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7/23/2019 Tich Phan Kep Phan 2
http://slidepdf.com/reader/full/tich-phan-kep-phan-2 1/28
ĐỔI BIẾN TRONG TÍCH PHÂN KÉP
7/23/2019 Tich Phan Kep Phan 2
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TỌA ĐỘ CỰC
M
y
r
ϕx
[0, 2 ] [ , ]ϕ π ϕ π π ∈ ∈ −ha y
cos , sin
x r y r ϕ= =
2 2 0r x y = + ≥
7/23/2019 Tich Phan Kep Phan 2
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TÍCH PHÂN KÉP TRONG TỌA ĐỘ CỰC
D
ϕ α =
ϕ β = Dij
j ϕ
1 j ϕ −
( )* *,i j r ϕ
ϕ ∆
:
a r b
D α ϕ β
≤ ≤
≤ ≤
7/23/2019 Tich Phan Kep Phan 2
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Tổng tích phân
* * * * *
,
( cos , sin )n i j i j i i j
S f r r r r = ∆ ∆∑ ϕ ϕ ϕ
0( , ) lim n
d D
f x y dxdy S →
=∫∫
0lim ( cos , sin )nd D
S f r r rdrd → = ∫∫ ϕ ϕ ϕ
7/23/2019 Tich Phan Kep Phan 2
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Công thức đổi bin !"ng t#" đ$ c%c
( , ) ( cos , sin )
D D
dxdy drd r f x y f r r =∫∫∫ ∫∫ ϕ ϕ
cos , sin x r y ϕ= =
7/23/2019 Tich Phan Kep Phan 2
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Một số !"ng c#ng $% &'(n ) t*#ng t+, ộ c-c
2 2 2 x y R
r R
+ =
⇔ =
cos , sin
x r y r ϕ
= =
2 2 2
0
0 2
r R
x y R
≤ ≤
≤
+
⇔ ≤
≤
ϕ π
R
R
&R D&R R
R
7/23/2019 Tich Phan Kep Phan 2
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•
2 2 2 x y Rx + = 2 2 2 x y Rx + ≤
•
R .R
2 cosr R =
ϕ
0 2 cos
2 2
r R ≤ ≤
−
≤ ≤
ϕ
π π
ϕ
7/23/2019 Tich Phan Kep Phan 2
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•
2 2 2 x y Ry + =
0 2 sin
0
r R ≤ ≤≤ ≤
ϕ
ϕ π
•R
.R2 sinr R = ϕ
2 2 2 x y Ry + ≤
7/23/2019 Tich Phan Kep Phan 2
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1( )r r = ϕ
2 ( )r r = ϕ
D 1 2( ) ( )
:r r r
Dϕ ϕ
ϕ
≤ ≤ ≤ ≤
(0 2 ) π< − ≤
( cos , sin )
D
f r r rdrd ∫∫ ϕ ϕ ϕ
2
1
( )
( )
( cos , sin )
r
r
d f r r rdr = ∫ ∫ ϕ
ϕ
ϕ ϕ ϕ
7/23/2019 Tich Phan Kep Phan 2
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/Í )0
2 2
D
I x y dxdy = +∫∫
cos , sin
x r y r ϕ= =
2 2
1:0
x y Dy
+ ≤≥
12 Tính3 $4'
'&'
( ) '
0 1
0:
r D
≤ ≤ ≤ ≤ ϕ π
1
2
0 0
.
D
I r rdrd d r dr = =∫∫ ∫ ∫ π
ϕ ϕ
0
1
3 3d = =∫
π
π
ϕ
7/23/2019 Tich Phan Kep Phan 2
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5 6
7 5 6
8 7
( ) *( ) '
1 2
34 4
:
r
D
≤ ≤ ≤ ≤π π
ϕ
cos , sin x r y r ϕ= =
( )
D
I x y dxdy = −∫∫ 2 21 4:,
x y Dy x y x
≤ + ≤≥ ≥ −
.2 Tính3
7/23/2019 Tich Phan Kep Phan 2
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( cos sin ).
D
r r rdrd = −∫∫ ϕ ϕ ϕ
3
242
1
4
(cos sin )d r dr = −∫ ∫
π
π
ϕ ϕ ϕ
( )
D
I x y dxdy = −∫∫
34
4
8 1(cos sin )
3 3d
= − − ÷ ∫
π
π
ϕ ϕ ϕ
72
3= −
1 2
3
4 4
:
r
D
≤ ≤
≤ ≤
π π
ϕ
7/23/2019 Tich Phan Kep Phan 2
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0 2sin
: 3
4
r
D
ϕ
π
ϕ π
≤ ≤
≤ ≤
* 6 .s'nϕ
2sin
2
3 0
cos cos
D
I r rdrd d r rdr = =∫∫ ∫ ∫ π ϕ
π
ϕ ϕ ϕ ϕ
1
6= −
cos , sin x r y r ϕ= =
D
I xdxdy = ∫∫ 2 2 2
: x y y
Dy x
+ ≤
≤ −
92 Tính3 $4'
7/23/2019 Tich Phan Kep Phan 2
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2 2 2 2
4 , 2 , , 0 x y x x y x y x y + = + = = =
:2 Tính ;'<n tích &'(n ) g'4' h=n >?'3
cos , sin x r y r ϕ= =* 6 :c#sϕ
* 6 .c#sϕ
5 6
7
2cos 4cosr ϕ ϕ ≤ ≤
04
π ϕ ≤ ≤
)
7/23/2019 Tich Phan Kep Phan 2
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( ) 1
D
S D dxdy = ∫∫
D
rdrd = ∫∫ ϕ
4cos4
0 2 cos
d rdr = ∫ ∫
π
ϕ
ϕ
ϕ
3 3
4 2
= +
2cos 4cosr ϕ ϕ ≤ ≤
04
π ϕ ≤ ≤
)
7/23/2019 Tich Phan Kep Phan 2
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D
I xydxdy = ∫∫ 2 2
:
3 0
x y x D
x y
+ ≤ −
≤ ≤
@2 Tính3 $4'
* 6 8 c#sϕ
3y x =
0 cosr ϕ ≤ ≤ −
40
3
π ϕ ≤ ≤
7/23/2019 Tich Phan Kep Phan 2
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ĐB BDN TNG EFT
( , ) ( , ) x y D u v D′∈ ⇔ ∈
D
+
,
7 6 7IJ$J 56 5IJ$
CLng thc ổ' >'n
1
( , )
( , )
J D u v
D x y
=
( , ) ( ( , ), ( , ))
D D
f x y dxdy f x u v y v v J dud u
′
=∫∫ ∫∫
( , )
( , )
u v
u v
x x D x y J
y y D u v
′ ′= =
′ ′
7/23/2019 Tich Phan Kep Phan 2
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p ;ng ổ' >'n tổng IQt
cos , sin
x r y r ϕ= =cos sin
sin cos
r
r
x x r J r
y y r
ϕ
ϕ
ϕ ϕ
ϕ ϕ
′ ′ −= = =
′ ′
( , ) ( cos , sin )
D D
f x y dxdy f r r rdrd
′
=∫∫ ∫∫ ϕ ϕ ϕ
T+, ộ c-c3
7/23/2019 Tich Phan Kep Phan 2
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2 2 2
( , ) ( , ).1
D u v R
f x y dxdy g u v dudv
+ ≤
=∫∫ ∫∫
cos , sinu r v r ϕ= =
>I
)3 7 ,. S 5 >. ≤ R.
)"' gốc t+, ộ ntâ& 7 6 I S ,J 5 6 $ S >
Đổ' t'p s,ng
t+, ộ c-c3
•
, 7
5
1 01
0 1
u v
u v
x x J
y y
′ ′
= = =′ ′
Hnh t*Un tâ& tV5 W3$
7/23/2019 Tich Phan Kep Phan 2
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•
,
>
7
5
I
$
)3 7 ,. S 5 >. ≤ R.
7 6 , S *c#sϕJ 5 6 > S *s'nϕ
X 6 *
TY& tZt3
* ϕ
0:0 2
r R Dϕ π
≤ ≤′ ≤ ≤
( , ) ( cos , sin )D D
f x y dxdy f a r b d r r rd ′
= + +∫∫ ∫∫ ϕ ϕ ϕ
7/23/2019 Tich Phan Kep Phan 2
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Đổ' >'n t*#ng [\\'pps[ 2 2
2 2: 1 x y
Da b
+ ≤
),
> 7 6 ,*c#sϕJ 5 6 >*s'nϕX 6 ,>*
( , ) ( cos , sin )
D D
f x y dxdy f ar br drd abr
′
=∫∫ ∫∫ ϕ ϕ ϕ
0:
0
1
2
r D
≤ ≤′ ≤ ≤ ϕ π
2 22
2 2
x y r
a b+ =
7/23/2019 Tich Phan Kep Phan 2
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7 6 . S *c#sϕJ 5 6 81 S *s'nϕ
X 6 *
(2 cos )( 1 sin )
D
I r r rdrd
′
= + − +
∫∫ ϕ ϕ ϕ
0 3
: 0
r
Dϕ π
≤ ≤′ ≤ ≤
-
.
D
I xydxdy = ∫∫ 12 Tính3
hnh t*Un3 7 ..
S 5 S 1.
≤ ]
$4' ) \% n^, t*_n c`,
7/23/2019 Tich Phan Kep Phan 2
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(2 cos )( 1 sin )
D
I r r rdrd
′
= + − +∫∫ ϕ ϕ ϕ
3
2
0 0
( 2 cos 2 sin sin cos )d r r r rdr= − − + +
∫ ∫
π
ϕ ϕ ϕ ϕ ϕ
9 18= − +
7/23/2019 Tich Phan Kep Phan 2
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/í ;
2 1
3 cos 0, 2 sin 0
r
r r ϕ
≤
≥ ≥
2 2
, : 1; 0; 09 4D
x y I xydxdy D y x = + ≤ ≥ ≥
∫∫ .2 Tính3
9
.7 6 9*c#sϕJ 5 6 .*s'nϕ
X 6 9a.a* 6 b*
M'(n ) !c $'t \='3
7/23/2019 Tich Phan Kep Phan 2
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12
0 0
3 cos .2 sin .6
D
xydxdy d r r rdr
=∫∫ ∫ ∫
π
ϕ ϕ ϕ
9
2
=
0 1:
0
2
r D
≤ ≤′ ≤ ≤
π
ϕ
0 1
cos 0,sin 0
r
ϕ ϕ
≤ ≤⇔ ≥ ≥
2 1
3 cos 0, 2 sin 0
r
r r ϕ
≤
≥ ≥
7/23/2019 Tich Phan Kep Phan 2
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•
92 Tính ;'<n tích &'(n g'4' h=n >?'2
2
1, 0, , 03
x
ellipse y y y x x + = = = ≥
3 cos , sin x r y r ϕ= =
M'(n ) !c $'t \='3
2
2 1, 03
x y y x + ≤ ≤ ≤
0 1,
0 sin 3 cos
r
r r
≤ ≤⇔
≤ ≤ ϕ ϕ
3J r =
7/23/2019 Tich Phan Kep Phan 2
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0 1,
sin0 tan 3cos
r ≤ ≤
⇔ ≤ = ≤ϕ
ϕ
ϕ
0 1,
0 sin 3 cos
r
r r
≤ ≤
≤ ≤ ϕ ϕ
0 1
03
r
π
ϕ
≤ ≤⇔
≤ ≤
13
0 0
( ) 3
D
S D dxdy d rdr = =∫∫ ∫ ∫
π
ϕ