(t)

q3(t) q1(t) , p1(t) p

2(t)

q1(t) q2(t) +

p2(t) p3(t)q2(t) q3(t) , p3(t) p1(t)q3(t) q1(t)

, p1(t) p2(t)q1(t) q2(t)

=p (t) dq (t)

dt +

dp (t)dt

q (t)

Bi tp 1.5. Vit phng trnh tip tuyn v php din ca ng:

12


14/115

2. Cc ng dng ca php tnh vi phn trong hnh hc khng gian 13

a.

x= a sin2 t

y= b sin t cos t

z= c cos2 t

ti im ng vit = 4 , (a, b, c > 0).

b. x= e

t sin t2

y= 1z= e

t cos t2

ti im ng vit = 2.

Li gii. a. Phng trnh tip tuyn:(d): xa2

a = y b2

0 = z c2c

Phng trnh php din: (P) : a

x a2 c z c2= 0.

b. Phng trnh tip tuyn: (d): x2

2

= y1

0 = z

2

22

2

.

Phng trnh php din: (P) :

2

2

x+

2

2 z

2

2 = 0.Bi tp 1.6. Vit phng trnh php tuyn v tip din ca mt cong:

a) x2 4y2 +2z2 =6ti im(2,2,3).

b) z= 2x2 +4y2 ti im(2,1,12).

c) z= ln (2x+y)ti im(1,3,0)

Li gii. a. Phng trnh php tuyn:(d): x

2

4 = y

2

16 = z

3

12

Phng trnh tip din: (P): 4 (x 2) 16 (y 2)+12 (z 3)= 0

b. Phng trnh php tuyn:(d): x28 = y1

8 = z12

1 Phng trnh tip din: (P): 8 (x 2)+8 (y 1) (z 12)= 0.

c. Phng trnh php tuyn:(d): x+12 = y3

1 = z1

Phng trnh tip din: (P): 2 (x+1)+(y 3) z= 0.

Bi tp 1.7. Vit phng trnh tip tuyn v php din ca ng:

a.

x2 +y2 =10

y2 +z2 =25ti im A (1,3,4)

b.

2x2 +3y2 +z2 =47

x2 +2y2 =zti imB (2,6,1)

13


15/115


Li gii. a. Ta c

f( x,y,z):= x2 +y2 10= 0g (x,y,z):=y2 +z2 25= 0 nn

nf =(2,6,0)

ng =(0,6,8) .

Do nf ng =2 (21, 8, 3). Vy:

Phng trnh tip tuyn(d): x121 = y38 =

z43

Phng trnh php din (P): 21 (x 1) 8 (y 3)+3 (z 4)= 0

b. Tng t,

nf =(8,6,12)ng =(4,4, 1)

,nf ng =2 (27,27,4)nn

Phng trnh tip tuyn(d): x+227 = y1

27 = z6

4

Phng trnh php din (P): 27 (x+2)+27 (y 1)+4 (z 6)= 0

14


16/115

CHNG

2TCH PHN BI1. TCH PHN KP

1.1 nh nghanh ngha 2.1. Cho hm s f(x,y) xc nh trong mt min ng, b chn D. Chiamin D mt cch tu thnh n mnh nh. Gi cc mnh v din tch ca chng l

S1,S2, ...,Sn. Trong mi mnh Sily mt im tu M(xi,yi)v thnh lp tng tch

phn In =n

i=1

f(xi,yi)Si. Nu khin sao chomax {Si0}m In tin ti mt gi

tr hu hn I, khng ph thuc vo cch chia minD v cch chn im M(xi,yi)th giihn y c gi l tch phn kp ca hm s f(x,y)trong minD, k hiu lD

f(x,y) dS

Khi ta ni rng hm s f(x,y)kh tch trong min D. Do tch phn kp khng phthuc vo cch chia minDthnh cc mnh nh nn ta c th chia Dthnh hai h ngthng song song vi cc trc to , khi dS= dxdyv ta c th vit

D

f(x,y) dS = D

f(x,y) dxdy

Tnh cht c bn:

Tnh cht tuyn tnh:D

[f(x,y)+g (x,y)] dxdy=D

f(x,y) dxdy+D

g (x,y) dxdy

15


17/115

16 Chng 2. Tch phn bi

D

k f(x,y) dxdy= kD

f( x,y) dxdy

Tnh cht cng tnh: Nu D = D1 D2vD1 D2= th

D

f(x,y) dxdy= D1

f(x,y) dxdy+ D2

f( x,y) dxdy

1.2 Tnh tch phn kp trong h to Descartes tnh cc tch phn hai lp, ta cn phi a v tnh cc tch phn lp.

1. Phc tho hnh dng ca min D.

2. Nu Dl min hnh ch nht( D) : a x b, c y dth ta c th s dng mttrong hai tch phn lp

D

f( x,y) dxdy=

ba

dx

dc

f(x,y) dy =

dc

dy

dc

f( x,y) dx

3. Nu Dl hnh thang cong c cch cnh song song vi Oy, (D) : a x b, (x) y (x)th dng tch phn lp vi th t dytrc,dxsau.

D

f(x,y) dxdy=

ba

dx

(x)(x)

f(x,y) dy

4. Nu Dl hnh thang cong c cch cnh song song vi O x, (D) : c y d, (y) x (y)th dng tch phn lp vi th t dxtrc,dysau.

D f(x,y) dxdy=d

c dy(y)

(y) f( x,y) dx

5. NuDl min c hnh dng phc tp, khng c dng 3,4 th thng thng ta s chiaminDthnh mt s hu hn min c dng 3 hoc 4 ri s dng tnh cht cng tnh a v vic tnh ton nhng tch phn lp trn min c dng 3, 4.

Cc dng bi tp c bn

16


18/115

1. Tch phn kp 17

Dng 1: i th t ly tch phn.

Trong phn trn, chng ta bit rng th t ly tch phn v hnh dng ca minDclin quan cht ch n nhau. Nu th t dytrc,dxsau th minD c dng hnh thangcong song song vi trcOy, v c biu din l(D) : a x b, (x) y (x). Ngc li,nu th t

dxtrc,

dysau th min

Dc dng hnh thang cong song song vi trc

Ox,

v c biu din l ( D) : c y d, (y) x (y). Do vy vic i th t ly tch phntrong tch phn lp chng qua l vic biu din minDt dng ny sang dng kia.

1. T biu thc tch phn lp, v phc tho min D.

2. NuDl min hnh thang cong c cc cnh song song viOyth ta chia Dthnh cchnh thang cong c cc cnh song song viOx. Tm biu din gii tch ca cc mincon, v d(Di): c i y di,i(y) x i(y), sau vit

ba

dx

y2(x)y1(x)

f(x,y) dy = i

dici

dy

i(y)i(y)

f( x,y) dx

3. Lm tng t trong trng hpDl hnh thang cong c cc cnh song song vi Ox.

Bi tp 2.1. Thay i th t ly tch phn ca cc tch phn sau:

a)1

0

dx

1x2

1

x2

f(x,y) dy

x1

y

1

O

D1

D2

Hnh2.1a)

Chia minDthnh hai min con D1,D2nh hnh v,D1 :

1 y 01 y2 x 1 y2 ,D2 :0 y 11 y x 1 y

I=

01

dy

1y2

1y2f( x,y) dx+

10

dy

1y

1yf(x,y) dx

17


19/115


b)1

0

dy

1+

1y22y

f(x,y) dx

x21

y

2

1

O

Hnh2.1b)

Li gii. Ta c: D :

1 x 2

2 x y 2x x2nn:

I=2

1

dx2xx22x

f(x,y) dy

c)2

0

dx

2x

2xx2

f(x,y) dx

x21

y

2

1

O

Hnh2.1c)

Li gii. ChiaDthnh 3 min nh hnh v,

D1: 0 y 1y22 x 1

1 y2

,D2 : 0 y 11+1 y2 x 2 ,D3: 1 y 2y22 x 2Vy:

I=

10

dy

1

1y2y2

2

f( x,y) dx+

10

dy

21+

1y2

f( x,y) dx+

21

dy

2y2

2

f( x,y) dx

18


20/115

1. Tch phn kp 19

d)

2

0

dy

y0

f(x,y) dx+

2

2

dy

4y2

0

f(x,y) dx

x2

y

2

O

Hnh2.1d)Li gii.

D : 0 x

2

x y

4 x2

nn:

I=

2

0

dx

4x2x

f(x,y) dy

Mt cu hi rt t nhin t ra l vic i th t ly tch phn trong cc bi ton tch

phn kp c ngha nh th no? Hy xt bi ton sau y:

Bi tp 2.2. Tnh I=1

0

dx

1x2

xey2dy.

x1

y

2

O

Hnh2.2

Li gii. Chng ta bit rng hm s f( x,y) = xey2 lin tc trn min Dnn chc chnkh tch trnD. Tuy nhin cc bn c th thy rng nu tnh tch phn trn m lm theo

19


21/115


th t dytrc th khng th tnh c, v hm sey2 khng c nguyn hm s cp! Cnnu i th t ly tch phn th:

I=

10

dy

y

0

xey2dx =

10

ey2x2

2 x=

y

x=0 dy =1

2

10

ey2.ydy =

1

4ey

2 |10 =1

4(e 1)

Dng 2: Tnh cc tch phn kp thng thng.

Bi tp 2.3. Tnh cc tch phn sau:

a)D

x sin (x+y) dxdy,D=

(x,y) R2 : 0 y 2 , 0 x 2

Li gii.

I=

2

0

dx

2

0

x sin (x+y) dy = ... =

2hoc I=

2

0

dy

2

0

x sin (x+y) dx = ... =

2

b) I=D

x2 (y x) dxdy,Dgii hn biy = x2&x= y2

x

y

O 1

1

y= x2

x= y2

Hnh2.3

Li gii.

I=

10

dx

x

x2

x2y x3

dy = ... = 1

504

20


22/115

1. Tch phn kp 21

Dng 3: Tnh cc tch phn kp c cha du gi tr tuyt i.

Mc ch ca chng ta l ph b c du gi tr tuyt i trong cc bi ton tnhtch phn kp c cha du gi tr tuyt i. V d, tnh cc tch phn kp dng

D

|f( x,y)| dxdy. Kho st du ca hm f( x,y), do tnh lin tc ca hm f(x,y) nn

ng cong f(x,y)= 0s chia min D thnh hai min,D +,D. TrnD+,f(x,y) 0, vtrnD,f(x,y) 0. Ta c cng thc:

D

|f(x,y)| dxdy=D+

f(x,y) dxdy D

f(x,y) dxdy (1) (1)

Cc bc lm bi ton tnh tch phn kp c cha du gi tr tuyt i:

1. V ng cong f(x,y)= 0 tm ng cong phn chia minD.

2. Gi s ng cong tm c chia minDthnh hai min. xc nh xem min nolD + , min no lD, ta xt mt im (x0,y0)bt k, sau tnh gi tr f(x0,y0).Nu f(x0,y0) > 0th min cha(x0,y0)l D+ v ngc li.

3. Sau khi xc nh c cc minD+,D, chng ta s dng cng thc (1) tnh tchphn.

Bi tp 2.4. TnhD

|x+y|dxdy,D : (x,y) R2 ||x 1| , |y| 1

O x

1

D+

y

1

D

Hnh2.4Li gii. Ta c:

D+ = D {x+y 0}={1 x 1, x y 1}

D= D {x+y 0}={1 x 1, 1 y x}

21


23/115


nn:

I=D+

(x+y) dxdy D

(x+y) dxdy= ... = 8

3

Bi tp 2.5. Tnh D

|y x2|dxdy,D: (x,y)R2 ||x| 1, 0 y 1

O x1

D+

y

1

D

Hnh2.5

Li gii.

D+ = D

(x,y)

y x2 0

=

1 x 1,x2 y 1

D= D

(x,y)

y x2 0={1 x 1, 0 y x}I=

D+

y x2dxdy+

D

x2 ydxdy = I1+ I2

trong

I1 =

1

1 dx1

x2y x2dy = 23

1

1 1 x232

dx

x=sin t

=

4

3

2

0

cos

4

tdt = ... =

4

I2 =

11

dx

x20

x2 ydy = 2

3

11

|x|3dx = 43

10

x3dx =1

3

VyI= 4 +13

22


24/115

1. Tch phn kp 23

Dng 4: Tnh cc tch phn kp trong trng hp min ly tch phn l min ixng.

nh l 2.2. Nu min Dl min i xng qua trcO x(hoc tng ngOy) v hm lhm l i viy (hoc tng ng i vix) th

D

f(x,y) dxdy= 0

nh l 2.3. Nu min Dl min i xng qua trcO x(hoc tng ngOy) v hm lhm chn i viy (hoc tng ng i vix) th

D

f(x,y) dxdy= 2D

f( x,y) dxdy

trong D l phn nm bn phi trcOxcaD(hoc tng ng pha trn ca trcOytng ng)

nh l 2.4. Nu minDl min i xng qua trc gc to Ov hm f(x,y)tho mnf(x, y)=f(x,y)th

D

f(x,y) dxdy= 0

Bi tp 2.6. Tnh

|x|+|y|1|x| + |y|dxdy.

O x

1

y

1

D1

Hnh2.6Li gii. DoDi xng qua cOxvOy, f(x,y)=|x| + |y|l hm chn vi x,ynn

I=4D1

f( x,y) dxdy= 4

10

dx

1x0

(x+y)dy =4

3

23


25/115


1.3 Php i bin s trong tch phn kpPhp i bin s tng qut

Php i bin s tng qut thng c s dng trong trng hp min D l giao ca

hai h ng cong. Xt tch phn kp:I= D f(x,y) dxdy, trong f(x,y)lin tc trnD.Thc hin php i bin s x = x(u, v) ,y= y (u, v) (1)tho mn:

x = x(u, v) ,y = y (u, v)l cc hm s lin tc v c o hm ring lin tc trongmin ngDuvca mt phngOuv.

Cc cng thc (1) xc nh song nh t DuvD.

nh thc Jacobi J= D(x,y)

D(u,v)=

xu x

v

y

u y

v=0

Khi ta c cng thc:

I=D

f( x,y) dxdy=Duv

f( x(u, v) ,y (u, v)) |J| dudv

Ch :

Mc ch ca php i bin s l a vic tnh tch phn t min Dc hnh dng

phc tp v tnh tch phn trn minDuvn gin hn nh l hnh thang cong hochnh ch nht. Trong nhiu trng hp, php i bin s cn c tc dng lm ngin biu thc tnh tch phn f(x,y).

Mt iu ht sc ch trong vic xc nh minDuv l php di bin s tng quts bin bin ca minDthnh bin ca minDuv, bin minDb chn thnh minDuvb chn.

C th tnh Jthng qua J1 = D(u,v)D(x,y)

=

u

x u

y

v

x

v

y.

Bi tp 2.7. Chuyn tch phn sau sang hai binu, v:

a)1

0

dx

xx

f( x,y) dxdy,nu t

u= x+yv= x yb) p dng tnh vi f(x,y)=(2 x y)2.

24


26/115

1. Tch phn kp 25

O x1

y

1

D

O u2

v

2

Hnh2.7Li gii. u= x+yv= x y

x= u+v2y= uv2 , |J|= D(x,y)

D (u, v)=

1 11 1=2

hn na

D

0 x 1

x

y

x

Duv

0 u 2

0

v

2 unn

I=1

2

20

du

2u0

f

u+v

2 ,

u v2

dv

Bi tp 2.8. Tnh I=D

4x2 2y2 dxdy,trong D :

1 xy 4x y 4x

x

y

O 1

1

y= 4x

y= x

xy = 4

xy = 1

Hnh2.8

25


27/115


Li gii. Thc hin php i bin

u= xy

v= yx

x=

uv

y=

uv,Duv :

1 u 4

1 v 4,J1 =

y xyx2

1x

= 2

y

x=2

uv

uv

=2v

khi

I=

41

du

41

4

u

v 2uv

.1

2vdv=

41

du

41

2u

v2 u

dv =

41

32

udu =454

Php i bin s trong to ccTrong rt nhiu trng hp, vic tnh ton tch phn kp trong to cc n gin

hn rt nhiu so vi vic tnh tch phn trong to Descartes, c bit l khi minDcdng hnh trn, qut trn, cardioids,. . . v hm di du tch phn c nhng biu thc

x2 +y2. To cc ca im M(x,y)l b(r,), trong

r=OM

=

OM ,Ox.

Cng thc i bin:

x= r cos

y= r sin

, trong min bin thin car,ph thuc vo hnh

dng ca minD. Khi J= D(x,y)D(r,)

=r, v I=Dr

f(r cos, r sin)rdrd

c bit, nuD :

1 2r1() r r2() , th

I=

21

d

r2()r1()

f(r cos, r sin) rdr

Bi tp 2.9. Tm cn ly tch phn trong to cc I =D

f(x,y) dxdy, trong Dl

min xc nh nh sau:

a) a2 x2 +y2 b2

26


28/115

1. Tch phn kp 27

x

y

O a

a

b

b

Hnh2.9a

Li gii.D:

0 2a r b I=20

d

ba

f(r cos, r sin) rdr

b) x2 +y2 4x,x2 +y2 8x,y x,y 2x

x42 8

y

O

Hnh2.9bLi gii. Ta c:

D : 4 34cos r 8cos I=3

4

d

8cos

4cos

f(r cos, r sin) rdr

Bi tp 2.10. Dng php i bin s trong to cc, hy tnh cc tch phn sau:

a)R

0

dx

R2x20

ln

1+x2 +y2

dy (R > 0).

27


29/115


xR

y

O

Hnh2.10a

T biu thc tnh tch phn ta suy ra biu thc gii tch ca minDl:

0 x R0 y R2 nn chuyn sang to cc, t:

x= r cos

y= r sinth

0 2

0 r R

I=

20

d

R0

ln

1+r2

rdr =

4

R0

ln

1+r2

d

1+r2

=

4

R2 +1

ln

R2 +1

R2

b) TnhD

xy2dxdy,Dgii hn bi

x2 +(y 1)2 =1x2 +y2 4y= 0

.

x

y

2

4

O

Hnh2.10b

t

x= r cosy= r sin 0 2sin r 4sin

28


30/115

1. Tch phn kp 29

I=

0

d

4sin2sin

r cos. (r sin)2 rdr

=0

Cch 2:VDi xng quaOyvxy2

l hm s l i vixnn I=0.


a)D

dxdy(x2+y2)

2 , trong D :

4y x2 +y2 8yx y x3

x

y

4

8

O

y= x

y= x3

Hnh2.11a

Li gii.

t

x= r cosy= r sin 4 34sin r 8sin

I=

3

4

d

8sin4sin

1r4

rdr =12

3

4

164sin2

116sin2

d= 3128

1 13

b)D

1x2y21+x2+y2

dxdytrong D : x2 +y2 1

29


31/115


x1

y

1

O

Hnh2.11b

t x= r cosy= r sin 0 20 r 1Ta c:

I=

20

d

10

1 r21+r2

rdru=r2

= 2

10

1

2

1 u1+u

du

t

t= 1 u1+u

du = 4t

(1+t2)2 dt0 t 1

I=

10

t

4t

(1+t2)2

dt =

10

4dt

1+t2+4

10

dt

(1+t2)2

=4arctg t

10 +4

1

2

t

t2 +1+

1

2arctg t

10

=

2

2

c)D

xyx2+y2

dxdytrong D :

x2 +y2 12

x2 +y2 2x

x2 +y2 2

3y

x 0,y 0

30


32/115

1. Tch phn kp 31

x2 2

3

y

2

3

O

D1

D2

Hnh2.11c

Li gii. Chia minDthnh hai min nh hnh v,

D= D1 D2,D1= 0 62cos r 23 ,D2=

6

2

2

3sin r 2

3

VyI= I1+ I2,trong

I1 =

6

0

d

2

32cos

r2 cos sin

r2 rdr =

1

2

6

0

cos sin

12 4cos2

d= ... =17

32

I2=

2

6

d

2

3

2

3sin

r2 cos sin

r2 rdr =

1

2

2

6

cos sin 12 12sin2 d= ... = 2732nn I= 118

Php i bin s trong to cc suy rng.

Php i bin trong to cc suy rng c s dng khi min Dc hnh dng ellipsehoc hnh trn c tm khng nm trn cc trc to . Khi s dng php bin i ny, btbuc phi tnh li cc Jacobian ca php bin i.

1. NuD : x2

a2+ y

2

b2 =1, thc hin php i bin

x= ar cosy= br sin ,J=abr2. NuD : (x a)2 +(y b)2 = R2, thc hin php i bin

x= a +r cosy= b+r sin ,J=r3. Xc nh min bin thin car,trong php i bin trong h to cc suy rng.

31


33/115


4. Thay vo cng thc i bin tng qut v hon tt qu trnh i bin.

Bi tp 2.12. TnhD

9x2 4y2 dxdy, trong D : x24 + y29 1.

x2

y

3

O

Hnh2.12

Li gii.t

x= 2r cosy= 3r sin J=6r,0 20 r 1

Ta c:

I=6Dr

36r2 cos2 36r2 sin2 rdrd= 6.36 20

|cos2| d1

0

r3dr =... = 216

Bi tp 2.13. TnhR

0

dxR2x2

R2x2

Rx x2 y2dy, (R > 0)

xR

y

O

Hnh2.13

Li gii. T biu thc tnh tch phn suy ra biu thc gii tch ca Dl:

D :

0 x RRx x2 y Rx x2

x R2

2+y2

R2

4

32


34/115

1. Tch phn kp 33

t

x= R2 +rcosy= r sin |J|= r,0 20 r R2

Vy

I=

2

0

d

R2

0R

2

4 r2

rdr=

2.1

2

R2

0R

2

4 r2

dR2

4 r2= R

3

12

Bi tp 2.14. TnhD

xydxdy, vi

a) Dl mt trn(x 2)2 +y2 1

x31

y

O

Hnh2.14a

Li gii.

t

x= 2+r cos

y= r sin

0 r 1

0 2

nnI=

20

d

10

(2+r cos) r sin.rdr =0

Cch 2.Nhn xt: Do Dl min i xng qua Ox, f(x,y)= xyl hm l i vi ynn I=0.

b) Dl na mt trn(x 2)2 +y2 1,y 0

x31

y

O

Hnh2.14b

33


35/115


Li gii.

t

x= 2+r cosy= r sin 0 r 10

nn

I=

0

d

10

(2+r cos) r sin.rdr = 43

34


36/115

2. Tch phn bi ba 35

2. TCH PHN BI BA

2.1 nh ngha v tnh chtnh ngha 2.2. Cho hm s f(x,y,z)xc nh trong mt min ng, b chn Vca khng

gianOxyz. Chia min Vmt cch tu thnh n min nh. Gi cc min v th tchca chng lV1,V2,...,Vn. Trong mi min ily mt im tu M(xi ,yi,zi)v thnh

lp tng tch phn In =n

i=1

f( xi,yi,zi)Vi. Nu khin+sao chomax {Vi0}mIntin ti mt gi tr hu hn I, khng ph thuc vo cch chia minVv cch chn im

M(xi ,yi,zi)th gii hn y c gi ltch phn bi baca hm s f(x,y,z)trong minV,k hiu l

V

f(x,y,z) dV.

Khi ta ni rng hm s f(x,y,z)kh tch trong minV.Do tch phn bi ba khng ph thuc vo cch chia min Vthnh cc min nh nn ta cth chiaVbi ba h mt thng song song vi cc mt phng to , khi dV= dxdydzv ta c th vit

V

f(x,y,z) dV=

V

f( x,y,z) dxdydz

Cc tnh cht c bn

Tnh cht tuyn tnh

V

[f(x,y,z)+g (x,y,z)] dxdydz= V

f( x,y,z) dxdydz+ V

g (x,y,z) dxdydz

V

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

V

f( x,y,z) dxdydz

Tnh cht cng tnh: Nu V=V1 V2vV1 V2 = th:V

f( x,y,z) dxdydz=

V1

f( x,y,z) dxdydz+

V2

f(x,y,z) dxdydz

2.2 Tnh tch phn bi ba trong h to DescartesCng ging nh vic tnh ton tch phn kp, ta cn phi a tch phn ba lp v tch

phn lp. Vic chuyn i ny s c thc hin qua trung gian l tch phn kp.

Tch phn ba lpTch phn hai lpTch phn lp

35


37/115


S trn cho thy vic tnh tch phn ba lp c chuyn v tnh tch phn kp (victnh tch phn kp c nghin cu bi trc). ng nhin vic chuyn i ny phthuc cht ch vo hnh dng ca minV. Mt ln na, k nng v hnh l rt quan trng.Nu minVc gii hn bi cc mt z = z1(x,y) ,z= z2(x,y), trong z1(x,y) ,z2(x,y)l cc hm s lin tc trn min D,Dl hnh chiu ca minVln mt phngOxyth ta

c:

I=

V

f( x,y,z) dxdydz=D

dxdy

z2(x,y)z1(x,y)

f(x,y,z) dz (2.1)

Thut ton chuyn tch phn ba lp v tch phn hai lp

1. Xc nh hnh chiu ca minVln mt phngOxy.

2. Xc nh bin diz = z1(x,y)v bin trnz = z2(x,y)caV.

3. S dng cng thc 2.1 hon tt vic chuyn i.

n y mi vic ch mi xong mt na, vn cn li by gi l:

Xc nh Dv cc binz = z1(x,y) ,z= z2(x,y) nh th no?

C hai cch xc nh: Dng hnh hc hoc l da vo biu thc gii tch ca min V.Mi cch u c nhng u v nhc im ring. Cch dng hnh hc tuy kh thc hinhn nhng c u im l rt trc quan, d hiu. Cch dng biu thc gii tch ca Vtuy

c th p dng cho nhiu bi nhng thng kh hiu v phc tp. Chng ti khuyn ccem sinh vin hy c gng th cch v hnh trc. Mun lm c iu ny, i hi cc bnsinh vin phi c k nng v cc mt cong c bn trong khng gian nh mt phng, mttr, mt nn, mt cu, ellipsoit, paraboloit, hyperboloit 1 tng, hyperboloit 2 tng, hnna cc bn cn c tr tng tng tt hnh dung ra s giao ct ca cc mt.Ch : Cng ging nh khi tnh tch phn kp, vic nhn xt c tnh i xng ca minVv tnh chn l ca hm ly tch phn f( x,y,z)i khi gip sinh vin gim c khilng tnh ton ng k.

nh l 2.5. NuVl min i xng qua mt phngz = 0(Oxy)v f(x,y,z)l hm s li viz th

V

f(x,y,z) dxdydz= 0.

nh l 2.6. NuVl min i xng qua mt phngz = 0(Oxy) v f(x,y,z) l hm schn i vizth

V

f( x,y,z) dxdydz = 2V+

f( x,y,z) dxdydz, trong V+ l phn pha

trn mt phngz = 0caV.

36


38/115

2. Tch phn bi ba 37

Tt nhin chng ta c th thay i vai tr ca ztrong hai nh l trn bng xhocy. Hainh l trn c th c chng minh d dng bng phng php i bin s.

Bi tp 2.15. Tnh V zdxdydztrong min Vc xc nh bi:

0 x 1

4

x y 2x

0 z

1 x2 y2

Li gii.

I=

14

0

dx

2xx

dy

1x2y2

0

zdz =

14

0

dx

2xx

1

2

1 x2 y2

dy =

1

2

14

0

x 10

3x3

dx = 43

3072

Bi tp 2.16. Tnh

V

x2 +y2

dxdydztrong V:

x2 +y2 +z2 =1

x2 +y2 z2 =0 .

y

z

x

O

z=

1 x2 y2

z=

x2 +y2

D

Hnh 2.16

37


39/115


Li gii. Do tnh cht i xng,

V

x2 +y2

dxdydz= 2

V1

x2 +y2

dxdydz= 2 I1, trong

V1l na pha trn mt phngOxyca V. Ta c

V1 :

x2 +y2 z

1 x2 y2

D: x2 +y2 1

2,

,

viDl hnh chiu ca V1ln Oxy. Ta c

I1 =D

x2 +y2dxdy

1x2y2

x2+y2

dz =D

x2 +y2

1 x2 y2

x2 +y2

dxdy

t

x= r cos

y= r sin J=r,

0 2

0 r 1

2

nn

I1 =

12

0

r3

1 r2 r

dr

20

d= 2

12

0

r3

1 r2 r

dr = (r=cos )... =2

5 .

8 5212

Vy

I=4

5 .

8 5212

2.3 Phng php i bin s trong tch phn bi baPhp i bin s tng qut

Php i bin s tng qut thng c s dng trong trng hp min Vl giao caba h mt cong. Gi s cn tnh I=

V

f(x,y,z) dxdydztrong f(x,y,z)lin tc trnV.

Thc hin php i bin s

x= x(u, v, w)

y= y (u, v, w)

z= z (u, v, w)

(2.2)

tho mn

x,y,zcng vi cc o hm ring ca n l cc hm s lin tc trn min ng Vuvwca mt phngOuvw.

Cng thc2.2xc nh song nhVuvwV.

38


40/115

2. Tch phn bi ba 39

J= D(x,y,z)D(u,v,w)=0trongVuvw. Khi

I=

V

f(x,y,z) dxdydz=Vuvw

f[ x (u, v, w) ,y (u, v, w) ,z (u, v, w)] |J| dudvdw

Cng ging nh php i bin trong tch phn kp, php i bin trong tch phn bi bacng bin bin ca min Vthnh bin ca minVuvw , bin minVb chn thnh minVuvwb chn.

Bi tp 2.17. Tnh th tch minVgii hn bi

x+y+z=3x+2y z=1x+4y+z=2

bitV=

V

dxdydz.

Li gii. Thc hin php i bin

u= x +y+zv= x+2y zw= x +4y+z

. V php i bin bin bin ca V

thnh bin caVuvwnn Vuvwgii hn bi:

u=3v=1w=2

J1

=

D (u, v, w)D (x,y,z) =

1 1 1

1 2 11 4 1 =6 J= 1

6V=1

6 Vuvw

dudvdw=

1

6.6.2.4= 8

Php i bin s trong to tr

Khi minVc bin l cc mt nh mt paraboloit, mt nn, mt tr, v c hnh chiuDlnOxyl hnh trn, hoc hm ly tch phn f(x,y,z)c cha biu thc(x2 +y2)th tahay s dng cng thc i bin trong h to tr. To tr ca im M(x,y,z)l b ba

(r,,z), trong (r,)chnh l to cc ca im M l hnh chiu ca im MlnOxy.

Cng thc i bin

x= r cos

y= r sin

z= z

. nh thc Jacobian ca php bin i l J= D(x,y,z)D(r,,z)

=r,

ta c:

I=

V

f( x,y,z) dxdydz=Vrz

f(rcos, r sin,z) rdrddz

39


41/115


Nu minV :

(x,y)D

z1(x,y) z z2(x,y), trong D :

1 2

r1() r r2()th:

I=

2

1d

r2()

r1()rdr

z2(r cos,r sin)

z1(r cos,r sin)f(r cos, r sin,z) dz

Bi tp 2.18. Tnh

V

x2 +y2

dxdydz, trong V :

x2 +y2 1

1 z 2.

y

z

x

O

V1

2

Hnh 2.18

Li gii. t

x= r cos

y= r sin

z= z

th

0 2

0 r 1

1 z 2

. Ta c

I=

20

d

10

r2dr

21

zdz = ... =3

4

Bi tp 2.19. Tnh

V

z

x2 +y2dxdydz, trong :

a) Vl min gii hn bi mt tr:x2 +y2 =2xv cc mt phngz = 0,z= a (a > 0).

b) Vl na ca hnh cu x2 +y2 +z2 a2,z 0 (a > 0)

40


42/115

2. Tch phn bi ba 41

y

z

x

O

Hnh 2.19a

Li gii. a) t

x= r cos

y= r sin

z= z

. T x2 +y2 =2xsuy rar = 2 cos. Do :

2 20 r 2cos

0 z a

.

Vy

I=

2

2

d

2cos0

r2dr

a0

zdz = ... = 16a2

9

y

z

x

O

Hnh 2.19b

41


43/115


Li gii. b) t

x= r cos

y= r sin

z= z

, ta c

0 2

0 r a

0 z

a2 r2. Vy

I=

20

d

a0

r2dr

a2r20

zdz = 2

a0

r2. a2

r2

2 dr = 2a

5

15

Bi tp 2.20. Tnh I=

V

ydxdydz, trong Vgii hn bi:

y=

z2 +x2

y= h.

y

z

x

O h

Hnh 2.20

Li gii. t

x= r cos

y= r sin

z= z

, ta c

0 2

0 r h

r y h

. Vy

I=

20

d

h0

rdr

hr

ydy = 2

h0

r.h2 r2

2 dr =

h4

4

Bi tp 2.21. Tnh I=

V

x2 +y2dxdydztrong Vgii hn bi:

x2 +y2 =z2

z= 1.

42


44/115

2. Tch phn bi ba 43

y

z

x

O

Hnh 2.21

Li gii. t

x= r cos

y= r sin

z= z

, ta c

0 2

0 r 1

r z 1

. Vy

I=

20

d

10

r2dr

1r

dz = 2

10

r2 (1 r) dr = 6

Bi tp 2.22. Tnh

V

dxdydzx2+y2+(z2)2

, trong V :

x2 +y2 =1|z| 1 .

43


45/115


y

z

x

O

Hnh 2.22

Li gii. t

x= r cos

y= r sin

z =z 2|J|= r, Vrz :

0 2

0 r 1

3 z 1, ta c

I=

20

d

10

rdr

13

dzr2 +z2

=

1

0 r. lnz+ r

2 +z2 z=1z=

3dr

=2

10

r ln

r2 +1 1

dr 1

0

r ln

r2 +9 3

dr

=2(I1 I2)

Vlimr0

r ln

r2 +1 1

= limr0

r ln

r2 +9 3

= 0nn thc cht I1,I2l cc tch phnxc nh.t

r2 +1= trdr= tdt, ta c

r ln

r2 +1 1 dr

=

t ln (t 1) dt

= t2

2ln (t 1) 1

2

t2

t 1 dt

= t2 1

2 ln (t 1) t

2

4 t

2+ C

44


46/115

2. Tch phn bi ba 45

nnI1 =

t2 1

2 ln (t 1) t

2

4 t

2

|

21 =

1

2ln

2 1

141

2

2 1

Tng t, I2 = t

292 ln (t 3) t

2

4 3t2 +Cnn

I2 = t2 92 ln (t 3) t24 3t2 |103 = 12ln 10 314 32 10 3Vy

I=2(I1 I2)=

ln

2 1

10 3 +3

10 8

2

Php i bin trong to cu

Trong trng hp minVc dng hnh cu, chm cu, mi cu,. . . v khi hm ly tch

phn f(x,y,z)c cha biu thcx2 +y2 +z2th ta hay s dng php i bin trong to cu.To cu ca im M(x,y,z)trong khng gian l b ba (r, ,), trong :

r=OM

= OM , Oz

=

OM , Ox

Cng thc ca php i bin l:

x= r sin cos

y= r sin sin

z= r cos

.

nh thc Jacobian J= D(x,y,z)D(r,,)

=r2 sin , ta c:V

f( x,y,z) dxdydz=Vr

f(r sin cos, r sin sin, r cos ) r2 sin drdd

c bit, nuVr :

1 2, (2

1 2)

1() 2()

r1(,) r r2(,)

th

I=

21

d

2()1()

sin d

r2(,)r1 (,)

f(r sin cos, r sin sin, r cos )r2dr

45


47/115


Bi tp 2.23. Tnh

V

x2 +y2 +z2

dxdydz, trong V :

1 x2 +y2 +z2 4

x2 +y2 z2

y

z

x

O

V1

Hnh 2.23

Li gii. t

x= r sin cos

y= r sin sin

z= r cos

. Do1 x2 +y2 +z2 4nn 1 r 2; trn mt nn c

phng trnhx2 +y2 =z2 nn = 4 . Vy0 2

0

4

1 r 2

nn

I=220

d

4

0

sin d2

1

r2.r2dr = 2.2. ( cos ) 40 . r55 21 = 4.315 1 22

Bi tp 2.24. Tnh

V

x2 +y2 +z2dxdydztrong V : x2 +y2 +z2 z.

46


48/115

2. Tch phn bi ba 47

y

z

x

O

Hnh 2.24

Li gii. t

x= r sin cos

y= r sin sin

z= r cos

. Nhn hnh v ta thy0 2, 0 2 .

Dox2 +y2 +z2 znn0 r cos . Vy

I=

20

d

2

0

sin d

cos 0

r.r2dr = 2.

2

0

sin .1

4cos4 d =

10

Php i bin trong to cu suy rng.

Tng t nh khi tnh tch phn kp, khi minVc dng hnh ellipsoit hoc hnh cuc tm khng nm trn cc trc to th ta s s dng php i bin s trong to cusuy rng. Khi ta phi tnh li Jacobian ca php bin i.

1. Nu minVc dnghnh ellipsoit hoc hnh cu c tm khng nm trn cc trc tonn ngh ti php i bin s trong to cu suy rng.

2. NuV : x2a2

+ y2

b2+ z

2

c2 =1th thc hin php i bin

x= arsin cos

y= brsin sin

z= cr cos

,J=abcr2 sin

47


49/115


NuV :(x a)2 +(y b)2 +(z c)2 = R2 th thc hin php i bin

x= a +r sin cos

y= b+r sin sin

z= c+r cos

,J=r2 sin

3. Xc nh min bin thin ca, , r.

4. Dng cng thc i bin tng qut hon tt vic i bin.

Bi tp 2.25. Tnh

V

z

x2 +y2dxdydz, trong Vl na ca khi ellipsoit x2+y2

a2 + z

2

b2

1,z 0, (a, b > 0)

Li gii. Cch 1: S dng php i bin trong to tr suy rng.t

z= bz

x= arcos

y= arsin

J= D(x,y,z)D(r,,z)

=a2br, Vrz =

0 2, 0 r 1, 0 z

1 r2

Vy

I=

2

0d

1

0dr

1r2

0bz.ar.a2brdz =2a3b2

1

0r2.

1 r22

dr =2a3b2

15

Cch 2: S dng php i bin trong to cu suy rng.t

x= arsin cos

y= ar sin sin

z= br cos

J= D (x,y,z)D(r, ,)

=a2br2 sin , Vrz =

0 2, 0

2, 0 r 1

Vy

I=

20

d

2

0

d

10

br cos .arsin .a2b sin =2a3b2

20

cos sin2 d

10

r4dr = 2a3b2

15

Bi tp 2.26. Tnh

V

x2

a2+ y

2

b2+ z

2

c2

dxdydz, V : x

2

a2+ y

2

b2+ z

2

c2 1, (a, b, c > 0).

48


50/115

2. Tch phn bi ba 49

Li gii. t

x= arsin cos

y= brsin sin

z= cr cos

J= D (x,y,z)D(r, ,)

=abcr2 sin , Vrz ={0 2, 0 , 0 r 1}

Vy

I= abc

20

d

0

d

10

r2.r2 sin =4

5 abc

49


51/115


3. CC NG DNG CA TCH PHN BI

3.1 Tnh din tch hnh phng

Cng thc tng qut: S= D

dxdy

Bi tp 2.27. Tnh din tch ca minDgii hn bi:

y= 2x

y= 2x

y= 4

.

xO

1

y

4y= 2xy= 2x

Hnh2.27

Li gii. Nhn xt:

D= D1 D2,D12 x 0

2x y 4,D2

0 x 22x y 4

nnS=

D

dxdy=D1

dxdy+D2

dxdy= 2D1

dxdy= ... = 2

8 3

ln 2

50


52/115

3. Cc ng dng ca tch phn bi 51

Bi tp 2.28. Tnh din tch ca minDgii hn bi:

y2 = x ,y2 =2x

x2 =y,x2 =2y

x

y

O

y= x2 x2 =2y

x= y2

2x= y2

Hnh2.28

Li gii. Ta cS =D

dxdy. Thc hin php i bin

u=

y2

xv=

x2

y

Duv : 1 u 21 v 2

,

th

J1 = D(u, v)D(x,y)

=

y2

x22yx

2xy x

2

y2

=3Vy

S= Duv

13

dudv= 13

Bi tp 2.29. Tnh din tch min Dgii hn bi

y= 0,y2 =4ax

x+y= 3a,y 0 (a > 0).

51


53/115


y

O x

3a

3a

6a

Hnh2.29

Li gii. Nhn hnh v ta thy D : 6a y 0y24a x 3a y

nn

S=D

dxdy=

06a

dy

3ayy2

4a

dx =

06a

3a y y

2

4a

dy = 18a2

Bi tp 2.30. Tnh din tch min Dgii hn bi

x2 +y2 =2x,x2 +y2 =4x

x= y,y= 0.

y= x

x2 4

y

O

Hnh2.30

Li gii. Ta cS =D

dxdy, t

x= r cos

y= r sinthD :

0

4

2cos r 4cosnn

S=

4

0

d

4cos2cos

rdr = 1

2

4

0

12cos2 d=3

4 +

3

2

52


54/115


Bi tp 2.31. Tnh din tch min Dgii hn bi ng trn r = 1, r= 23cos.

Ch :

r= al phng trnh ng trn tm O(0, 0), bn knha.

r= a cosl phng trnh ng trn tm (a, 0), bn knha.

x

y

O

Hnh2.31

Li gii. Giao ti giao im ca 2 ng trn:

r= 1 = 2

3cos=

6

nn

S= 2

6

0

d

23

cos

1

rdr = 2.

1

2

6

043cos2 1 d=

3

6

18

Bi tp 2.32. Tnh din tch minDgii hn bi ng

x2 +y22

=2a2xy (a > 0)(ng)

x

y

O

r= a

sin2

Hnh2.32

53


55/115


Li gii. Tham s ho ng cong cho, t

x= r cos

y= r sin, phng trnh ng cong

tng ng vir2 = a2 sin2. Kho st v v ng cong cho trong h to cc (xemhnh v2.32). Ta c

D: 0

2,

3

20 r a

sin2

Do tnh i xng ca hnh v nn

S= 2

2

0

d

a

sin20

rdr =

2

0

a2 sin2d= a2

Bi tp 2.33. Tnh din tch minDgii hn bi ngx3 +y3 = axy (a > 0)(L Descartes)

x

y

O 1

2

12

TCX: y =x 13Hnh2.33

Tham s ho ng cong cho, t

x= r cos

y= r sin, phng trnh ng cong tng ng

vir= a sin cos

sin3 +cos3

Kho st v v ng cong cho trong h to cc (xem hnh v 2.33). Ta c

D:

0

2

0 r a sin cos

sin3 +cos3

54


56/115


nn

S=

2

0

d

a sin cos

sin3 +cos3 0

rdr =a2

2

2

0

sin2 cos2

sin3 +cos3

2

dt=tg

= a2

2.

1

3

+0

d

t3 +1

(t3 +1)2

= a2

6

Bi tp 2.34. Tnh din tch min Dgii hn bi ng r = a (1+cos) (a > 0), (ngCardioids hay ng hnh tim)

x

y

O2a

a

a

Hnh2.34

Li gii. Ta cD={0 2, 0 r a (1+cos)}

nn

S= 2

0

d

a(1+cos)0

rdr = a2

0

(1+cos)2 d= ... = 3a22

3.2 Tnh th tch vt thCng thc tng qut:

V=

V

dxdydz

Cc trng hp c bit

1. Vt th hnh tr, mt xung quanh l mt tr c ng sinh song song vi trcOz, y l min D trong mt phng Oxy, pha trn gii hn bi mt cong z =f( x,y) ,f(x,y) 0v lin tc trn Dth V =

D

f( x,y) dxdy. (Xem hnh v di

y).

55


57/115


y

z

x

O

z= f(x,y)

D

2. Vt th l khi tr, gii hn bi cc ng sinh song song vi trc Oz, hai mtz = z1(x,y) ,z = z2(x,y). Chiu cc mt ny ln mt phng Oxyta c min D,z1(x,y) ,z2(x,y)l cc hm lin tc, c o hm ring lin tc trn D . Khi :

V=D

|z1(x,y) z2(x,y)|dxdy

y

z

x

O

z= f(x,y)

z= g(x,y)

D

Bi tp 2.35. Tnh din tch min gii hn bi

3x+y 1

3x+2y 2

y 0, 0 z 1 x y.

56


58/115


y

z

x

O

Hnh2.35Li gii.

V=D

f(x,y) dxdy=

10

dy

22y3

1y3

(1 x y) dx = 16

10

1 2y+y2

dy =

1

18

Bi tp 2.36. Tnh th tch ca minVgii hn bi z= 4 x2 y2

2z= 2+x2

+y2.

y

z

x

O

2z= 2+x2 +y2

z= 4 x2 y2

Hnh2.36

57


59/115


Li gii. Giao tuyn ca hai mt cong:

x2 +y2 =2

z= 2, nn hnh chiu caVln mt phng

OxylD : x2 +y2 2. Hn na trnDth 4 x2 y2 2+x2+y22 nn ta c:

V= D4 x2 y2 2+x

2 +y2

2 dxdyt

x= r cos

y= r sinth

0 2

0 r

2, do

V=

20

d

2

0

3 3

2r2

rdr = ... = 3

Bi tp 2.37. Tnh th tch ca V :

0 z 1 x2 y2y x,y

3x

.

y

z

x

O 1

1

Hnh2.37

Li gii. Doxy 3xnn x ,y0. Ta c

V=D

1 x2 y2

dxdy

58


60/115


t

x= r cos

y= r sinth

4

3

0 r 1. Vy

V=

3

4

d

1

01 r2 rdr = . . .= 48

Bi tp 2.38. Tnh th tchV :

x2 +y2 +z2 4a2

x2 +y2 2ay 0 .

y

z

x

O

2a

2a2a

Hnh2.38

Li gii. Do tnh cht i xng ca minVnn

V=4D

4a2 x2 y2dxdy,

trongDl na hnh trnD :

x2 +y2 2ay 0x 0

.t

x= r cos

y= r sin 0

2

0 r 2a sin

59


61/115


Vy

V=4

2

0

d

2a sin0

4a2 r2rdr

=4. 12

2

0

234a2 r2 32 r=2a sinr=0 d

=4

3

2

0

8a3 8a3 cos3

d

=32a3

3

2 2

3

Bi tp 2.39. Tnh th tch ca minVgii hn bi

z= 0

z= x2

a2+

y2

b2

x2

a2+

y2

b2 =

2x

a

.

x

z

O

z= x2

a2+

y2

b2

a

1

Hnh2.39

Li gii. Ta c hnh chiu caVln mt phngOxyl minD : x2

a2+

y2

b2 = 2xa. Do tnh cht

i xng ca minVnn:

V=2D+

x2

a2+

y2

b2

dxdy,

60


62/115


trong D+ l na ellipseD + : x2

a2+ y

2

b2 = 2xa ,y 0

t

x= arcos

y= brsinth |J|= abr,

0

2

0 r 2cos. Vy

V=2

2

0

d

2cos0

r2rdr =... = 32

Bi tp 2.40. Tnh th tch ca minV :

az = x2 +y2

z=

x2 +y2.

y

z

O aa

a

Hnh2.40

Li gii. Giao tuyn ca hai ng cong:

z=

x2 +y2 = x2 +y2

a

x2 +y2 =a2

z= a

Vy hnh chiu caVln mt phngOxyl

D: x2 +y2 =a2

Nhn xt rng, trong minDth mt nn pha trn mt paraboloit nn:

V=D

x2 +y2 x

2 +y2

a

dxdy

61


63/115


t

x= r cos

y= r sinth

0 2

0 r a. Vy

V=

2

0d

a

0 r r

2

a rdr = ... =

a3

6

3.3 Tnh din tch mt congMt z = f( x,y)gii hn bi mt ng cong kn, hnh chiu ca mt cong ln mt

phngOxylD. f( x,y)l hm s lin tc, c cc o hm ring lin tc trn D . Khi :

=D

1+p2 +q2dxdy,p= fx, q= fy

y

z

x

O

z= f(x,y)

D

62


64/115

CHNG

3TCH PHN PH THUC THAM S.1. TCH PHN XC NH PH THUC THAM S.

1.1 Gii thiu

Xt tch phn xc nh ph thuc tham s: I(y) =b

a

f( x,y) dx, trong f( x,y)kh

tch theo x trn [a, b]vi miy [c, d]. Trong bi hc ny chng ta s nghin cu mt stnh cht ca hm sI(y)nh tnh lin tc, kh vi, kh tch.

1.2 Cc tnh cht ca tch phn xc nh ph thuctham s.

1) Tnh lin tc.

nh l 3.7. Nu f( x,y)l hm s lin tc trn [a, b] [c, d]th I(y)l hm s lintc trn[c, d]. Tc l:

limyy0

I(y)= I(y0) limyy0

ba

f( x,y) dx =b

a

f(x,y0) dx

2) Tnh kh vi.

nh l 3.8. Gi s vi miy[c, d], f( x,y)l hm s lin tc theoxtrn[a, b]vf

y(x,y)l hm s lin tc trn [a, b] [c, d]th I(y)l hm s kh vi trn (c, d)v

63


65/115

64 Chng 3. Tch phn ph thuc tham s.

I(y)=b

a

f

y(x,y) dx , hay ni cch khc chng ta c th a du o hm vo trong

tch phn.

3) Tnh kh tch.nh l 3.9. Nu f(x,y)l hm s lin tc trn [a, b] [c, d]th I(y)l hm s khtch trn[c, d], v:

dc

I(y) dy :=

dc

ba

f(x,y) dx

dy = ba

dc

f(x,y) dy

dxBi tp

Bi tp 3.1. Kho st s lin tc ca tch phn I(y) =1

0

y f(x)x2+y2

dx , vi f(x)l hm s

dng, lin tc trn[0, 1].

Li gii. Nhn xt rng hm sg (x,y)= y f(x)x2+y2

lin tc trn mi hnh ch nht[0, 1] [c, d]v [0, 1] [d, c]vi 0 < c < dbt k, nn theo nh l 3.7, I(y) lin tc trn mi[c, d] , [d, c], hay ni cch khc I(y)lin tc vi miy=0.By gi ta xt tnh lin tc ca hm sI(y)ti imy = 0. Do f(x)l hm s dng, lintc trn[0, 1]nn tn ti m > 0sao cho f( x) m > 0

x

[0, 1]. Khi vi > 0th:

I()=

10

f(x)

x2 + 2dx

10

.m

x2 + 2dx = m.arctg

x

I()=1

0

f(x)x2 + 2

dx

10

.mx2 +2

dx =m.arctg x

Suy ra|I() I()| 2m.arctg x 2m.2khi0, tc l|I() I()|khng tin ti0 khi

0, I(y)gin on tiy = 0.


a) In()=1

0

x lnn xdx, n l s nguyn dng.

Li gii. Vi mi > 0, hm s fn(x, )= x lnn x, n = 0,1, 2, ...lin tc theo xtrn[0, 1]

64


66/115

1. Tch phn xc nh ph thuc tham s. 65

V limx0+

x lnn+1 x= 0nn fn(x,) = x lnn+1 xlin tc trn[0, 1] (0, +).

Ngha l hm s fn(x, )= x lnn xtho mn cc iu kin ca nh l 3.8nn:

In1()=

d

d

1

0x lnn1 xdx=

1

0d

d x lnn1 x

dx =

1

0x lnn xdx= In()

Tng t, In2 = In1, ...,I2 = I1,I

1 = I0 , suy ra In() = [I0()]

(n). M I0() =1

0

xdx = 1+1 In()=

1+1

(n)= (1)

nn!

(+1)n+1.

b)

2

0

ln

1+ y sin2 x

dx, viy > 1.

Li gii. Xt hm s f( x,y)= ln 1+ y sin2 xtho mn cc iu kin sau: f(x,y)= ln 1+ y sin2 xxc nh trn0, 2 (1, +)v vi miy >1cho

trc, f(x,y)lin tc theo xtrn

0, 2.

Tn ti fy(x,y)= sin2 x

1+y sin2 xxc nh, lin tc trn

0, 2

(1, +).Theo nh l3.8, I(y)=

2

0

sin2 x1+y sin2 x

dx =

2

0

dx1

sin2 x+y

.

tt =tgxthdx= dt1+t2

, 0 t + .

I(y)=+0

t2dt

(t2 +1) (1+t2 +yt2)=

+0

1

y

1

t2 +1 1

1+(y+1) t2

dt

=1

y

arctgt|+0

1y+1

arctg

t

y+1

|+0

=

2y

1 1

1+y

=

2

1+y.

1

1+

1+y

Suy ra

I(y)=

I(y) dy =

2

1+y.

1

1+

1+ydy = ln

1+

1+y

+C

Do I(0)= 0nnC =ln 2vI(y)= ln 1+1+y ln 2.Bi tp 3.3. Xt tnh lin tc ca hm s I(y)=

10

y2x2(x2+y2)

2 dx.

65


67/115


Li gii. Tiy = 0, I(0)=1

0

1x2

dx =, nn hm s I(y)khng xc nh ti y = 0.

Tiy= 0, I(y) =1

0

(x2+y2)2x.x(x2+y2)

2 dx =

10

d

xx2+y2

= 1

1+y2, nn I(y)xc nh v lin tc

vi miy=0.

1.3 Cc tnh cht ca tch phn ph thuc tham s vicn bin i.

Xt tch phn ph thuc tham s vi cn bin i

J(y)=

b(y)

a(y)f(x,y) dx, viy[c, d] , a a (y) , b (y) by[c, d]

1) Tnh lin tc

nh l 3.10. Nu hm s f(x,y)lin tc trn [a, b] [c, d], cc hm sa (y) , b (y)lin tc trn [c, d] v tho mn iu kin a a (y) , b (y) by [c, d]th J(y) lmt hm s lin tc i viytrn[c, d].

2) Tnh kh vi

nh l 3.11. Nu hm s f( x,y) lin tc trn [a, b] [c, d] , fy(x,y) lin tc trn[a, b]

[c, d], va (y) , b (y)kh vi trn[c, d]v tho mn iu kin a a (y) , b (y)

by[c, d]thJ(y)l mt hm s kh vi i vi y trn [c, d], v ta c:

J(y)=b(y)

a(y)

f

y(x,y) dx+ f(b (y) ,y) b

y(y) f(a (y) ,y)ay(y)

.

Bi tp

Bi tp 3.4. Tm limy0

1+y

y

dx

1+x2+y2 .

Li gii. D dng kim tra c hm sI(y)=1+yy

dx1+x2+y2

lin tc tiy = 0da vo nh

l3.10, nnlimy0

1+yy

dx1+x2+y2

= I(0)=

10

dx1+x2

= 4 .

66


68/115

2. TCH PHN SUY RNG PH THUC THAM S.

2.1 Cc tnh cht ca tch phn suy rng ph thuctham s.

Xt tch phn suy rng ph thuc tham s I(y)=+a

f(x,y)dx, y [c, d]. Cc kt qu

di y tuy pht biu i vi tch phn suy rng loi II (c cn bng v cng) nhng uc th p dng mt cch thch hp cho trng hp tch phn suy rng loi I (c hm didu tch phn khng b chn).

1) Du hiu hi t Weierstrass

nh l 3.12. Nu|f( x,y)| g (x) (x,y) [a, +] [c, d] v nu tch phn suyrng

+a

g (x) dx hi t, th tch phn suy rngI(y) =

+a

f( x,y)dx hi t u i vi

y[c, d].

2) Tnh lin tc

nh l 3.13. Nu hm s f( x,y)lin tc trn[a, +] [c, d]v nu tch phn suy

rng I(y)=

+

a f(x,y)dx hi t u i viy [c, d]th I(y)l mt hm s lin tctrn[c, d].

3) Tnh kh vi

nh l 3.14. Gi s hm s f(x,y)xc nh trn[a, +] [c, d]sao cho vi miy[c, d], hm s f(x,y)lin tc i vix trn[a, +]v f

y(x,y)lin tc trn[a, +]

[c, d]. Nu tch phn suy rngI(y)=

+a

f(x,y)dx hi t v

+a

f

y(x,y)dx hi t u

i viy[c, d]thI(y)l hm s kh vi trn [c, d]vI(y)=+a

f

y(x,y) dx.

4) Tnh kh tch

nh l 3.15. Nu hm s f( x,y)lin tc trn[a, +] [c, d]v nu tch phn suyrng I(y)hi t u i viy [c, d]th I(y)l hm s kh tch trn [c, d]v ta c

67


69/115


th i th t ly tch phn theo cng thc:

dc

I(y) dy :=

dc

+a

f( x,y) dx

dy = +a

dc

f( x,y) dy

dx.

2.2 Bi tpDng 1. Tnh tch phn suy rng ph thuc tham s bng cch i th t ly

tch phn

Gi s cn tnh I(y)=+a

f(x,y)dx.

B1. Biu din f(x,y)=d

c F(x,y) dy.B2. S dng tnh cht i th t ly tch phn:

I(y)=

+a

f( x,y)dx =

+a

dc

F(x,y) dy

dx = dc

+a

F(x,y) dx

dyCh :Phi kim tra iu kin i th t ly tch phn trong nh l3.15i vi tchphn suy rng ca hm s F(x,y).


a)1

0

xbxaln x dx, (0 < a < b).

Li gii. Ta c:

xb xalnx

= F(x, b) F(x, a) =b

aF

y(x,y) dy=

b

axydy; F(x,y):=

xy

lnxnn:

10

xb xalnx

dx =

10

ba

xydy

dx = ba

10

xydx

dy = ba

1

y+1dy = ln

b+1

a+1

Kim tra iu kin v i th t ly tch phn:

68


70/115

2. Tch phn suy rng ph thuc tham s. 69

b)+0

exexx dx, (, > 0).

Li gii. Ta c:

ex

ex

xF(x,y):=eyxx = F(x, ) F(x,) =

Fy(x,y)=

eyx dy

nn:

+0

ex exx

dx =

+0

eyx dy

dx =

+0

eyx dx

dy =

dy

y =ln

.


c)+

0

ex2ex2x2

dx, (, > 0).

Li gii. Ta c:

ex2 ex2x2

F(x,y):=e

yx2x2

= F(x, ) F(x,) =

F

y(x,y) dy=

eyx2dy

nn:+0

ex2

ex2

x2 dx =

+0

ex2ydy dx =

+0

ex2ydx dyVi iu kin bit

+0

ex2 dx =

2 ta c+0

ex2ydx =

2

y .

Suy ra I=

2

y dy =

.Kim tra iu kin v i th t ly tch phn:

e)+0

eaxsin bxsin cxx , (a, b, c > 0).

Li gii. Ta c:

eaxsin bx sin cx

x

F(x,y)=

eax sinyxx

= F(x, b) F(x, c)=

bc

F

y(x,y) dy =

bc

eax cosyxdx

69


71/115


nn:

I=

+0

bc

eax cosyxdy

dx = bc

+0

eax cosyxdx

dy

M eax cosyxdx = a

a2+y2 eax

cosyx +

y

a2+y2 eax

sinyx,suyra

+

0

eax

cosyxdx =

a

a2+y

v I=b

c

aa2+y2

dy = arctg ba arctg ca .


Dng 2. Tnh tch phn bng cch o hm qua du tch phn.

Gi s cn tnh I(y)=+a

f(x,y)dx.

B1. Tnh I(y)bng cch I(y)=+a

f

y(x,y) dx.

B2. Dng cng thc Newton-Leibniz khi phc liI(y)bng cch I(y)=

I(y) dy.

Ch :Phi kim tra iu kin chuyn du o hm qua tch phn trong nh l3.14.

Bi tp 3.6. Chng minh rng tch phn ph thuc tham sI(y)=+

arctg(x+y)

1+x2 dxl mt

hm s lin tc kh vi i vi biny. Tnh I(y)ri suy ra biu thc ca I(y).

Li gii. Ta c:

f( x,y)= arctg(x+y)1+x2

lin tc trn[, +] [, +].

arctg(x+y)1+x2

2 .

11+x2

, m+

11+x2

= hi t, nn I(y)=+

arctg(x+y)1+x2

dxhi t u

trn[, +].Theo nh l3.13, I(y)lin tc trn[, +].

Hn nafy(x,y) = 1(1+x2)[1+(x+y)2] 11+x2 ,y; do

+

f

y(x,y)dx hi t u trn

[, +]. Theo nh l3.14,I(y)kh vi trn[, +], v: I(y)=+

1

(1+x2)[1+(x+y)2]dx.

70


72/115


t 1(1+x2)[1+(x+y)2]

= Ax+B1+x2

+ Cx+D1+(x+y)2

, dng phng php ng nht h s ta thu c:A=2

y(y2+4),B= 2

y(y2+4), C= 1

y2+4,D= 3

y2+4. Do :

I(y)= 1

y2 +4

+

2x+y

1+x2 +

2x+3y

1+(x+y)2=

1

y2 +4

ln

1+x2

+y arctgx+ln

1+(x+y)2

+y arctg (x+y)

|+x=

= 4

y2 +4

Suy ra I(y) =

I(y) dy = 2 arctgy2 +C, mt khc I(0)=+

arctg x1+x2

dx = 0nn C = 0v

I(y)= 2 arctgy2


a)1

0

xbxaln x dx, (0 < a < b).

Li gii. t I(a)=1

0

xbxaln x dx,f( x, a)=

xbxaln x . Ta c:

f(x, a)= xbxaln x lin tc trn theoxtrn[0, 1]vi mi0 < a < b. fa(x, a)=xa lin tc trn[0, 1] (0, +).

1

0

fa(x, a)dx =

10

xadx = 1a+1hi t u trn[0, 1]v n l TPX.

Do theo nh l3.14,

I(a)=1

0

fa(x, a) dx =

1

a+1 I(a)=

I(a) da = ln (a+1)+C.

Mt khc I(b)= 0nnC = ln (b+1)v do I(a)= ln b +1a+1 .

b)+0

exexx dx, (, > 0).

Li gii. t I()=+0

exexx dx,f( x, )=

exexx . Ta c:

71


73/115


f(x, )= exexx lin tc theoxtrn[0, +)vi mi, > 0. f(x, )=ex lin tc trn[0, +) (0, +).

+0

f(x, ) dx =

+0

exdx =1hi t u i vi trn mi khong[, +)

theo tiu chun Weierstrass, tht vy,|ex| ex, m+0

exdx = 1 hi t.

Do theo nh l3.14,

I()=+0

f(x, ) dx =

1

I()=

I() d= ln +C.

Mt khc, I()= 0nnC = lnv I=ln .

c)+0

ex2ex2x2

dx, (, > 0).

Li gii. t I()=+0

ex2ex2x2

dx,f(x, )= ex2ex2

x2 . Ta c:

f(x, )= ex2ex2x2

lin tc theoxtrn[0, +)vi mi, > 0.

f

(x, )=

ex2 lin tc trn[0, +)

(0, +).

+0

f(x, ) dx =

+0

ex2 dx x=y

= +0

ey2 dy

=

2 . 1

hi t u theo

trn mi [, +) theo tiu chun Weierstrass, tht vy,ex2 ex2 m

+0

ex2 dxhi t.

Do theo nh l3.14,

I()=+

0

f(x, ) dx =2 .

1 I()= I() d=.+C.

Mt khc, I()= 0nnC =

.v I()=

.d)

+0

dx

(x2+y)n+1

72


74/115


Li gii. t In(y)=+0

dx

(x2+y)n+1 ,fn(x,y)=

1

(x2+y)n+1 . Khi :

[In

1(y)]y =

+

0dx

(x2

+y)n

y

=

n

+

0dx

(x2 +y)n+1

=

n.In(y)

In=

1

n

( In

1)

.

Tng t, In1= 1n1( In2)

,In2= 1n2( In3)

, ...,I1= (I0).

Do , In(y) = (1)n

n! [I0(y)](n). M I0(y) =

+0

1x2+y

dx = 1y arctg x

y |+0 = 2y nn

In(y)=

2 .(2n1)!!

(2n)!! . 1

y2n+1.

Vn cn li l vic kim tra iu kin chuyn o hm qua du tch phn.

Cc hm s f(x,y) = 1

x2+y ,f

y(x,y) = 1

(x2+y)2 , ...,f

(n)

yn (x,y) =

(

1)n

(x2+y)n+1 lin tctrong[0, +) [, +)vi mi > 0cho trc. 1

x2+y 1

x2+,

1(x2+y)2 1(x2+)2 , ...,

(1)n(x2+y)n+1 1(x2+)n+1

M cc tch phn+0

1x2+

dx, ...,

+0

1

(x2+)n+1 dxu hi t, do

+0

f(x,y) dx,

+0

f

y(x,y) dx,...,

+0

f(n)

yn (x,y) dxhi t u trn[, +)vi mi >

0.

e)+0

eaxsin bxsin cxx dx (a, b, c > 0) .

Li gii. t I(b)=+0

eaxsin bxsin cxx dx,f( x, b)= eaxsin bxsin cx

x . Ta c:

f(x, b)= eaxsin bxsin cxx lin tc theoxtrn[0, +)vi mia, b, c > 0. fb(x, b)= eax cos bxlin tc trn[0, +) (0, +).

+0

fb(x, b) dx =

+0

eax cos bx = a

a2+b2eax cos bx+ b

a2+b2eax sin bx

+0 = aa2+b2hi t u theobtrn mi(0, +)theo tiu chun Weierstrass, tht vy,

|eax cos bx| eax2 m+0

eax2 dxhi t.

73


75/115


Do theo nh l3.14, Ib(x, b)= aa2+b2

, I=

aa2+b2

db = arctg ba+C.

Mt khc I(c)= 0nnC =arctg cavI=arctg ba arctg ca .

f)

+

0

ex2

cos (yx) dx.

Li gii. t I(y)=+0

ex2 cos (yx) dx,f(x,y)= ex2 cos (yx) .Ta c:

f(x,y)lin tc trn[0, +) (, +).

f

y(x,y)=

xex2 sinyxlin tc trn[0, +)

(

, +).

+0

f

y(x,y) dx =

+0

xex2 sinyxdx = 12 ex2

sinyx+0 12 +

0

yex2 cosyxdx = y2 I

hi t u theo tiu chun Weierstrass, tht vy,fy(x,y) xex2, m +

0

xex2 dx

12hi t.

Do theo nh l3.14, I(y)

I(y)

=

y2

I=Ce

y2

4.

M I(0)= C =

2 nn I(y)=

2 ey24.

Nhn xt:

Vic kim tra cc iu kin o hm qua du tch phn hay iu kin i th tly tch phn i khi khng d dng cht no.

Cc tch phn +0

f(x, ) dx cu b, c, d ch hi t u trn khong[, +)vi mi

>0, m khng hi t u trn(0, +). Tuy nhin iu cng khng nh

rng I =+0

f(x, ) dxtrn(0, +).

74


76/115

3. Tch phn Euler 75

3. TCH PHNEULER

3.1 Hm Gamma

(p)=

+

0

xp1exdxxc nh trn (0, +)

Cc cng thc

1. H bc: (p+1)= p (p) , ( n)= (1)n()(1)(2)...(n) .

ngha ca cng thc trn l nghin cu (p)ta ch cn nghin cu (p)vi0 < p 1m thi, cn vip > 1chng ta s s dng cng thc h bc.

2. c bit, (1)= 1nn (n)=(n 1)!nN.

12=

nn n+

12=

(2n1)!!22

.

3. o hm ca hm Gamma: (k) (p)=+0

xp1

lnk x

.exdx.

4. (p) . (1 p)= sinp0 < p < 1.

3.2 Hm Beta

Dng 1:B (p, q)=1

0

xp1 (1 x)q1dx.

Dng 2:B (p, q)=+0

xp1(1+x)p+q

dx.

Dng lng gic:B (p, q) = 2

2

0

sin2p1 t cos2q1 tdt, B

m+12 ,

n +12

= 2

2

0

sinm t cosm tdt.

Cc cng thc:

1. Tnh i xng: B (p, q)=B (q,p).2. H bc: B (p, q)=

p1p+q1B (p 1, q) , nup > 1

B (p, q)= q1p+q1B (p, q 1) , nuq > 1

ngha ca cng thc trn ch mun nghin cu hm bta ta ch cn nghin cun trong khong(0, 1] (0, 1]m thi.

75


77/115


78/115


79/115


g)1

0

1n1xn dx, nN

Li gii. tx n =tdx= 1n t1n 1dt

I=1

0

1n t

1n 1dt

(1 t) 1n=

1

n

10

t1n 1. (1 t) 1n dt = 1

nB 1

n, 1 1

n

=

1

n

sin n

78


80/115

CHNG

4TCH PHN NG1. TCH PHN NG LOII

1.1 nh ngha

Cho hm s f(x,y)xc nh trn mt cung phngAB . Chia cungAB thnh ncungnh, gi tn v di ca chng ln lt l s1,s2,...sn.Trn mi cung sily mt imMibt k. Gii hn, nu c, ca tng

n

i=1

f(Mi)sikhi nsao cho max si 0khng

ph thuc vo cch chia cungABv cch chn cc im Mic gi l tch phn ngloi mt ca hm s f( x,y)dc theo cungAB, k hiu l AB

f(x,y) ds.

Ch :

Tch phn ng loi mt khng ph thuc vo hng ca cungAB. Nu cung

AB c khi lng ring ti M(x,y) l (x,y) th khi lng ca n l

AB (x,y) ds. nu tch phn tn ti.

Chiu di ca cungABc tnh theo cng thc l = AB

ds.

Tch phn ng loi mt c cc tnh cht ging nh tch phn xc nh.

79


81/115

80 Chng 4. Tch phn ng

1.2 Cc cng thc tnh tch phn ng loi I1. Nu cungABcho bi phng trnhy = y (x) , a x bth

AB f(x,y) ds =b

a f(x,y (x))1+y2 (x)dx. (1)2. Nu cungABcho bi phng trnh x = x (y) , c y dth

AB

f(x,y) ds =

dc

f(x (y) ,y)

1+x2 (y)dy. (2)

3. Nu

ABcho bi phng trnh x = x(t),y= y(t), t1tt2, th

AB

f(x,y)ds =t2

t1

f(x(t),y(t))

x2(t) +y2(t)dt (3)

4. Nu cungABcho bi phng trnh trong to cc r = r() ,1 2th coi nnh l phng trnh di dng tham s, ta c ds=

r2 ()+r2 ()dv

AB

f(x,y) ds =

21

f(r() cos, r () sin)

r2 ()+r2 ()d (4)

1.3 Bi tp

Bi tp 4.1. TnhC

(x y) ds, Cl ng trn c phng trnhx2 +y2 =2x.

Li gii. t

x= 1+cos t

y= sin t, 0 t 2

I=20

(1+cos t sin t)

( sin t)2 +cos2 tdt = 2

Bi tp 4.2. TnhC

y2ds, Cl ng cong

x= a (t sin t)y= a (1 cos t) , 0 t 2, a > 0.80


82/115

1. Tch phn ng loi I 81

Li gii. x(t)= a (1 cos t)y(t)= a sin t

x2 (t)+y2 (t)= 2a sin t

2

I=

2

0

a2

(1 cos t)2

.2a sin

t

2 dt =

256a3

15 .

Bi tp 4.3. TnhC

x2 +y2ds, Cl ng

x= a (cos t+t sin t)y= a (sin t t cos t) , 0 t 2, a > 0.Li gii.

x(t)= at cos t

y(t)= at sint

x2 (t)+y2 (t)= at

I=20

a2

(cos t+t sin t)2 +(sin t t cos t)2

.atdt= a3

3

(1+42)

3 1

81


83/115


2. TCH PHN NG LOIII

2.1 nh nghaChohaihmsP (x,y) , Q (x,y)xc nh trn cungAB.ChiacungABthnh ncung nhsibi cc im chiaA0= A,A1,A2, ...,An =B.Gi to ca vect Ai1Ai =(xi,yi)v

ly imMibt k trn mi cungsi.Giihn,nuc,catng n

i=1

[P (Mi)xi+Q (Mi)yi]

sao chomax xi 0, khng ph thuc vo cch chia cungABv cch chn cc im Mic gi l tch phn ng loi hai ca cc hm sP (x,y) , Q (x,y)dc theo cungAB, khiu l

AB

P (x,y) dx+Q (x,y) dy.

Ch :

Tch phn ng loi hai ph thuc vo hng ca cungAB,nuichiutrnngly tch phn th tch phn i du,

ABP (x,y) dx+Q (x,y) dy =

BAP (x,y) dx+Q (x,y

Tch phn ng loi hai c cc tnh cht ging nh tch phn xc nh.

2.2 Cc cng thc tnh tch phn ng loi II1. Nu cung

AB c cho bi phng trnh y = y (x), im u v im cui ng vi

x= a,x= bth AB

Pdx +Qdy =

ba

P (x,y (x))+Q (x,y (x)) .y(x)

dx. (5)

2. Nu cungAB c cho bi phng trnh x = x(y), im u v im cui ng viy= c,y= dth

AB

Pdx+Qdy =

dc

P

x(y) .x(y)

dy,y

+Q (x(y) ,y). (6)

3. Nu cungABc cho bi phng trnh x= x (t)y= y (t) , im u v im cui tngng vit = t1, t= t2th

ABPdx +Qdy =

t2t1

P (x (t) ,y (t)) .x(t)+Q (x(t) ,y (t))y(t)

dt (7)

82


84/115

2. Tch phn ng loi II 83

Bi tp

Bi tp 4.4. Tnh

AB

x2 2xy dx+ 2xy y2 dy, trong AB l cung paraboly = x2 tA (1, 1)n B (2, 4).

Li gii. p dng cng thc (5) ta c:

I=

21

x2 2x3

+

2x3 x4

.2x

dx =4130

.

Bi tp 4.5. TnhC

x2 2xy dx + 2xy y2 dytrong Cl ng cong

x= a(t sin t)y= a(1 cos t)theo chiu tng cat, 0t2, a > 0.

Li gii. Ta cx(t) =a(1 cos t)y(t) =a sin t nn:

I=

20

{[2a(t sin t) a(1 cos t)]a(1 cos t) +a(t sin t).a sint} dt

= a220

[(2t 2)+sin 2t+(t 2) sin t (2t 2) cos t]dt

= a220

[(2t 2)+t sin t 2t cos t]dt

= a2

42 6

.

Bi tp 4.6. Tnh

ABC A

2

x2 +y2

dx+x (4y+3) dy ABCAl ng gp khc i qua

A(0, 0),B(1, 1), C(0, 2).

x

y

AO

B

1

C

1

Hnh4.6

83


85/115


Li gii. Ta c

phng trnh ng thng AB : x = y

phng trnh ng thng BC: x = 2 yphng trnh ng thng C A: x = 0

nn

I= AB

...+ BC

...+ C A

...

=

10

2

y2 +y2

+y (4y+3)

dy+

21

2

(2 y)2 +y2

. (1)+(2 y) (4y+3) dy+0

=3

Bi tp 4.7. Tnh

ABC DA

dx+dy|x|+|y|trong ABCDAl ng gp khc quaA(1, 0),B(0, 1), C(1

x

y

OA1

B

C

D

1

Hnh4.7Li gii. Ta c

AB : x +y= 1 dx+dy = 0BC : x y=1 dx = dyCD : x +y=1 dx+dy = 0DA: x y= 1 dx = dy

nn

I=

AB

...+

BC

...+

CD

...+

DA

=0+ BC

2dxx+y

+ 0+ DA

2dxx y

=

10

2dx+

10

2dx

=0

84


86/115


Bi tp 4.8. TnhC

4

x2+y2

2 dx+dytrong

x= t sin

t

y= t cos

t

0t 24theo chiu tng cat.

Li gii. tu = t0u,x= u2 sin uy= u2 cos u x(u)= 2u sin u+u2 cos uy(u)= 2u cos u u2 sin uI=

2

0

u2

2u sin u+u2 cos u

+2u cos u u2 sin u

du

=

0

u3

2 +2u

cos udu

=3

22 +2

2.3 Cng thc Green.Hng dng ca ng cong kn:Nu ng ly tch phn l ng cong kn th

ta quy c hng dng ca ng cong l hng sao cho mt ngi i dc theo ngcong theo hng y s nhn thy min gii hn bi n gn pha mnh nht nm v pha

bn tri.

x

y

O

C

D

Gi sDR2 l min n lin, lin thng, b chn vi bin gii Dl ng cong kn vihng dng, hn na P, Qcng cc o hm ring cp mt ca chng lin tc trn D.Khi

C

Pdx+Qdy =D

Qx

Py

dxdy

Ch :

85


87/115


NuDc hng m thC

Pdx +Qdy =D

Qx

Py

dxdy

Trong nhiu bi ton, nu Cl ng cong khng kn, ta c th b sung C cng cong kn v p dng cng thc Green.

Bi tp 4.9. Tnh cc tch phn sau C

(xy+x+y) dx+(xy+x y) dy bng hai cch:

tnh trc tip, tnh nh cng thc Green ri so snh cc kt qu, viC l ng:

a) x2 +y2 = R2

x

y

O

Hnh4.9a

Cch 1: Tnh trc tip

t

x= R cos t

y= R sin t

0t

I=...

= R3

2

20

(cos t cos2t+sin t cos2t) dt

=0

Cch 2: S dng cng thc Green

P(x,y) = xy+x+y

Q(x,y) = xy+x

y

Qx Py =y x

I=

x2+y2R2

(y x) dxdy

=

x2+y2R2

ydxdy

x2+y2R2

xdxdy

=0

b) x2 +y2 =2x

x

y

O

Hnh4.9b

86


88/115


Cch 1: Tnh trc tip.Ta cx2 +y2 =2x(x 1)2 +y2 =1nn

t

x= 1+cos ty= sin t , 0t2

I=

20

{[(1+cos t) sin t+1+cos t+sin t] ( sin t)+[(1+cos t) sin t+1+cos t sin t] co

=

20

2sin2 t+cos2 t cos t sin t+cos t sin t cos t sin2 t+cos2 t sin t

dt

=...

=

Cch 2: S dng cng thc Green.

Ta c:

P(x,y) = xy+x+yQ(x,y) = xy+x y Qx Py =y x

I= (x1)2+y21

(y x) dxdy, tx= r cosy= r sin , 2 2I=

2

2

d

2cos0

(r sin 1 r cos)rdr

=

2

2

1

2(sin cos) .4cos2 2cos

d

=

c) x2

a2+ y

2

b2 =1, (a, b > 0)

87


89/115


Cch 1: Tnh trc tip

t

x= a cos t

y= b sin t

0t2x(t) =a sin t

y(t) =b cos t

I=...

=

20

ab sin2 t+ab cos2 t

dt

=0

Cch 2: S dng cng thc GreenP(x,y) = xy+x+yQ(x,y) = xy+x y Qx Py =y x

I= x2

a2+

y2

b21

(y x) dxdy

=

x2

a2+y

2

b21

ydxdy

x2

a2+y

2

b21

xdxdy

=0

Bi tp 4.10. Tnh

x2+y2=2x

x2

y+ x4

dy y2

x+ y4

dx.

x

y

O

Hnh4.10Li gii. p dng cng thc Green ta c:

I=D

Q

x P

y

dxdy=

D

4xy+

3

4x2 +

3

4y2

dxdy=3

4

D

x2 +y2

dxdyv

D

4xydxdy= 0

t

x= r cos

y= r sin, ta c2 2 , 0r2 cos. Vy

I=3

4

22

d

2cos0

r2.rdr =3

4

22

4cos4 =9

8

Bi tp 4.11. Tnh OABO

ex [(1 cosy) dx (y siny) dy] trong OABO l ng gpkhcO(0, 0),A(1, 1),B(0, 2)

88


90/115


x

y

O

A

1

B

1

Hnh4.11

Li gii. t

P (x,y)= ex (1 cosy)Q (x,y)=ex (y siny) Qx Py =exy.p dng cng thc Green ta c:

I=

Dexydxdy

=

10

dx

2xx

exydy

=1

2

10

ex (4x 4) dx

=4 2e

Bi tp 4.12. Tnh

x2+y2=2x

(xy+ex sinx+x+y) dx (xy ey +x siny) dy

x

y

O

Hnh4.12

Li gii. t

P (x,y)= xy+ex sinx+x+yQ (x,y)= xy ey +x siny Qx Py =y x 2.89


91/115


p dng cng thc Green ta c:

I=D

y x 2dxdy

= D

x 2dxdyv D

ydxdy = 0

t

x= r cosy= r sin 2 2, 0r2 cos=

2

2

d

2cos0

(r cos 2) rdr

=3

Bi tp 4.13. TnhC

xy4 +x2 +y cos xy

dx+

x3

3 +xy2 x+x cos xy

dy

trong C

x= a cos t

y= a sin t(a > 0).

x

y

O

Hnh4.13

Li gii. t

P (x,y)= xy4 +x2 +y cos xyQ (x,y)= x33 +xy2 x+x cos xy Qx Py = x2 +y2 4xy3 1.90


92/115


p dng cng thc Green ta c:

I=D

x2 +y2 4xy3 1dxdy

=

Dx2 +y2 1dxdyv

D4xy3dxdy= 0

t

x= r cosy= r sin 0 2, 0ra=

20

d

a0

r2 1

rdr

=

Bài Giảng GIẢI TÍCH II

Documents

Transcript of Bài Giảng GIẢI TÍCH II