Lý thuyết và bài tập môn Toán cao cấp (Tập 3): Phần 2

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Chu.o.ng 12
`eu biˆ
T´ıch phˆ
an h`
am nhiˆ
e´n

12.1 T´ıch phˆ
an 2-l´
o.p . . . . . . . . . . . . . . . . 118
`en ch˜
o.ng ho..p miˆ
u. nhˆ
a.t . . . . . . . . . 118
12.1.1 Tru.`
`en cong . . . . . . . . . . . . 118
o.ng ho..p miˆ
12.1.2 Tru.`
12.1.3 Mˆ
o.t v`
ai u
´.ng du.ng trong h`ınh ho.c . . . . . . 121
12.2 T´ıch phˆ
an 3-l´
o.p . . . . . . . . . . . . . . . . 133
`en h`ınh hˆ
o.ng ho..p miˆ
o.p . . . . . . . . . 133
12.2.1 Tru.`
`en cong . . . . . . . . . . . . 134
o.ng ho..p miˆ

12.2.2 Tru.`
12.2.3

. . . . . . . . . . . . . . . . . . . . . . . . 136

12.2.4 Nhˆ
a.n x´et chung . . . . . . . . . . . . . . . . 136
o.ng . . . . . . . . . . . . . . . 144
12.3 T´ıch phˆ
an d u.`
12.3.1 C´
ac di.nh ngh˜ıa co. ba’n . . . . . . . . . . . . 144
o.ng . . . . . . . . . . . . 146
12.3.2 T´ınh t´ıch phˆ
an du.`
12.4 T´ıch phˆ
an m˘
a.t . . . . . . . . . . . . . . . . . 158
12.4.1 C´
ac di.nh ngh˜ıa co. ba’n . . . . . . . . . . . . 158
ap t´ınh t´ıch phˆ
an m˘
a.t . . . . . . 160
12.4.2 Phu.o.ng ph´

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

118

12.4.3 Cˆ
ong th´
u.c Gauss-Ostrogradski . . . . . . . 162
12.4.4 Cˆ
ong th´
u.c Stokes . . . . . . . . . . . . . . . 162

12.1

T´ıch phˆ
an 2-l´
o.p

12.1.1

`en ch˜
o.ng ho..p miˆ
u. nhˆ
a.t
Tru.`

Gia’ su’.
D = [a, b] × [c, d] = {(x, y) : a

x

b, c

y

d}

`en D. Khi d´o t´ıch phˆan 2-l´o.p cu’a
v`a h`am f(x, y) liˆen tu.c trong miˆ
`en ch˜
h`am f (x, y) theo miˆ
u. nhˆa.t
D = {(x, y) : a

x

b; c

y

d}

u.c
du.o..c t´ınh theo cˆong th´
b

f(M)dxdy =

d

dx
a

D

f (M)dy;
c

d

f(M)dxdy =
D

(12.1)

b

dy
c

f (M)dx,

M = (x, y).

(12.2)

a

`au tiˆen t´ınh t´ıch phˆan trong I(x) theo y xem x l`a h˘`ang
Trong (12.1): dˆ
sˆo´, sau d´o t´ıch phˆan kˆe´t qua’ thu du.o..c I(x) theo x. Dˆo´i v´o.i (12.2) ta
c˜

ung tiˆe´n h`anh tu.o..ng tu.. nhu.ng theo th´
u. tu.. ngu.o..c la.i.

12.1.2

`en cong
Tru.`
o.ng ho..p miˆ

`en bi. ch˘a.n
Gia’ su’. h`am f (x, y) liˆen tu.c trong miˆ
D = {(x, y) : a

x

b; ϕ1(x)

y

ϕ2 (x)}

12.1. T´ıch phˆan 2-l´o.p

119

trong d´o y = ϕ1 (x) l`a biˆen du.´o.i, y = ϕ2(x) l`a biˆen trˆen, ho˘a.c
D = {(x, y) : c

y

d; g1 (y)

x

g2 (y)}

trong d´o x = g1 (y) l`a biˆen tr´ai c`on x = g2 (y) l`a biˆen pha’i, o’. dˆay
`eu liˆen tu.c trong c´ac khoa’ng
ta luˆon gia’ thiˆe´t c´ac h`am ϕ1, ϕ2 , g1 , g2 dˆ
`en D luˆon luˆon tˆ
`on ta.i.
tu.o.ng u
´.ng. Khi d´o t´ıch phˆan 2-l´o.p theo miˆ
.
Dˆe’ t´ınh t´ıch phˆan 2-l´o p ta c´o thˆe’ ´ap du.ng mˆo.t trong hai phu.o.ng
ph´ap sau.
`e viˆe.c du.a t´ıch
1+ Phu.o.ng ph´ap Fubini du..a trˆen di.nh l´
y Fubini vˆ
`e t´ıch phˆan l˘a.p. Phu.o.ng ph´ap n`ay cho ph´ep ta du.a t´ıch
phˆan 2-l´o.p vˆ
`e t´ıch phˆan l˘a.p theo hai th´
phˆan 2-l´o.p vˆ
u. tu.. kh´ac nhau:
b

ϕ2 (x)

f (M)dxdy =

f(M)dy dx =
a

D

g2 (y)

f (M)dxdy =
g1 (y)

f (M)dy, (12.3)

ϕ1 (x)
g2 (y)

d

f(M)dx dy =
c

D

dx
a

ϕ1 (x)
d

ϕ2 (x)

b

dy
c

f (M)dx.

(12.4)

g1 (y)

a.n cu’a c´
ac t´ıch phˆ
an trong biˆe´n thiˆen
T`
u. (12.3) v`a (12.4) suy r˘a`ng cˆ
v`
a phu. thuˆ
o.c v`
ao biˆe´n m`
a khi t´ınh t´ıch phˆ
an trong, n´
o du.o..c xem l`
a
`
khˆ
ong dˆ

o’i. Cˆ
a.n cu’a t´ıch phˆ
an ngo`
ai luˆ
on luˆ
on l`
a h˘
ang sˆ
o´.
`an biˆen du.´o.i
Nˆe´u trong cˆong th´
u.c (12.3) (tu.o.ng u
´.ng: (12.4)) phˆ
`an biˆen trˆen (tu.o.ng u
`an biˆen tr´ai hay pha’i) gˆ
`om t`
hay phˆ
´.ng: phˆ
u. mˆo.t
`an v`a mˆ˜o i phˆ
`an c´o phu.o.ng tr`ınh riˆeng th`ı miˆ
`en D cˆ
`an chia th`anh
sˆo´ phˆ
`en con bo’.i c´ac du.`o.ng th˘a’ng song song v´o.i tru.c Oy (tu.o.ng
nh˜
u.ng miˆ
`en con d´o c´ac phˆ
`an biˆen
u

´.ng: song song v´o.i tru.c Ox) sao cho mˆo˜ i miˆ
`an biˆen tr´ai, pha’i) dˆ
`eu chı’ du.o..c biˆe’u
du.´o.i hay trˆen (tu.o.ng u
´.ng: phˆ
˜e n bo’.i mˆo.t phu.o.ng tr`ınh.
diˆ
2+ Phu.o.ng ph´ap dˆo’i biˆe´n. Ph´ep dˆo’i biˆe´n trong t´ıch phˆan 2-l´o.p
du.o..c thu..c hiˆe.n theo cˆong th´
u.c
D(x, y)
dudv
(12.5)
f(M)dxdy =
f[ϕ(u, v), ψ(u, v)]
D(u, v)
D

D∗

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

120

`en biˆe´n thiˆen cu’a to.a dˆo. cong (u, v) tu.o.ng u
´.ng

trong d´o D∗ l`a miˆ
khi c´ac diˆe’m (x, y) biˆe´n thiˆen trong D: x = ϕ(u, v), y = ψ(u, v);
(u, v) ∈ D∗ , (x, y) ∈ D; c`on
∂x
D(x, y)
= ∂u
J=
∂y
D(u, v)
∂u

∂x
∂v = 0
∂y
∂v

(12.6)

l`a Jacobiˆen cu’a c´ac h`am x = ϕ(u, v), y = ψ(u, v).
To.a dˆo. cong thu.`o.ng d`
ung ho.n ca’ l`a to.a dˆo. cu..c (r, ϕ). Ch´
ung
liˆen hˆe. v´o.i to.a dˆo. Dˆecac bo’.i c´ac hˆe. th´
u.c x = r cos ϕ, y = r sin ϕ,
0
r < +∞, 0
ϕ < 2π. T`
u. (12.6) suy ra J = r v`a trong to.a dˆo.
cu..c (12.5) c´o da.ng
f(M )dxdy =

f (r cos ϕ, r sin ϕ)rdrdϕ.

(12.7)

D∗

D

K´
y hiˆe.u vˆe´ pha’i cu’a (12.7) l`a I(D∗). C´o c´ac tru.`o.ng ho..p cu. thˆe’ sau
dˆay.
(i) Nˆe´u cu..c cu’a hˆe. to.a dˆo. cu..c n˘`am ngo`ai D th`ı
r2 (ϕ)

ϕ2

I(D∗ ) =

dϕ
ϕ1

f (r cos ϕ, r sin ϕ)rdr.

(12.8)

r1 (ϕ)

u. cu..c c˘a´t biˆen ∂D
(ii) Nˆe´u cu..c n˘a`m trong D v`a mˆo˜ i tia di ra t`

khˆong qu´a mˆo.t diˆe’m th`ı
r(ϕ)

2π

I(D∗ ) =

dϕ
0

f (r cos ϕ, r sin ϕ)rdr.

(12.9)

0

(iii) Nˆe´u cu..c n˘`am trˆen biˆen ∂D cu’a D th`ı
r(ϕ)

ϕ2
∗

I(D ) =

dϕ
ϕ1

f (r cos ϕ, r sin ϕ)rdr.

(12.10)

0

12.1. T´ıch phˆan 2-l´o.p

12.1.3

121

Mˆ
o.t v`
ai u
´.ng du.ng trong h`ınh ho.c

`en ph˘a’ng D du.o..c t´ınh theo cˆong th´
u.c
1+ Diˆe.n t´ıch SD cu’a miˆ
SD =

dxdy ⇒ SD =

rdrdϕ.

(12.11)

D∗

D

`en D (thuˆo.c
u.ng c´o d´ay l`a miˆ
2+ Thˆe’ t´ıch vˆa.t thˆe’ h`ınh tru. th˘a’ng d´
m˘a.t ph˘a’ng Oxy) v`a gi´o.i ha.n ph´ıa trˆen bo’.i m˘a.t z = f (x, y) > 0 du.o..c
t´ınh theo cˆong th´
u.c
V =

f (x, y)dxdy.

(12.12)

D

3+ Nˆe´u m˘a.t (σ) du.o..c cho bo’.i phu.o.ng tr`ınh z = f (x, y) th`ı diˆe.n
˜e n bo’.i t´ıch phˆan 2-l´o.p
t´ıch cu’a n´o du.o..c biˆe’u diˆ
1 + (fx )2 + (fy )2dxdy,

Sσ =

(12.13)

D(x,y)

trong d´o D(x, y) l`a h`ınh chiˆe´u vuˆong g´oc cu’a m˘a.t (σ) lˆen m˘a.t ph˘a’ng
to.a dˆo. Oxy.
´ V´I DU

CAC
.
V´ı du. 1. T´ınh t´ıch phˆan
xydxdy,

D = {(x, y) : 1

x

2; 1

y

2}.

D

Gia’i. Theo cˆong th´
u.c (12.2):
2

xydxdy =
D

2

dy
1

xydx.

1

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

122

T´ınh t´ıch phˆan trong (xem y l`a khˆong dˆo’i) ta c´o
2

xydx = y

I(x) =

x2
2

2
1

1
= 2y − y.
2

1

Bˆay gi`o. t´ınh t´ıch phˆan ngo`ai:

2

9
1
2y − y dy = ·
2
4

xydxdy =
1

D

xydxdy nˆe´u D du.o..c gi´o.i ha.n bo’.i c´ac

V´ı du. 2. T´ınh t´ıch phˆan
D

du.`o.ng cong y = x − 4, y 2 = 2x.
Gia’i. B˘`ang c´ach du..ng c´ac du.`o.ng gi˜
u.a c´ac giao diˆe’m A(8, 4) v`a
`en lˆa´y t´ıch phˆan D.
B(2, −2) cu’a ch´
ung, ba.n do.c s˜e thu du.o..c miˆ
`au tiˆen lˆa´y t´ıch phˆan theo x v`a tiˆe´p dˆe´n lˆa´y t´ıch phˆan theo
Nˆe´u dˆ
˜e n bo’.i mˆo.t t´ıch phˆan bˆo.i
`en D du.o..c biˆe’u diˆ
y th`ı t´ıch phˆan theo miˆ
y4

4

I=

xydxdy =

ydy
−2

D

xdx,

y 2 /2

`en D lˆen tru.c Oy. T`
u. d´o
trong d´o doa.n [−2, 4] l`a h`ınh chiˆe´u cu’a miˆ
4

I=

x2
y
2

4

y4

y 2 /2

1
dy =
2

−2

y (y + 4)2 −

y4
dy = 90.
4

−2

`au tiˆen theo y, sau d´o theo
Nˆe´u t´ınh t´ıch phˆan theo th´
u. tu.. kh´ac: dˆ
`an chia miˆ
`en D th`anh hai miˆ
`en con bo’.i du.`o.ng th˘a’ng qua B v`a
x th`ı cˆ
song song v´o.i tru.c Oy v`a thu du.o..c
√
2

I=

+

D1

=

xdx

8

ydy +

√
− 2x

0

D2

√
2x

8

xdx · 0 +
0

xdx
2

2

=

2x

y2
x
2

ydy

x−4
√
2x

dx = 90.
x−4

2

12.1. T´ıch phˆan 2-l´o.p

123

Nhu. vˆa.y t´ıch phˆan 2-l´o.p d˜a cho khˆong phu. thuˆo.c th´
u. tu.. t´ınh t´ıch
`an cho.n mˆo.t th´
phˆan. Do vˆa.y, cˆ

u. tu.. t´ıch phˆan dˆe’ khˆong pha’i chia
`en.
miˆ
`en D du.o..c
(y − x)dxdy. trong d´o miˆ

V´ı du. 3. T´ınh t´ıch phˆan
D

7
1
gi´o.i ha.n bo’.i c´ac du.`o.ng th˘a’ng y = x + 1, y = x − 3, y = − x + ,
3
3
1
y = − x + 5.
3
Gia’i. Dˆe’ tr´anh su.. ph´
u.c ta.p, ta su’. du.ng ph´ep dˆo’i biˆe´n u = −y − x;
1
u.c (12.5). Qua ph´ep dˆo’i biˆe´n d˜a cho.n,
v = y + x v`a ´ap du.ng cˆong th´
3
du.`o.ng th˘a’ng y = x + 1 biˆe´n th`anh du.`o.ng th˘a’ng u = 1; c`on y = x − 3
biˆe´n th`anh u = −3 trong m˘a.t ph˘a’ng Ouv; tu.o.ng tu.., c´ac du.`o.ng th˘a’ng
7
1
7
1
y = − x + , y = − x + 5 biˆe´n th`anh c´ac du.`o.ng th˘a’ng v = , v = 5.

3
3
3
3
7
∗
∗
.
˜e d`ang thˆa´y
`en D = [−3, 1] × , 5 . Dˆ
`en D tro’ th`anh miˆ
Do d´o miˆ
3
D(x, y)
3
r˘`ang
= − . Do d´o theo cˆong th´
u.c (12.5):
D(u, v)
4
3
1
u+ v −
4
4

(y − x)dxdy =

3
3

− u+ v
4
4

3
dudv
4

D∗

D

5

3
ududv =
4

=
D∗

4

3
udu = −8.
4

dv
7/3

−3

Nhˆ
a.n x´et. Ph´ep dˆo’i biˆe´n trong t´ıch phˆan hai l´o.p nh˘`am mu.c d´ıch
`en lˆa´y t´ıch phˆan. C´o thˆe’ l´
do.n gia’n h´oa miˆ
uc d´o h`am du.´o.i dˆa´u t´ıch
phˆan tro’. nˆen ph´
u.c ta.p ho.n.
(x2 + y 2)dxdy, trong d´o D l`a h`ınh tr`on

V´ı du. 4. T´ınh t´ıch phˆan
D

gi´o.i ha.n bo’.i du.`o.ng tr`on x2 + y 2 = 2x.
u.c (12.7).
Gia’i. Ta chuyˆe’n sang to.a dˆo. cu..c v`a ´ap du.ng cˆong th´
Cˆong th´
u.c liˆen hˆe. (x, y) v´o.i to.a dˆo. cu..c (r, ϕ) v´o.i cu..c ta.i diˆe’m O(0, 0)

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

124
c´o da.ng

x = r cos ϕ,

y = r sin ϕ.

(12.14)

Thˆe´ (12.14) v`ao phu.o.ng tr`ınh du.`o.ng tr`on ta thu du.o..c r2 = 2r cos ϕ ⇒
r = 0 ho˘a.c r = 2 cos ϕ (dˆay l`a phu.o.ng tr`ınh du.`o.ng tr`on trong to.a dˆo.
cu..c). Khi d´o
D∗ = (r, ϕ) : −

π
2

π
,0
2

ϕ

r

2 cos ϕ

T`
u. d´o thu du.o..c
π/2
2

I=

2

3

(x + y )dxdy =

r drdϕ =
D∗

D
π/2

r3 dr

dϕ
0

−π/2
π/2

r4
4

=

2 cos ϕ

2 cos ϕ

cos4 ϕf ϕ =

dϕ = 4
0

−π/2

3π
·
2

−π/2

Nhˆ
a.n x´et. Nˆe´u lˆa´y cu..c ta.i tˆam h`ınh tr`on th`ı
x − 1 = r cos ϕ
y = r sin ϕ
D∗ = (r, ϕ) : 0

r

1, 0

ϕ

2π}

v`a do x2 + y 2 = 1 + 2r cos ϕ + r2 nˆen
r(1 + 2r cos ϕ + r2 )drdϕ

I=

D∗
2π

1

(r + 2r2 cos ϕ + r3 )dr =

dϕ

=
0

3π
·
2

0

V´ı du. 5. T´ınh thˆe’ t´ıch vˆa.t thˆe’ T gi´o.i ha.n bo’.i paraboloid z = x2 + y 2,
m˘a.t tru. y = x2 v`a c´ac m˘a.t ph˘a’ng y = 1, z = 0.

12.1. T´ıch phˆan 2-l´o.p

125

Gia’i. H`ınh chiˆe´u cu’a vˆa.t thˆe’ T lˆen m˘a.t ph˘a’ng Oxy l`a
D(x, y) = (x, y) : −1

x

1, x2

y

1 .

Do d´o ´ap du.ng (12.12) ta c´o
1
2

zdxdy =

V (T ) =
D(x,y)

2

(x + y )dxdy =

x2 y +

=

y3
3

1

x2

dx =

(x2 + y 2)dy

dx
−1

D(x,y)

1

1

x2

88
·
105

−1

`au b´an k´ınh R v´o.i tˆam ta.i gˆo´c to.a dˆo..
V´ı du. 6. T`ım diˆe.n t´ıch m˘a.t cˆ
`au d˜a cho c´o da.ng
Gia’i. Phu.o.ng tr`ınh m˘a.t cˆ
x2 + y 2 + z 2 = R2 .
`au l`a
Do d´o phu.o.ng tr`ınh nu’.a trˆen m˘a.t cˆ

R2 − x2 − y 2.

z=

u.ng nˆen ta chı’ t´ınh diˆe.n t´ıch nu’.a trˆen l`a du’. Ta c´o
Do t´ınh dˆo´i x´
ds =

1 + zx2 + zy 2 dxdy =

Rdxdy
R2

− x2 − y 2

`en lˆa´y t´ıch phˆan D(x, y) = {(x, y) : x2 + y 2
Miˆ

·

R2 }. Do d´o

x = r cos ϕ
dxdy = y = r sin ϕ
R2 − x2 − y 2
J =r
R

S=2
D(x,y)

2π

= 2R

R

rdr
√
R2 − r2

dϕ
0

0

√
= 4πR − R2 − r2

R
0

= 4πR2 .

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

126

`an m˘a.t tru. x2 = 2z gi´o.i ha.n bo’.i giao
V´ı du. 7. T´ınh diˆe.n t´ıch phˆ
√
tuyˆe´n cu’a m˘a.t tru. d´o v´o.i c´ac m˘a.t ph˘a’ng x − 2y = 0, y = 2x, x = 2 2.
˜e thˆa´y r˘a`ng h`ınh chiˆe´u cu’a phˆ
`an m˘a.t d˜a nˆeu l`a tam gi´ac
Gia’i. Dˆ
.
v´o i c´ac ca.nh n˘a`m trˆen giao tuyˆe´n cu’a m˘a.t ph˘a’ng Oxy v´o.i c´ac m˘a.t
ph˘a’ng d˜a cho.
x2
T`
u. phu.o.ng tr`ınh m˘a.t tru. ta c´o z = , do vˆa.y
2
∂z
= x,
∂x

√
∂z
= 0 → dS = 1 + x2dxdy.
∂y

T`
u. d´o suy r˘a`ng
√
2 2

√

1 + x2dx

S=
0

√
2 2

2x

√
x 1 + x2dx = 13.

3
2

dy =

0

x/2

` TA
ˆ. P
BAI
T`ım cˆa.n cu’a t´ıch phˆan hai l´o.p

`en D gi´o.i
f (x, y)dxdy theo miˆ
D

ha.n bo’.i c´ac du.`o.ng d˜a chı’ ra . (Dˆe’ ng˘´an go.n ta k´
y hiˆe.u f (x, y) = f (−)).
1. x = 3, x = 5, 3x − 2y + 4 = 0, 3x − 2y + 1 = 0.
3x+4
5

5

(DS.

dx
3

f (−)dy)
3x+1
5

2. x = 0, y = 0, x + y = 2
2

(DS.

2−x

dx
0

f (−)dy)
0

12.1. T´ıch phˆan 2-l´o.p
3. x2 + y 2

1, x

127

0, y

0.
√
1

(DS.

dx

f (−)dy)

0

4. x + y

1, x − y

1, x

0

0.
1

1−x

dx

(DS.
0

5. y

x2, y

f (−)dy)

x−1

4 − x2 .
√

4−x2

2

(DS.

dx
√
− 2

x2 y 2
+
6.
4
9

1−x2

f (−)dy)
x2

1.
3
2

+2

(DS.
√

4−x2

dx
−2

7. y = x2, y =

√

f (−)dy)

− 32

√

4−x2

x.
√
1

dx

(DS.
0

x

f (−)dy)
x2

8. y = x, y = 2x, x + y = 6.
2

2x

dx

(DS.
0

3

f(−)dy +
x

6−x

dx
2

f (−)dy)
x

Thay dˆo’i th´
u. tu.. t´ıch phˆan trong c´ac t´ıch phˆan

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

128
4

9.

4

4

dy
0

f(−)dx.

(DS.

dx

y
0

1

11.

(DS.

f (−)dx)
√

0

−

2−x2

1

y

f(−)dy.

(DS.

dy

x

0

y

1

dy
1

dy

x+1

2

12.

y−1

1

f(−)dy.

dx
0

2

1−x2

dx
−1

f (−)dy)

2
√

10.

x

fdx.

(DS.

1/y

1−y 2
√
2

f dx +
0

dx
1/2

dy
1

2

2

dx
1

1/x

f dx)
0

2

f dy +

2−y

f dy)
x

T´ınh c´ac t´ıch phˆan l˘a.p sau
1

13.

2x

dx
0

(x − y + 1)dy.

14.

y

y3
dx.
x2 + y 2

(DS. 6π)

(x + 2y)dx.

(DS. −11, 2)

dy
−2

0
y2

0

15.

dy
2

0
5

16.

5−x

dx
0

4 + x + ydy.

(DS.

506
)
15

0
4

17.

2

dy
.
(x + y)2

dx
3

(DS.

25
)
24

1
√
2 ax

a

dx
0

1
)
3

x
4

18.

(DS.

√
−2 ax

(x2 + y 2)dy.

(DS.

344 4
a)
105

12.1. T´ıch phˆan 2-l´o.p
2π

19.

129

a

dϕ
0

rdr.

(DS.

πa2
)
2

a sin ϕ
√
1

20∗.

1−x2

1 − x2 − y 2dy.

dx
0

(DS.

π
)
6

0

T´ınh c´ac t´ıch phˆan 2-l´o.p theo c´ac h`ınh ch˜
u. nhˆa.t d˜a chı’ ra.
21.

(x + y 2)dxdy; 2

x

3, 1

y

2.

5
(DS. 4 )
6

(x2 + y)dxdy; 1

x

2, 0

y

1.

5
(DS. 2 )
6

D

22.
D

23.

(x2 + y 2 )dxdy; 0

x

3y 2 dxdy
;0
1 + x2

1, 0

1, 0

y

1.

(DS.

y

1.

(DS.

π
)
4

2
)
3

D

24.

x

D

25.

sin(x + y)dxdy; 0

x

π
,0
2

y

π
.
2

(DS. 2)

y

0.

(DS.

1
)
e

D

26.

xexy dxdy; 0

x

1, −1

dxdy
;1
(x − y)2

x

2, 3

D

27.

y

4
(DS. ln )
3

4.

D

`en D gi´o.i ha.n c´ac du.`o.ng d˜a chı’
T´ınh c´ac t´ıch phˆan 2-l´o.p theo miˆ

ra
28.

xydxdy; y = 0, y = x, x = 1.

(DS.

1
)
8

D

xydxdy; y = x2 , x = y 2.

29.

(DS.

1
)
12

D

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

130

xdxdy; y = x3, x + y = 2, x = 0.

30.

7
)
15

(DS.

D

31.

5
(DS. 20 )
6

xdxdy; xy = 6, x + y − 7 = 0.
D

3
(DS. 1 )
5

y 2xdxdy; x2 + y 2 = 4, x + y − 2 = 0.

32.
D

33.

(x + y)dxdy; 0

y

π, 0

x

sin y.

(DS.

5π
)
4

D

34.

sin(x + y)dxdy; x = y, x + y =

π
, y = 0.
2

(DS.

1
)
2

D
2
e−y dxdy; D l`a tam gi´ac v´o.i dı’nh O(0, 0), B(0, 1), A(1, 1).

35.
D

(DS. −

1
1
+ )
2e 2

xydxdy; D l`a h`ınh elip 4x2 + y 2

36.

4.

(DS. 0)

D

x2ydxdy; y = 0, y =

37.

√
2ax − x2 .

(DS.

4a5
)
5

D

xdxdy
; y = x, x = 2, x = 2y.
x2 + y 2

38.

(DS.

1
π
− 2arctg )
2
2

D

√
x + ydxdy; x = 0, y = 0, x + y = 1.

39.

(DS.

2
)
5

(DS. 4

4
)
15

D

(x − y)dxdy; y = 2 − x2, y = 2x − 1.

40.
D

41.

(x + 2y)dxdy; y = x, y = 2x, x = 2, x = 3.

1
(DS. 25 )
3

D

12.1. T´ıch phˆan 2-l´o.p

42.

131

xdxdy; x = 2 + sin y, x = 0, y = 0, y = 2π.

(DS.

9π
)
2

D

xydxdy; (x − 2)2 + y 2 = 1.

43.

(DS.

4
)
3

D

dxdy
√
; D l`a h`ınh tr`on b´an k´ınh a n˘`am trong g´oc vuˆong I
2a − x
D
8 √
v`a tiˆe´p x´
uc v´o.i c´ac tru.c to.a dˆo.. (DS. a 2a)
3
44.

45.

ydxdy; x = R(t − sin t), y = R(1 − cos t), 0

t

`en
2π (l`a miˆ

D

5

gi´o.i ha.n bo’.i v`om cu’a xicloid.) (DS. πR3 )
2
y=f (x)

2πR

˜
Chı’ dˆ
a n.

ydxdy =

dx
0

D

ydy
0

Chuyˆe’n sang to.a dˆo. cu..c v`a t´ınh t´ıch phˆan trong to.a dˆo. m´o.i
πR4
46.
)
(x2 + y 2 )dxdy; D : x2 + y 2 R2 , y 0. (DS.
4
D

47.

π
(e − 1))
4

ex

2 +y 2

dxdy; D : x2 + y 2

1, x

0, y

ex

2 +y 2

dxdy; D : x2 + y 2

R2 .

(DS. 2π(eR − 1))

0. (DS.

D

48.

2

D

49.

1 − x2 − y 2dxdy; D : x2 + y 2

x.

1 − x2 − y 2
dxdy; D : x2 + y 2
1 + x2 + y 2

1, x

(DS.

4
1
π− )
4
3

D

50.

0, y

0.

D

(DS.

π(π − 2)
)
2

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

132

ln(x2 + y 2)
dxdy; D : 1
x2 + y 2

51.

x2 + y 2

e.

(DS. 2π)

D

(x2 + y 2)dxdy; D gi´o.i ha.n bo’.i c´ac du.`o.ng tr`on

52.
D

x2 + y 2 + 2x − 1 = 0,

x2 + y 2 + 2x = 0. (DS.

5π
)
2

˜
Chı’ dˆ
a n. D˘a.t x − 1 = r cos ϕ, y = r sin ϕ.
T´ınh thˆe’ t´ıch cu’a vˆa.t thˆe’ gi´o.i ha.n bo’.i c´ac m˘a.t d˜a chı’ ra.
1
53. x = 0, y = 0, z = 0, x + y + z = 1. (DS. )
6
1
54. x = 0, y = 0, z = 0, x + y = 1, z = x2 + y 2 . (DS. )
6
88
)
55. z = x2 + y 2, y = x2, y = 1, z = 0. (DS.
105
2

56. z = x2 + y 2 , x2 + y 2 = a2, z = 0. (DS. πa3)
3
πa4
)
57. z = x2 + y 2, x2 + y 2 = a2, z = 0. (DS.
2
4a3
58. z = x, x2 + y 2 = a2, z = 0. (DS.
)
3
1
59. z = 4 − x2 − y 2, x = ±1, y = ±1.
(DS. 13 )
3
11
)
60. 2 − x − y − 2z = 0, y = x2 , y = x.
(DS.
120
61. x2 + y 2 = 4x, z = x, z = 2x.
(DS. 4π)
`an m˘a.t d˜a chı’ ra.
T´ınh diˆe.n t´ıch c´ac phˆ
`an m˘a.t ph˘a’ng 6x + 3y + 2z = 12 n˘a`m trong g´oc phˆ
`an t´am I.
62. Phˆ
(DS. 14)
`an m˘a.t ph˘a’ng x + y + z = 2a n˘`am trong m˘a.t tru. x2 + y 2 = a2.
63. Phˆ
√

(DS. 2a2 3)

12.2. T´ıch phˆan 3-l´o.p

133

`an m˘a.t paraboloid z = x2 + y 2 n˘a`m trong m˘a.t tru. x2 + y 2 = 4.
64. Phˆ
π √
(DS. (17 17 − 1))
6
`an m˘a.t 2z = x2 + y 2 n˘a`m trong m˘a.t tru. x2 + y 2 = 1.
65. Phˆ
2 √
(DS. (2 2 − 1)π)
3
`an m˘a.t n´on z = x2 + y 2 n˘a`m trong m˘a.t tru. x2 + y 2 = a2 .
66. Phˆ
√
(DS. πa2 2)
`an m˘a.t cˆ
`au x2 + y 2 + z 2 = R2 n˘`am trong m˘a.t tru. x2 + y 2 = Rx.
67. Phˆ
(DS. 2R2 (π − 2))
`an m˘a.t n´on z 2 = x2 + y 2 n˘`am trong m˘a.t tru. x2 + y 2 = 2x.
68. Phˆ
√

(DS. 2 2π)
`an t´am th´
`an m˘a.t tru. z 2 = 4x n˘a`m trong g´oc phˆ
u I v`a gi´o.i ha.n
69. Phˆ
4 √
bo’.i m˘a.t tru. y 2 = 4x v`a m˘a.t ph˘a’ng x = 1. (DS. (2 2 − 1))
3
`an m˘a.t cˆ
`au x2 + y 2 + z 2 = R2 n˘`am trong m˘a.t tru. x2 + y 2 = a2
70. Phˆ
√
(a R). (DS. 4πa(a − a2 − R2 ))

12.2

T´ıch phˆ
an 3-l´
o.p

12.2.1

`en h`ınh hˆ
Tru.`
o.ng ho..p miˆ
o.p

`en D ⊂ R3:
Gia’ su’. miˆ
D = [a, b] × [c, d] × [e, g] = {(x, y, z) : a

x

b, c

y

d, e

z

g}

v`a h`am f (x, y, z) liˆen tu.c trong D. Khi d´o t´ıch phˆan 3-l´o.p cu’a h`am
`en D du.o..c t´ınh theo cˆong th´
f (x, y, z) theo miˆ
u.c
b

g

d

f(x, y, z)dxdydz =

f (x, y, z)dz dy dx
a

D

c

e

b

=

dx
a

g

d

dy
c

f (M)dx.

(12.15)

e

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

134

T`
u. (12.15) suy ra c´ac giai doa.n t´ınh t´ıch phˆan 3-l´o.p:
g

`au tiˆen t´ınh I(x, y) =
(i) Dˆ

f (M)dz;
e
d

(ii) Tiˆe´p theo t´ınh I(x) =

I(x, y)dy;
c
b

(iii) Sau c`
ung t´ınh t´ıch phˆan I =

I(x)dx.
a

u. tu.. kh´ac th`ı c´ac giai doa. n
Nˆe´u t´ıch phˆan (12.15) du.o..c t´ınh theo th´
`au tiˆen t´ınh t´ıch phˆan trong, tiˆe´p dˆe´n t´ınh t´ıch
t´ınh vˆa˜ n tu.o.ng tu..: dˆ
phˆan gi˜

u.a v`a sau c`
ung l`a t´ınh t´ıch phˆan ngo`ai.

12.2.2

`en cong
Tru.`
o.ng ho..p miˆ

`en bi. ch˘a.n
1+ Gia’ su’. h`am f(M) liˆen tu.c trong miˆ
D = (x, y, z) : a

x

b, ϕ1(x)

y

ϕ2 (x), g1 (x, y)

z

g2 (x, y) .

`en D du.o..c t´ınh theo
Khi d´o t´ıch phˆan 3-l´o.p cu’a h`am f(M) theo miˆ
cˆong th´
u.c
ϕ2 (x)

b

g2 (x,y)

f(M )dxdydz =

f (M)dx dy dx
a

D

ϕ1 (x)

(12.16)

g1 (x,y)

ho˘a.c
g2 (x,y)

f(M )dxdydz =
D

D(x,y)

dxdy

f (M)dz,

(12.17)

g1 (x,y)

trong d´o D(x, y) l`a h`ınh chiˆe´u vuˆong g´oc cu’a D lˆen m˘a.t ph˘a’ng Oxy.
`e t´ınh liˆen tiˆe´p ba t´ıch phˆan thˆong
Viˆe.c t´ınh t´ıch phˆan 3-l´o.p du.o..c quy vˆ

12.2. T´ıch phˆan 3-l´o.p

135

u. t´ıch phˆan trong, tiˆe´p dˆe´n t´ıch phˆan gi˜
thu.`o.ng theo (12.16) t`
u.a v`a
sau c`
ung l`a t´ınh t´ıch phˆan ngo`ai. Khi t´ınh t´ıch phˆan 3-l´o.p theo cˆong
`au tiˆen t´ınh t´ıch phˆan trong v`a sau d´o c´o thˆe’ t´ınh t´ıch
th´
u.c (12.17): dˆ
`en D(x, y) theo c´ac phu.o.ng ph´ap d˜a c´o trong 12.1.
phˆan 2-l´o.p theo miˆ
2+ Phu.o.ng ph´ap dˆo’i biˆe´n. Ph´ep dˆo’i biˆe´n trong t´ıch phˆan 3-l´o.p
du.o..c tiˆe´n h`anh theo cˆong th´
u.c
f(M )dxdydz =

f ϕ(u, v, w), ψ(u, v, w), χ(u, v, w) ×
D∗

D

×

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

(12.18)

`en biˆe´n thiˆen cu’a to.a dˆo. cong u, v, w tu.o.ng u
trong d´o D∗ l`a miˆ
´.ng khi
c´ac diˆe’m (x, y, z) biˆe´n thiˆen trong D: x = ϕ(u, v, w), y = ψ(u, v, w),
D(x, y, z)
l`a Jacobiˆen cu’a c´ac h`am ϕ, ψ, χ
z = χ(u, v, w),
D(u, v, w)
∂ϕ
∂u
D(x, y, z)
∂ψ
=
J=
D(u, v, w)
∂u
∂χ

∂u

∂ϕ
∂v
∂ψ
∂v
∂χ
∂v

∂ϕ
∂w
∂ψ
= 0.
∂w
∂χ
∂w

(12.19)

`au.
Tru.`o.ng ho..p d˘a.c biˆe.t cu’a to.a dˆo. cong l`a to.a dˆo. tru. v`a to.a dˆo. cˆ
u. to.a dˆo. Dˆec´ac sang to.a dˆo. tru. (r, ϕ, z) du.o..c thu..c
(i) Bu.´o.c chuyˆe’n t`
hiˆe.n theo c´ac hˆe. th´
u.c x = r cos ϕ, y = r sin ϕ, z = z; 0
r < +∞,
.
0 ϕ < 2π, −∞ < z < +∞. T`
u (12.19) suy ra J = r v`a trong to.a
dˆo. tru. ta c´o

f(M )dxdydz =
D

f r cos ϕ, r sin ϕ, z rdrdϕdz,

(12.20)

D∗

`en biˆe´n thiˆen cu’a to.a dˆo. tru. tu.o.ng u
trong d´o D∗ l`a miˆ
´.ng khi diˆe’m
(x, y, z) biˆe´n thiˆen trong D.

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

136

`au (r, ϕ, θ) du.o..c
u. to.a dˆo. Dˆec´ac sang to.a dˆo. cˆ
(ii) Bu.´o.c chuyˆe’n t`
thu..c hiˆe.n theo c´ac hˆe. th´
u.c x = r sin θ cos ϕ, y = r sin θ sin ϕ, z =
r cos θ, 0
r < +∞, 0
ϕ < 2π, 0

θ
π. T`
u. (12.19) ta c´o
`au ta c´o
J = r2 sin θ v`a trong to.a dˆo. cˆ
f(M)dxdydz =
D

f r sin θ cos ϕ, r sin θ sin ϕ, r cos θ r2 sin θdrdϕdθ,

=

(12.21)

D∗

`en biˆe´n thiˆen cu’a to.a dˆo. cˆ
`au tu.o.ng u
´.ng khi diˆe’m
trong d´o D∗ l`a miˆ
(x, y, z) biˆe´n thiˆen trong D.

12.2.3
`en D ⊂ R3 du.o..c t´ınh theo cˆong
Thˆe’ t´ıch cu’a vˆa.t thˆe’ cho´an hˆe´t miˆ
th´
u.c
VD =

dxdydz.

(12.22)

D

12.2.4

Nhˆ
a.n x´
et chung

B˘a`ng c´ach thay dˆo’i th´
u. tu.. t´ınh t´ıch phˆan trong t´ıch phˆan 3-l´o.p ta s˜e
thu du.o..c c´ac cˆong th´
u.c tu.o.ng tu.. nhu. cˆong th´
u.c (12.16) dˆe’ t´ınh t´ıch
phˆan. Viˆe.c t`ım cˆa.n cho t´ıch phˆan do.n thˆong thu.`o.ng khi chuyˆe’n t´ıch
`e t´ıch phˆan l˘a.p du.o..c thu..c hiˆe.n nhu. dˆo´i v´o.i tru.`o.ng ho..p
phˆan 3-l´o.p vˆ
t´ıch phˆan 2-l´o.p.
´ V´I DU
CAC
.
V´ı du. 1. T´ınh t´ıch phˆan l˘a.p
1

I=

1

dx
−1

2

dy
x2

(4 + z)dx.
0

12.2. T´ıch phˆan 3-l´o.p

137

Gia’i. Ta t´ınh liˆen tiˆe´p ba t´ıch phˆan x´ac di.nh thˆong thu.`o.ng b˘´at
`au t`
dˆ
u. t´ıch phˆan trong
2

I(x, y) =

(4 + z)dz = 4z

2
0

+

z2
2

2

= 10;
0

0
1

1

dy = 10(1 − x2);

I(x, y)dy = 10

I(x) =
x2
1

I=

x2
1

10(1 − x2 )dx =

I(x)dx =
−1

40
·
3

−1

V´ı du. 2. T´ınh t´ıch phˆan
I=

(x + y + z)dxdydz,
D

`en D du.o..c gi´o.i ha.n bo’.i c´ac m˘a.t ph˘a’ng to.a dˆo. v`a m˘a.t
trong d´o miˆ
ph˘a’ng x + y + z = 1.
`en D d˜a cho l`a mˆo.t t´
Gia’i. Miˆ
u. diˆe.n c´o h`ınh chiˆe´u vuˆong g´oc trˆen
m˘a.t ph˘a’ng Oxy l`a tam gi´ac gi´o.i ha.n bo’.i c´ac du.`o.ng th˘a’ng x = 0,
y = 0, x + y = 1. R˜o r`ang l`a x biˆe´n thiˆen t`
u. 0 dˆe´n 1 (doa.n [0, 1] l`a
h`ınh chiˆe´u cu’a D lˆen tru.c Ox). Khi cˆo´ di.nh x, 0 x 1 th`ı y biˆe´n
thiˆen t`
u. 0 dˆe´n 1 − x. Nˆe´u cˆo´ di.nh ca’ x v`a y (0 x 1, 0 y 1 − x)
th`ı diˆe’m (x, y, z) biˆe´n thiˆen theo du.`o.ng th˘a’ng d´
u. m˘a.t ph˘a’ng

u.ng t`
z = 0 dˆe´n m˘a.t ph˘a’ng x + y + z = 1, t´
u. 0 dˆe´n
u.c l`a z biˆe´n thiˆen t`
1 − x − y. Theo cˆong th´
u.c (12.16) ta c´o
1

I=

1−x

dx
0

1−x−y

dy
0

(x + y + z)dz.
0

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

138

˜e d`ang thˆa´y r˘`ang
Dˆ
1

I=

1−x

dx
0

xz + yz +

z2
2

1−x−y

dy
0

0
1

=

1
2

y − yx2 − xy 2 −

y3
3

1−x

dx
0

0
1

1
=
6

(2 − 3x + x3)dx =

1
·
8

0

dxdydz
`en D du.o..c gi´o.i
, trong d´o miˆ
(x + y + z)3

V´ı du. 3. T´ınh I =

D

ha.n bo’.i c´ac m˘a.t ph˘a’ng x + z = 3, y = 2, x = 0, y = 0, z = 0.
`en D d˜a cho l`a mˆo.t h`ınh l˘ang tru. c´o h`ınh chiˆe´u vuˆong
Gia’i. Miˆ
g´oc lˆen m˘a.t ph˘a’ng Oxy l`a h`ınh ch˜
u. nhˆa.t D(x, y) = (x, y) : 0
x
3, 0
y
2 . V´o.i diˆe’m M(x, y) cˆo´ di.nh thuˆo.c D(x, y) diˆe’m
(x, y, z) ∈ D biˆe´n thiˆen trˆen du.`o.ng th˘a’ng d´
u. m˘a.t ph˘a’ng Oxy
u.ng t`
(z = 0) dˆe´n m˘a.t ph˘a’ng x + z = 3, t´
u. 0 dˆe´n 3 − x:
u.c l`a z biˆe´n thiˆen t`
0 z 3 − x. T`
u. d´o theo (12.17) ta c´o
z=3−x

f(M )dxdydz =
D

=

(x + y + z + 1)−3 dz

dxdy
z=0

D(x,y)

(x + y + z + 1)−2
−2

3−x

dxdy = · · · =
0

4 ln 2 − 1
·
8

D(x,y)

`en
(x2 + y 2 + z 2 )dxdydz, trong d´o miˆ

V´ı du. 4. T´ınh t´ıch phˆan
D

D du.o..c gi´o.i ha.n bo’.i m˘a.t 3(x2 + y 2) + z 2 = 3a2 .
Gia’i. Phu.o.ng tr`ınh m˘a.t biˆen cu’a D c´o thˆe’ viˆe´t du.´o.i da.ng
z2
x2 y 2
√
+
+

= 1.
a2 b2
(a 3)2

12.2. T´ıch phˆan 3-l´o.p

139

D´o l`a m˘a.t elipxoid tr`on xoay, t´
u.c l`a D l`a h`ınh elipxoid tr`on xoay.
H`ınh chiˆe´u vuˆong g´oc D(x, y) cu’a D lˆen m˘a.t ph˘a’ng Oxy l`a h`ınh tr`on
x2 + y 2
a2. Do d´o ´ap du.ng c´ach lˆa.p luˆa.n nhu. trong c´ac v´ı du. 2
v`a 3 ta thˆa´y r˘a`ng khi diˆe’m M(x, y) ∈ D(x, y) du.o..c cˆo´ di.nh th`ı diˆe’m
`en D biˆe´n thiˆen trˆen du.`o.ng th˘a’ng d´
(x, y, z) cu’a miˆ
u.
u.ng M(x, y) t`
m˘a.t biˆen du.´o.i cu’a D
z = − 3(a2 − x2 − y 2)
dˆe´n m˘a.t biˆen trˆen
3(a2 − x2 − y 2).

z=+
T`
u. d´o theo (12.17) ta c´o
√

3(a2 −x2 −y 2 )

+

I=

(x2 + y 2 + z 2 )dz

dxdy
√

D(x,y)

−

3(a2 −x2 −y 2 )

√
= 2a2 3

a2 − x2 − y 2dxdy = |chuyˆe’n sang to.a dˆo. cu..c|

x2 +y 2 a2
2

√

√
√
a2 − r2 rdrdϕ = a2 3

3

= 2a

r a

2π

a

(a2 − r2 )1/2rdr

dϕ
0

0

5

4πa
= √ ·
3
V´ı du. 5. T´ınh thˆe’ t´ıch cu’a vˆa.t thˆe’ gi´o.i ha.n bo’.i c´ac m˘a.t ph˘a’ng
x + y + z = 4, x = 3, y = 2, x = 0, y = 0, z = 0.
`en D d˜a cho l`a mˆo.t h`ınh lu.c diˆe.n trong khˆong gian. N´o
Gia’i. Miˆ
c´o h`ınh chiˆe´u vuˆong g´oc D(x, y) lˆen m˘a.t ph˘a’ng Oxy l`a h`ınh thang
vuˆong gi´o.i ha.n bo’.i c´ac du.`o.ng th˘a’ng x = 0, y = 0, x = 3, y = 2 v`a

`eu biˆe´n
Chu.o.ng 12. T´ıch phˆan h`am nhiˆ

140

x + y = 4. Do d´o ´ap du.ng (12.17) ta c´o
4−x−y

VD =

dxdydz =
D

dz =
0

D(x,y)

1

3

dy

=

dxdy

0

0

dy
1

1

(4 − y)x −

=

x2
2

(4 − x − y)dx
0
2

x2
2

3

dy +

(4 − y)x −

0

0

4−y

dy
0

1
1

2

1
15
− 3y dy +
2
2

=

D(x,y)
4−y

2

(4 − x − y)dx +

(4 − x − y)dxdy

0

(4 − y)2dy =

55
·
6

1

V´ı du. 6. T´ınh t´ıch phˆan
I=

x2 + y 2dxdydz,

z
D

`en D gi´o.i ha.n bo’.i m˘a.t ph˘a’ng y = 0, z = 0, z = a v`a m˘a.t
trong d´o miˆ
tru. x2 + y 2 = 2x (x 0, y 0, a > 0).
Gia’i. Chuyˆe’n sang to.a dˆo. tru. ta thˆa´y phu.o.ng tr`ınh m˘a.t tru. x2 +
π
(h˜ay v˜e h`ınh
y 2 = 2x trong to.a dˆo. tru. c´o da.ng r = 2 cos ϕ, 0 ϕ
2
.
!). Do d´o theo cˆong th´
u c (12.20) ta c´o

π/2

I=

2 cos ϕ

a
2

dϕ

r dr

0

0

π/2

a2
zdz =
2

0

2 cos ϕ

r2 dr

dϕ

0

0

π/2

=

4a2
3

8
cos3 ϕdϕ = a2.
9
0

V´ı du. 7. T´ınh t´ıch phˆan
(x2 + y 2 )dxdydz,

I=
D

12.2. T´ıch phˆan 3-l´o.p

141

`au x2 + y 2 + z 2 R2 , z 0.

`en D l`a nu’.a trˆen cu’a h`ınh cˆ
nˆe´u miˆ
`au, miˆ
`en biˆe´n thiˆen D∗ cu’a c´ac to.a dˆo.
Gia’i. Chuyˆe’n sang to.a dˆo. cˆ
`au tu.o.ng u
´.ng khi diˆe’m (x, y, z) biˆe´n thiˆen trong D l`a c´o da.ng
cˆ
D∗ : 0

ϕ < 2π, 0

θ

π
, 0
2

r

2π

π/2

R.

T`
u. d´o
2