bài giảng giẢi tÍch ii
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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
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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
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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
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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
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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:
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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
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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.
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8 Chng 1. Cc ng dng ca php tnh vi phn trong hnh hc
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
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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 .
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10 Chng 1. Cc ng dng ca php tnh vi phn trong hnh hc
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.
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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
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12 Chng 1. Cc ng dng ca php tnh vi phn trong hnh hc
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:
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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)
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14 Chng 1. Cc ng dng ca php tnh vi phn trong hnh hc
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
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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
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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
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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
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18 Chng 2. Tch phn bi
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
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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
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20 Chng 2. Tch phn bi
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
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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}
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22 Chng 2. Tch phn bi
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
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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
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24 Chng 2. Tch phn bi
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.
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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
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26 Chng 2. Tch phn bi
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
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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).
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28 Chng 2. Tch phn bi
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
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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.
Bi tp 2.11. Tnh cc tch phn sau:
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
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30 Chng 2. Tch phn bi
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
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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.
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32 Chng 2. Tch phn bi
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
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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
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34 Chng 2. Tch phn bi
Li gii.
t
x= 2+r cosy= r sin 0 r 10
nn
I=
0
d
10
(2+r cos) r sin.rdr = 43
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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
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36 Chng 2. Tch phn bi
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.
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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
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38 Chng 2. Tch phn bi
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.
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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
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40 Chng 2. Tch phn bi
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)
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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
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42 Chng 2. Tch phn bi
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.
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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 .
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44 Chng 2. Tch phn bi
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
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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
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46 Chng 2. Tch phn bi
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.
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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
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48 Chng 2. Tch phn bi
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).
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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
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50 Chng 2. Tch phn bi
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
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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).
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52 Chng 2. Tch phn bi
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
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3. Cc ng dng ca tch phn bi 53
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
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54 Chng 2. Tch phn bi
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
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3. Cc ng dng ca tch phn bi 55
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).
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56 Chng 2. Tch phn bi
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.
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3. Cc ng dng ca tch phn bi 57
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
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58 Chng 2. Tch phn bi
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
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3. Cc ng dng ca tch phn bi 59
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
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60 Chng 2. Tch phn bi
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,
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3. Cc ng dng ca tch phn bi 61
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
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62 Chng 2. Tch phn bi
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
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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
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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.
Bi tp 3.2. Tnh cc tch phn sau:
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]
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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.
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66 Chng 3. Tch phn ph thuc tham s.
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 .
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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
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68 Chng 3. Tch phn ph thuc tham s.
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).
Bi tp 3.5. Tnh cc tch phn sau:
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:
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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
.
Kim tra iu kin v i th t ly tch phn:
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
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70 Chng 3. Tch phn ph thuc tham s.
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 .
Kim tra iu kin v i th t ly tch phn:
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.
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2. Tch phn suy rng ph thuc tham s. 71
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
Bi tp 3.7. Tnh cc tch phn sau:
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:
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72 Chng 3. Tch phn ph thuc tham s.
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
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2. Tch phn suy rng ph thuc tham s. 73
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.
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74 Chng 3. Tch phn ph thuc tham s.
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, +).
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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.
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78 Chng 3. Tch phn ph thuc tham s.
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
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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.
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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
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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
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82 Chng 4. Tch phn ng
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)
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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
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84 Chng 4. Tch phn ng
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
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2. Tch phn ng loi II 85
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 :
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86 Chng 4. Tch phn ng
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
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2. Tch phn ng loi II 87
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)
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88 Chng 4. Tch phn ng
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)
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2. Tch phn ng loi II 89
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
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90 Chng 4. Tch phn ng
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
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2. Tch phn ng loi II 91
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
=