Dvilypiai integralai skirti pagreitinti integralų (tūrio , ploto ir kt.) skaičiavimą, kad, užuot dalinus į kelias dalis ir integruojant kiekvieną dalį atskirai, būtų galimą greičiau suintegruoti.
[ keisti ] Dvilypio integralo skaičiavimo pavyzdžiai.
1. Dvilypį integralą
∬
D
f
(
x
;
y
)
d
x
d
y
{\displaystyle \iint _{D}f(x;y)dxdy}
D riboja ašis Oy , parabolė
y
=
x
{\displaystyle y={\sqrt {x}}}
x
+
y
=
2.
{\displaystyle x+y=2.}
x
+
y
=
2
{\displaystyle x+y=2}
y
=
x
{\displaystyle y={\sqrt {x}}}
A . Tuo tikslu išsprendžiame lygčių sistemą {
x
+
y
=
2
,
{\displaystyle x+y=2,}
{
y
=
x
,
{\displaystyle y={\sqrt {x}},}
iš kurios randame: x=1; y=1. Sritį D Gausime, kai x kis nuo 0 iki 1, o y - nuo apatinės kreivės
y
=
x
{\displaystyle y={\sqrt {x}}}
y
=
2
−
x
.
{\displaystyle y=2-x.}
∬
D
f
(
x
;
y
)
d
x
d
y
=
∫
0
1
d
x
∫
x
2
−
x
f
(
x
;
y
)
d
y
.
{\displaystyle \iint _{D}f(x;y)dxdy=\int _{0}^{1}dx\int _{\sqrt {x}}^{2-x}f(x;y)dy.}
Kad gautume D srities plotą reikia skaičiuoti šitaip:
∫
0
1
d
x
∫
x
2
−
x
d
y
=
∫
0
1
y
|
x
2
−
x
d
x
=
∫
0
1
(
2
−
x
−
x
)
d
x
=
(
2
x
−
x
2
2
−
2
3
x
3
2
)
|
0
1
=
2
−
1
2
−
2
3
=
5
6
.
{\displaystyle \int _{0}^{1}dx\int _{\sqrt {x}}^{2-x}dy=\int _{0}^{1}y|_{\sqrt {x}}^{2-x}dx=\int _{0}^{1}(2-x-{\sqrt {x}})dx=(2x-{\frac {x^{2}}{2}}-{\frac {2}{3}}x^{3 \over 2})|_{0}^{1}=2-{1 \over 2}-{2 \over 3}={5 \over 6}.}
(
x
=
y
)
{\displaystyle ({\sqrt {x}}=y)}
x
=
y
2
{\displaystyle x=y^{2}}
(
−
1
)
{\displaystyle (-1)}
S
Δ
=
1
2
1
⋅
1
=
1
2
,
{\displaystyle S_{\Delta }={1 \over 2}1\cdot 1={1 \over 2},}
S
p
a
r
a
b
=
∫
0
1
y
2
d
y
=
y
3
3
|
0
1
=
1
3
,
{\displaystyle S_{parab}=\int _{0}^{1}y^{2}dy={y^{3} \over 3}|_{0}^{1}={1 \over 3},}
S
=
S
Δ
+
S
p
a
r
a
b
=
1
2
+
1
3
=
5
6
.
{\displaystyle S=S_{\Delta }+S_{parab}={1 \over 2}+{1 \over 3}={5 \over 6}.}
Tą patį plotą galima gauti dvilypį integralą skaičiuojant šitaip:
∫
0
1
d
y
∫
0
y
2
d
x
+
∫
1
2
d
y
∫
0
2
−
y
d
x
=
∫
0
1
x
|
0
y
2
d
y
+
∫
1
2
x
|
0
2
−
y
d
x
=
∫
0
1
y
2
d
y
+
∫
1
2
(
2
−
y
)
d
y
=
{\displaystyle \int _{0}^{1}dy\int _{0}^{y^{2}}dx+\int _{1}^{2}dy\int _{0}^{2-y}dx=\int _{0}^{1}x|_{0}^{y^{2}}dy+\int _{1}^{2}x|_{0}^{2-y}dx=\int _{0}^{1}y^{2}dy+\int _{1}^{2}(2-y)dy=}
=
y
3
3
|
0
1
+
(
2
y
−
y
2
2
)
|
1
2
=
1
3
+
4
−
2
−
2
+
1
2
=
5
6
.
{\displaystyle ={y^{3} \over 3}|_{0}^{1}+(2y-{y^{2} \over 2})|_{1}^{2}={1 \over 3}+4-2-2+{1 \over 2}={5 \over 6}.}
Apskaičiuosime srities D plotą, kurį apriboja parabolė
y
=
x
{\displaystyle y={\sqrt {x}}}
y
=
x
2
,
{\displaystyle y=x^{2},}
x kinta nuo 0 iki 1.
∫
0
1
d
x
∫
x
2
x
d
y
=
∫
0
1
y
|
x
2
x
d
x
=
∫
0
1
(
x
−
x
2
)
d
x
=
(
2
3
x
3
2
−
x
3
3
)
|
0
1
=
2
3
−
1
3
=
1
3
.
{\displaystyle \int _{0}^{1}dx\int _{x^{2}}^{\sqrt {x}}dy=\int _{0}^{1}y|_{x^{2}}^{\sqrt {x}}dx=\int _{0}^{1}({\sqrt {x}}-x^{2})dx=({2 \over 3}x^{3 \over 2}-{x^{3} \over 3})|_{0}^{1}={2 \over 3}-{1 \over 3}={1 \over 3}.}
2. Tą patį atsakymą galima gauti iš pradžių apskaičiavus plotą po parabole
y
=
x
{\displaystyle y={\sqrt {x}}}
y
=
x
2
.
{\displaystyle y=x^{2}.}
y
=
x
2
{\displaystyle y=x^{2}}
S
1
=
∫
0
1
x
2
d
x
=
x
3
3
|
0
1
=
1
3
.
{\displaystyle S_{1}=\int _{0}^{1}x^{2}dx={x^{3} \over 3}|_{0}^{1}={1 \over 3}.}
Plotas po parabole
y
=
x
{\displaystyle y={\sqrt {x}}}
S
2
=
∫
0
1
x
d
x
=
2
3
x
3
2
|
0
1
=
2
3
.
{\displaystyle S_{2}=\int _{0}^{1}{\sqrt {x}}dx={2 \over 3}x^{3 \over 2}|_{0}^{1}={2 \over 3}.}
Atėmus plotą po parabole
y
=
x
{\displaystyle y={\sqrt {x}}}
y
=
x
2
{\displaystyle y=x^{2}}
S
=
S
2
−
S
1
=
2
3
−
1
3
=
1
3
.
{\displaystyle S=S_{2}-S_{1}={2 \over 3}-{1 \over 3}={1 \over 3}.}
Apskaičiuosime integralą
∬
D
x
d
x
d
y
{\displaystyle \iint _{D}xdxdy}
D , apribota linija
y
=
−
x
;
{\displaystyle y=-x;}
y
=
1
;
{\displaystyle y=1;}
y
=
x
2
.
{\displaystyle y=x^{2}.}
a lygiagreti x ašiai ir lygi 1, o kita krašinėb sutampa su y ašimi ir taip pat lygi 1, o trikampio įžambinė lygi
1
2
+
1
2
=
2
{\displaystyle {\sqrt {1^{2}+1^{2}}}={\sqrt {2}}}
x ašiai ir lygi 1 (nes
1
=
x
2
=
1
2
{\displaystyle 1=x^{2}=1^{2}}
y ašimi ir taip pat lygi 1, šoną riboja parabolė. 3.
∬
D
d
x
d
y
=
∫
0
1
d
y
∫
−
y
y
d
x
=
∫
0
1
x
|
−
y
y
d
y
=
∫
0
1
(
y
+
y
)
d
y
=
{\displaystyle \iint _{D}dxdy=\int _{0}^{1}dy\int _{-y}^{\sqrt {y}}dx=\int _{0}^{1}x|_{-y}^{\sqrt {y}}dy=\int _{0}^{1}({\sqrt {y}}+y)dy=}
=
(
2
3
y
3
2
+
y
2
2
)
|
0
1
=
2
3
+
1
2
=
7
6
.
{\displaystyle =({2 \over 3}y^{3 \over 2}+{y^{2} \over 2})|_{0}^{1}={2 \over 3}+{1 \over 2}={7 \over 6}.}
Šios dvimatės figuros plotą taip pat galima apskaičiuoti šitaip:
S
Δ
=
1
2
;
{\displaystyle S_{\Delta }={1 \over 2};}
S
1
=
1
2
−
∫
0
1
x
2
d
x
=
1
−
x
3
3
|
0
1
=
1
−
1
3
=
2
3
;
{\displaystyle S_{1}=1^{2}-\int _{0}^{1}x^{2}dx=1-{x^{3} \over 3}|_{0}^{1}=1-{1 \over 3}={2 \over 3};}
S
=
S
Δ
+
S
1
=
1
2
+
2
3
=
7
6
.
{\displaystyle S=S_{\Delta }+S_{1}={1 \over 2}+{2 \over 3}={7 \over 6}.}
4. Apskaičiuosime ploksčios figuros plotą D apribotą
y
2
=
x
+
1
;
{\displaystyle y^{2}=x+1;}
x
+
y
=
1.
{\displaystyle x+y=1.}
D apribota iš kairės parabole
x
=
y
2
−
1
,
{\displaystyle x=y^{2}-1,}
x
=
1
−
y
.
{\displaystyle x=1-y.}
y
2
−
1
=
1
−
y
,
{\displaystyle y^{2}-1=1-y,}
y
2
+
y
−
2
=
0
;
{\displaystyle y^{2}+y-2=0;}
D
=
b
2
−
4
a
c
=
1
+
8
=
9
;
{\displaystyle D=b^{2}-4ac=1+8=9;}
x
1
;
2
=
−
b
±
D
2
a
=
−
1
±
3
2
=
−
2
;
1.
{\displaystyle x_{1;2}={-b\pm {\sqrt {D}} \over 2a}={-1\pm 3 \over 2}=-2;1.}
∬
D
d
x
d
y
=
∫
−
2
1
d
y
∫
y
2
−
1
1
−
y
d
x
=
∫
−
2
1
(
2
−
y
−
y
2
)
d
y
=
(
2
y
−
y
2
2
−
y
3
3
)
|
−
2
1
=
2
−
1
2
−
1
3
+
4
+
2
−
8
3
=
9
2
.
{\displaystyle \iint _{D}dxdy=\int _{-2}^{1}dy\int _{y^{2}-1}^{1-y}dx=\int _{-2}^{1}(2-y-y^{2})dy=(2y-{y^{2} \over 2}-{y^{3} \over 3})|_{-2}^{1}=2-{1 \over 2}-{1 \over 3}+4+2-{8 \over 3}={9 \over 2}.}
S
1
=
∫
−
2
1
(
1
−
y
)
d
y
=
(
y
−
y
2
2
|
−
2
1
=
1
−
1
2
−
(
−
2
−
2
)
=
5
−
1
2
=
9
2
;
{\displaystyle S_{1}=\int _{-2}^{1}(1-y)dy=(y-{y^{2} \over 2}|_{-2}^{1}=1-{1 \over 2}-(-2-2)=5-{1 \over 2}={9 \over 2};}
S
2
=
∫
−
1
1
(
y
2
−
1
)
d
y
=
(
y
3
3
−
y
)
|
−
1
1
=
1
3
−
1
−
(
−
1
3
+
1
)
=
2
3
−
2
=
−
4
3
;
|
S
2
|
=
4
3
.
{\displaystyle S_{2}=\int _{-1}^{1}(y^{2}-1)dy=({y^{3} \over 3}-y)|_{-1}^{1}={1 \over 3}-1-(-{1 \over 3}+1)={2 \over 3}-2=-{4 \over 3};\;|S_{2}|={4 \over 3}.}
S
3
=
∫
−
2
−
1
(
y
2
−
1
)
d
y
=
(
y
3
3
−
y
)
|
−
2
−
1
=
−
8
3
+
2
−
(
−
1
3
+
1
)
=
−
7
3
+
1
=
−
4
3
;
|
S
3
|
=
4
3
.
{\displaystyle S_{3}=\int _{-2}^{-1}(y^{2}-1)dy=({y^{3} \over 3}-y)|_{-2}^{-1}=-{8 \over 3}+2-(-{1 \over 3}+1)=-{7 \over 3}+1=-{4 \over 3};\;|S_{3}|={4 \over 3}.}
S
=
S
1
−
|
S
3
|
+
|
S
2
|
=
9
2
−
4
3
+
4
3
=
9
2
.
{\displaystyle S=S_{1}-|S_{3}|+|S_{2}|={9 \over 2}-{4 \over 3}+{4 \over 3}={9 \over 2}.}
Vaizdas:Integral379380.jpg 379. Rasime tūrį kūno V , apriboto paviršiais
y
=
x
2
;
{\displaystyle y=x^{2};}
y
=
1
;
{\displaystyle y=1;}
z
=
0
;
{\displaystyle z=0;}
z
=
x
2
+
y
2
.
{\displaystyle z=x^{2}+y^{2}.}
D , apribotas iš viršaus paraboloidu
z
=
x
2
+
y
2
,
{\displaystyle z=x^{2}+y^{2},}
V
=
∬
D
(
x
2
+
y
2
)
d
x
d
y
=
2
∫
0
1
d
y
∫
0
y
(
x
2
+
y
2
)
d
x
=
2
∫
0
1
(
x
3
3
+
x
y
2
)
|
0
y
=
2
∫
0
1
(
1
3
y
3
2
+
y
5
2
)
d
y
=
{\displaystyle V=\iint _{D}(x^{2}+y^{2})dxdy=2\int _{0}^{1}dy\int _{0}^{\sqrt {y}}(x^{2}+y^{2})dx=2\int _{0}^{1}({x^{3} \over 3}+xy^{2})|_{0}^{\sqrt {y}}=2\int _{0}^{1}({1 \over 3}y^{3 \over 2}+y^{5 \over 2})dy=}
=
2
(
2
3
⋅
5
y
5
2
+
2
7
y
7
2
)
|
0
1
=
2
(
2
15
+
2
7
)
=
88
105
.
{\displaystyle =2({2 \over 3\cdot 5}y^{5 \over 2}+{2 \over 7}y^{7 \over 2})|_{0}^{1}=2({2 \over 15}+{2 \over 7})={88 \over 105}.}
Vaizdas:Integral379380.jpg 380. Rasime tūrį kūno V , išpjauto iš begalines prizmės (stačiakampio gretasienio , kurio plotis ir aukštis
a
=
b
=
2
{\displaystyle a=b=2}
c
=
∞
{\displaystyle c=\infty }
x
=
±
1
;
{\displaystyle x=\pm 1;}
y
=
±
1
{\displaystyle y=\pm 1}
x
2
+
y
2
=
4
−
z
;
{\displaystyle x^{2}+y^{2}=4-z;}
x
2
+
y
2
=
4
(
z
+
2
)
.
{\displaystyle x^{2}+y^{2}=4(z+2).}
V randame kaip sumą tūrių
V
1
{\displaystyle V_{1}}
V
2
{\displaystyle V_{2}}
XOY . Tokiu budu
V
1
=
∬
D
(
4
−
x
2
−
y
2
)
d
x
d
y
=
4
∫
0
1
d
x
∫
0
1
(
4
−
x
2
−
y
2
)
d
y
=
4
∫
0
1
(
4
−
x
2
−
1
3
)
d
x
=
40
3
=
13.33333333
;
{\displaystyle V_{1}=\iint _{D}(4-x^{2}-y^{2})dxdy=4\int _{0}^{1}dx\int _{0}^{1}(4-x^{2}-y^{2})dy=4\int _{0}^{1}(4-x^{2}-{1 \over 3})dx={40 \over 3}=13.33333333;}
V
2
=
−
∬
D
(
x
2
+
y
2
4
−
2
)
d
x
d
y
=
−
4
∫
0
1
d
x
∫
0
1
(
x
2
+
y
2
4
−
2
)
d
y
=
−
4
∫
0
1
(
x
2
4
+
1
12
−
2
)
d
x
=
{\displaystyle V_{2}=-\iint _{D}({x^{2}+y^{2} \over 4}-2)dxdy=-4\int _{0}^{1}dx\int _{0}^{1}({x^{2}+y^{2} \over 4}-2)dy=-4\int _{0}^{1}({x^{2} \over 4}+{1 \over 12}-2)dx=}
=
−
4
(
1
12
+
1
12
−
2
)
=
22
3
=
7.33333333
;
{\displaystyle =-4({1 \over 12}+{1 \over 12}-2)={22 \over 3}=7.33333333;}
V
=
V
1
+
V
2
=
40
3
+
22
3
=
62
3
=
20.66666667.
{\displaystyle V=V_{1}+V_{2}={40 \over 3}+{22 \over 3}={62 \over 3}=20.66666667.}
Vaizdas:Cilpav139.jpg 13.9. Kūną riboja koordinačių plokštumos
x
=
0
,
{\displaystyle x=0,}
y
=
0
,
{\displaystyle y=0,}
z
=
0
,
{\displaystyle z=0,}
z
=
9
−
x
2
{\displaystyle z=9-x^{2}}
x
+
y
=
3
{\displaystyle x+y=3}
D yra projekcija plokštumoje xOy . Žinome, kad kūno tūris
V
=
∬
D
(
9
−
x
2
)
d
x
d
y
=
∫
0
3
d
x
∫
0
3
−
x
(
9
−
x
2
)
d
y
=
∫
0
3
(
9
y
−
x
2
y
)
|
0
3
−
x
d
x
=
{\displaystyle V=\iint _{D}(9-x^{2})dxdy=\int _{0}^{3}dx\int _{0}^{3-x}(9-x^{2})dy=\int _{0}^{3}(9y-x^{2}y)|_{0}^{3-x}dx=}
=
∫
0
3
(
27
−
9
x
−
3
x
2
+
x
3
)
d
x
=
(
27
x
−
9
x
2
2
−
x
3
+
x
4
4
)
|
0
3
=
81
−
40.5
−
27
+
20.25
=
33.75.
{\displaystyle =\int _{0}^{3}(27-9x-3x^{2}+x^{3})dx=(27x-{9x^{2} \over 2}-x^{3}+{x^{4} \over 4})|_{0}^{3}=81-40.5-27+20.25=33.75.}
Apitiksliu budu patikrinsime ar atsakymas teisingas. Ant xOy plokštumos yra pagrindas, kuris yra ligiakraštis trikampis su kraštinėmis a=b=3;
c
=
3
2
+
3
2
=
18
=
3
2
=
4.242640687
{\displaystyle c={\sqrt {3^{2}+3^{2}}}={\sqrt {18}}=3{\sqrt {2}}=4.242640687}
Ox ašis padalinama į 30 dalių. O trikampio plotas gali būti gautas tokiu budu:
S
Δ
=
3
30
⋅
(
3
+
(
3
−
1
⋅
0
,
1
)
+
(
3
−
2
⋅
0
,
1
)
+
(
3
−
3
⋅
0
,
1
)
+
(
3
−
4
⋅
0
,
1
)
+
(
3
−
5
⋅
0
,
1
)
+
.
.
.
+
(
3
−
30
⋅
0
,
1
)
)
=
{\displaystyle S_{\Delta }={\frac {3}{30}}\cdot (3+(3-1\cdot 0,1)+(3-2\cdot 0,1)+(3-3\cdot 0,1)+(3-4\cdot 0,1)+(3-5\cdot 0,1)+...+(3-30\cdot 0,1))=}
=
0.1
(
3
∗
30
−
0.1
−
0.2
−
0.3
−
0.4
−
0.5
−
0.6
−
0.7
−
0.8
−
0.9
−
1
−
1.1
−
1.2
−
1.3
−
1.4
−
1.5
−
1.6
−
1.7
−
1.8
−
1.9
−
2
−
2.1
−
2.2
−
2.3
−
2.4
−
2.5
−
2.6
−
2.7
−
2.8
−
2.9
)
=
{\displaystyle =0.1(3*30-0.1-0.2-0.3-0.4-0.5-0.6-0.7-0.8-0.9-1-1.1-1.2-1.3-1.4-1.5-1.6-1.7-1.8-1.9-2-2.1-2.2-2.3-2.4-2.5-2.6-2.7-2.8-2.9)=}
=
0.1
(
90
−
5.5
−
15.5
−
22.5
)
=
0.1
(
90
−
43.5
)
=
0.1
(
46.5
)
=
4.65.
{\displaystyle =0.1(90-5.5-15.5-22.5)=0.1(90-43.5)=0.1(46.5)=4.65.}
S
Δ
=
a
⋅
b
2
=
3
⋅
3
2
=
4.5
{\displaystyle S_{\Delta }={\frac {a\cdot b}{2}}={\frac {3\cdot 3}{2}}=4.5}
O viso kūno tūrį padaliname į 30 stačiakampių gretasienių:
V
=
3
30
⋅
[
3
⋅
9
+
(
3
−
1
⋅
0.1
)
(
9
−
(
1
⋅
0.1
)
2
)
+
(
3
−
2
⋅
0.1
)
(
9
−
(
2
⋅
0.1
)
2
)
+
(
3
−
3
⋅
0.1
)
(
9
−
(
3
⋅
0.1
)
2
)
+
.
.
.
+
(
3
−
30
⋅
0.1
)
(
9
−
(
30
⋅
0.1
)
2
)
]
=
{\displaystyle V={\frac {3}{30}}\cdot [3\cdot 9+(3-1\cdot 0.1)(9-(1\cdot 0.1)^{2})+(3-2\cdot 0.1)(9-(2\cdot 0.1)^{2})+(3-3\cdot 0.1)(9-(3\cdot 0.1)^{2})+...+(3-30\cdot 0.1)(9-(30\cdot 0.1)^{2})]=}
=
0.1
(
18
+
2.9
∗
8.99
+
2.8
∗
8.96
+
2.7
∗
8.91
+
2.6
∗
8.84
+
2.5
∗
8.75
+
2.4
∗
8.64
+
2.3
∗
8.51
+
2.2
∗
8.36
+
2.1
∗
8.19
+
2
∗
8
+
{\displaystyle =0.1(18+2.9*8.99+2.8*8.96+2.7*8.91+2.6*8.84+2.5*8.75+2.4*8.64+2.3*8.51+2.2*8.36+2.1*8.19+2*8+}
+
1.9
∗
7.79
+
1.8
∗
7.56
+
1.7
∗
7.31
+
1.6
∗
7.04
+
1.5
∗
6.75
+
1.4
∗
6.44
+
1.3
∗
6.11
+
1.2
∗
5.76
+
1.1
∗
5.39
+
1
∗
(
9
−
2
2
)
+
{\displaystyle +1.9*7.79+1.8*7.56+1.7*7.31+1.6*7.04+1.5*6.75+1.4*6.44+1.3*6.11+1.2*5.76+1.1*5.39+1*(9-2^{2})+}
+
0.9
∗
4.59
+
0.8
∗
4.16
+
0.7
∗
3.71
+
0.6
∗
3.24
+
0.5
∗
2.75
+
0.4
∗
2.24
+
0.3
∗
1.71
+
0.2
∗
1.16
+
0.1
∗
0.59
)
=
{\displaystyle +0.9*4.59+0.8*4.16+0.7*3.71+0.6*3.24+0.5*2.75+0.4*2.24+0.3*1.71+0.2*1.16+0.1*0.59)=}
=
0.1
(
18
+
211.975
+
97.025
+
15.075
)
=
0.1
(
18
+
324.075
)
=
34.2075.
{\displaystyle =0.1(18+211.975+97.025+15.075)=0.1(18+324.075)=34.2075.}
Tūrių atsakymai yra panašūs, o padalinus į daugiau dalių atsakymai taptų dar panašesni.
Pagal kai kuriuos samprotavimus, iš esmės čia yra plotas po parabolės šaka atimtas iš stačiakampiom kurio plotas
y
x
=
x
2
x
{\displaystyle yx=x^{2}x}
V
m
a
x
=
h
(
x
2
x
−
∫
0
3
x
2
d
x
)
=
3
(
3
2
⋅
3
−
∫
0
3
x
2
d
x
)
=
3
(
27
−
x
3
3
|
0
3
)
=
3
(
27
−
(
3
3
3
−
0
)
)
=
3
(
27
−
27
3
)
=
3
(
27
−
9
)
=
3
⋅
18
=
54.
{\displaystyle V_{max}=h(x^{2}x-\int _{0}^{3}x^{2}\;dx)=3(3^{2}\cdot 3-\int _{0}^{3}x^{2}\;dx)=3(27-{\frac {x^{3}}{3}}|_{0}^{3})=3(27-({\frac {3^{3}}{3}}-0))=3(27-{\frac {27}{3}})=3(27-9)=3\cdot 18=54.}
Palyginimui, stačiakampio gretasienio, kurio kraštinės a=3, b=3, c=9, tūris yra
V
s
t
=
a
b
c
=
3
⋅
3
⋅
9
=
81.
{\displaystyle V_{st}=abc=3\cdot 3\cdot 9=81.}
5. Apskaičiuosime tūrį kūno, apriboto paviršiais
x
=
0
,
{\displaystyle x=0,}
y
=
0
,
{\displaystyle y=0,}
z
=
0
{\displaystyle z=0}
x
+
y
+
z
=
1.
{\displaystyle x+y+z=1.}
V
=
∬
D
(
1
−
x
−
y
)
d
x
d
y
,
{\displaystyle V=\iint _{D}(1-x-y)dxdy,}
D - trikampė integravimo sritis, apribota tiesėmis
x
=
0
,
{\displaystyle x=0,}
y
=
0
,
{\displaystyle y=0,}
x
+
y
=
1.
{\displaystyle x+y=1.}
V
=
∫
0
1
d
x
∫
0
1
−
x
(
1
−
x
−
y
)
d
y
=
∫
0
1
(
y
−
x
y
−
y
2
2
)
|
0
1
−
x
d
x
=
{\displaystyle V=\int _{0}^{1}dx\int _{0}^{1-x}(1-x-y)dy=\int _{0}^{1}(y-xy-{y^{2} \over 2})|_{0}^{1-x}dx=}
=
∫
0
1
(
1
−
x
−
x
+
x
2
−
1
2
+
x
−
x
2
2
)
d
x
=
∫
0
1
(
1
2
−
x
+
x
2
2
)
d
x
=
{\displaystyle =\int _{0}^{1}(1-x-x+x^{2}-{1 \over 2}+x-{x^{2} \over 2})dx=\int _{0}^{1}({1 \over 2}-x+{x^{2} \over 2})dx=}
=
(
x
2
−
x
2
2
+
x
3
6
)
|
0
1
=
1
2
−
1
2
+
1
6
=
1
6
.
{\displaystyle =({x \over 2}-{x^{2} \over 2}+{x^{3} \over 6})|_{0}^{1}={1 \over 2}-{1 \over 2}+{1 \over 6}={1 \over 6}.}
Šį tūrį galima rasti ir taikant piramidės formulę:
V
=
1
3
⋅
B
⋅
h
=
1
3
⋅
1
2
⋅
1
⋅
1
⋅
1
=
1
6
,
{\displaystyle V={\frac {1}{3}}\cdot B\cdot h={\frac {1}{3}}\cdot {\frac {1}{2}}\cdot 1\cdot 1\cdot 1={\frac {1}{6}},}
6. Kūną riboja tiesės
x
=
y
+
2
{\displaystyle x=y+2}
x
=
y
2
{\displaystyle x=y^{2}}
diskriminantą .
y
+
2
=
y
2
,
{\displaystyle y+2=y^{2},}
y
2
−
y
−
2
=
0
,
{\displaystyle y^{2}-y-2=0,}
D
=
1
−
4
(
−
2
)
=
9
;
{\displaystyle D=1-4(-2)=9;}
x
1
,
2
=
1
±
3
2
=
−
1
;
2.
{\displaystyle x_{1,2}={1\pm 3 \over 2}=-1;2.}
S
=
∬
D
d
x
d
y
=
∫
−
1
2
d
y
∫
y
2
y
+
2
d
x
=
∫
−
1
2
(
y
+
2
−
y
2
)
d
y
=
{\displaystyle S=\iint _{D}dxdy=\int _{-1}^{2}dy\int _{y^{2}}^{y+2}dx=\int _{-1}^{2}(y+2-y^{2})dy=}
=
(
y
2
2
+
2
y
−
y
3
3
)
|
−
1
2
=
2
+
4
−
8
3
−
(
1
2
−
2
+
1
3
)
=
8
−
1
2
−
3
=
9
2
.
{\displaystyle =({y^{2} \over 2}+2y-{y^{3} \over 3})|_{-1}^{2}=2+4-{8 \over 3}-({1 \over 2}-2+{1 \over 3})=8-{1 \over 2}-3={9 \over 2}.}
Šį plotą galimą buvo lengvai apskaičiuoti be dvilypio integralo:
S
1
=
∫
−
1
2
(
y
+
2
)
d
y
=
(
y
2
2
+
2
y
)
|
−
1
2
=
2
+
4
−
(
1
2
−
2
)
=
8
−
1
2
=
15
2
;
{\displaystyle S_{1}=\int _{-1}^{2}(y+2)dy=({y^{2} \over 2}+2y)|_{-1}^{2}=2+4-({1 \over 2}-2)=8-{1 \over 2}={15 \over 2};}
S
2
=
∫
−
1
2
y
2
d
y
=
y
3
3
|
−
1
2
=
8
3
+
1
3
=
3
;
{\displaystyle S_{2}=\int _{-1}^{2}y^{2}dy={y^{3} \over 3}|_{-1}^{2}={8 \over 3}+{1 \over 3}=3;}
S
=
S
1
−
S
2
=
15
2
−
3
=
15
−
6
2
=
9
2
.
{\displaystyle S=S_{1}-S_{2}={15 \over 2}-3={15-6 \over 2}={9 \over 2}.}
[ keisti ] Polinėje koordinačių sistemoje
∬
D
f
(
x
,
y
)
d
x
d
y
=
∬
D
f
(
ρ
cos
ϕ
,
ρ
sin
ϕ
)
ρ
d
ρ
d
ϕ
.
{\displaystyle \iint _{D}f(x,y)dxdy=\iint _{D}f(\rho \cos \phi ,\rho \sin \phi )\rho d\rho d\phi .}
x
=
ρ
cos
ϕ
;
y
=
ρ
sin
ϕ
.
{\displaystyle x=\rho \cos \phi ;\;y=\rho \sin \phi .}
x
2
+
y
2
=
ρ
2
cos
2
ϕ
+
ρ
2
sin
2
ϕ
=
ρ
2
.
{\displaystyle x^{2}+y^{2}=\rho ^{2}\cos ^{2}\phi +\rho ^{2}\sin ^{2}\phi =\rho ^{2}.}
Paraboloidas. Kūną riboja plokštuma xOy , cilindrinis paviršius
x
2
+
y
2
=
1
{\displaystyle x^{2}+y^{2}=1}
z
=
x
2
+
y
2
.
{\displaystyle z=x^{2}+y^{2}.}
D yra skritulio dalis, esanti I ketvirtyje, tai
0
≤
ϕ
≤
π
2
,
0
≤
ρ
≤
1.
{\displaystyle 0\leq \phi \leq {\pi \over 2},\;0\leq \rho \leq 1.}
∬
D
(
x
2
+
y
2
)
d
x
d
y
=
∬
D
ρ
2
⋅
ρ
d
ρ
d
ϕ
=
∫
0
π
2
d
ϕ
∫
0
1
ρ
3
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
1
d
ϕ
=
1
4
∫
0
π
2
d
ϕ
=
1
4
ϕ
|
0
π
2
=
π
8
.
{\displaystyle \iint _{D}(x^{2}+y^{2})dxdy=\iint _{D}\rho ^{2}\cdot \rho d\rho d\phi =\int _{0}^{\pi \over 2}d\phi \int _{0}^{1}\rho ^{3}d\rho =\int _{0}^{\pi \over 2}{\rho ^{4} \over 4}|_{0}^{1}d\phi ={1 \over 4}\int _{0}^{\pi \over 2}d\phi ={1 \over 4}\phi |_{0}^{\pi \over 2}={\pi \over 8}.}
V
=
4
⋅
π
8
=
π
2
.
{\displaystyle V=4\cdot {\pi \over 8}={\pi \over 2}.}
Kūną riboja plokštuma xOy , cilindrinis paviršius
x
2
+
y
2
=
4
{\displaystyle x^{2}+y^{2}=4}
z
=
x
2
+
y
2
.
{\displaystyle z=x^{2}+y^{2}.}
D yra skritulis ant plokštumos xOy , tai
0
≤
ϕ
≤
2
π
,
0
≤
ρ
≤
4.
{\displaystyle 0\leq \phi \leq 2\pi ,\;0\leq \rho \leq 4.}
V
=
∬
D
(
x
2
+
y
2
)
d
x
d
y
=
∬
D
ρ
2
⋅
ρ
d
ρ
d
ϕ
=
∫
0
2
π
d
ϕ
∫
0
4
ρ
3
d
ρ
=
∫
0
2
π
ρ
4
4
|
0
4
d
ϕ
=
{\displaystyle V=\iint _{D}(x^{2}+y^{2})dxdy=\iint _{D}\rho ^{2}\cdot \rho d\rho d\phi =\int _{0}^{2\pi }d\phi \int _{0}^{4}\rho ^{3}d\rho =\int _{0}^{2\pi }{\rho ^{4} \over 4}|_{0}^{4}d\phi =}
=
(
4
4
4
−
0
4
4
)
∫
0
2
π
d
ϕ
=
256
4
ϕ
|
0
2
π
=
64
⋅
(
2
π
−
0
)
=
128
π
=
402.1238597.
{\displaystyle =({4^{4} \over 4}-{\frac {0^{4}}{4}})\int _{0}^{2\pi }d\phi ={256 \over 4}\phi |_{0}^{2\pi }=64\cdot (2\pi -0)=128\pi =402.1238597.}
Patikrinsime paraboloido
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
0
≤
x
≤
4
{\displaystyle 0\leq x\leq 4}
0
≤
y
≤
4
{\displaystyle 0\leq y\leq 4}
Ox ašį ilgis yra 0,4 (y reikšmės tuomet visada būna 0). Todėl reikiau gauti visas x reikšmes:
x
0
=
0
;
{\displaystyle x_{0}=0;}
x
1
=
0.4
;
{\displaystyle x_{1}=0.4;}
x
2
=
2
⋅
0.4
=
0.8
;
{\displaystyle x_{2}=2\cdot 0.4=0.8;}
x
3
=
3
⋅
0.4
=
1.2
;
{\displaystyle x_{3}=3\cdot 0.4=1.2;}
x
4
=
4
⋅
0.4
=
1.6
;
{\displaystyle x_{4}=4\cdot 0.4=1.6;}
x
5
=
5
⋅
0.4
=
2
;
{\displaystyle x_{5}=5\cdot 0.4=2;}
x
6
=
6
⋅
0.4
=
2.4
;
{\displaystyle x_{6}=6\cdot 0.4=2.4;}
x
7
=
7
⋅
0.4
=
2.8
;
{\displaystyle x_{7}=7\cdot 0.4=2.8;}
x
8
=
8
⋅
0.4
=
3.2
;
{\displaystyle x_{8}=8\cdot 0.4=3.2;}
x
9
=
9
⋅
0.4
=
3.6
;
{\displaystyle x_{9}=9\cdot 0.4=3.6;}
x
10
=
10
⋅
0.4
=
4.
{\displaystyle x_{10}=10\cdot 0.4=4.}
Dabar toliau reikia surasti visas z reikšmes, įstačius x reikšmes:
z
0
=
x
0
2
=
0
2
=
0
;
{\displaystyle z_{0}=x_{0}^{2}=0^{2}=0;}
z
1
=
x
1
2
=
0.4
2
=
0.16
;
{\displaystyle z_{1}=x_{1}^{2}=0.4^{2}=0.16;}
z
2
=
x
2
2
=
0.8
2
=
0.64
;
{\displaystyle z_{2}=x_{2}^{2}=0.8^{2}=0.64;}
z
3
=
x
3
2
=
1.2
2
=
1.44
;
{\displaystyle z_{3}=x_{3}^{2}=1.2^{2}=1.44;}
z
4
=
x
4
2
=
1.6
2
=
2.56
;
{\displaystyle z_{4}=x_{4}^{2}=1.6^{2}=2.56;}
z
5
=
x
5
2
=
2
2
=
4
;
{\displaystyle z_{5}=x_{5}^{2}=2^{2}=4;}
z
6
=
x
6
2
=
2.4
2
=
5.76
;
{\displaystyle z_{6}=x_{6}^{2}=2.4^{2}=5.76;}
z
7
=
x
7
2
=
2.8
2
=
7.84
;
{\displaystyle z_{7}=x_{7}^{2}=2.8^{2}=7.84;}
z
8
=
x
8
2
=
3.2
2
=
10.24
;
{\displaystyle z_{8}=x_{8}^{2}=3.2^{2}=10.24;}
z
9
=
x
9
2
=
3.6
2
=
12.96
;
{\displaystyle z_{9}=x_{9}^{2}=3.6^{2}=12.96;}
z
10
=
x
10
2
=
4
2
=
16.
{\displaystyle z_{10}=x_{10}^{2}=4^{2}=16.}
Dabar sudėsime 10 diskų, kai kiekvieno disko aukštis yra
h
1
=
z
1
−
z
0
{\displaystyle h_{1}=z_{1}-z_{0}}
h
2
=
z
2
−
z
1
{\displaystyle h_{2}=z_{2}-z_{1}}
h
3
=
z
3
−
z
2
{\displaystyle h_{3}=z_{3}-z_{2}}
V
=
π
(
r
1
2
(
z
1
−
z
0
)
+
r
2
2
(
z
2
−
z
1
)
+
r
3
2
(
z
3
−
z
2
)
+
r
4
2
(
z
4
−
z
3
)
+
r
5
2
(
z
5
−
z
4
)
+
r
6
2
(
z
6
−
z
5
)
+
r
7
2
(
z
7
−
z
6
)
+
r
8
2
(
z
8
−
z
7
)
+
r
9
2
(
z
9
−
z
8
)
+
r
10
2
(
z
10
−
z
9
)
)
=
{\displaystyle V=\pi (r_{1}^{2}(z_{1}-z_{0})+r_{2}^{2}(z_{2}-z_{1})+r_{3}^{2}(z_{3}-z_{2})+r_{4}^{2}(z_{4}-z_{3})+r_{5}^{2}(z_{5}-z_{4})+r_{6}^{2}(z_{6}-z_{5})+r_{7}^{2}(z_{7}-z_{6})+r_{8}^{2}(z_{8}-z_{7})+r_{9}^{2}(z_{9}-z_{8})+r_{10}^{2}(z_{10}-z_{9}))=}
=
π
(
0.4
2
(
0.16
−
0
)
+
0.8
2
(
0.64
−
0.16
)
+
1.2
2
(
1.44
−
0.64
)
+
1.6
2
(
2.56
−
1.44
)
+
2
2
(
4
−
2.56
)
+
2.4
2
(
5.76
−
4
)
+
2.8
2
(
7.84
−
5.76
)
+
3.2
2
(
10.24
−
7.84
)
+
3.6
2
(
12.96
−
10.24
)
+
4
2
(
16
−
12.96
)
)
=
{\displaystyle =\pi (0.4^{2}(0.16-0)+0.8^{2}(0.64-0.16)+1.2^{2}(1.44-0.64)+1.6^{2}(2.56-1.44)+2^{2}(4-2.56)+2.4^{2}(5.76-4)+2.8^{2}(7.84-5.76)+3.2^{2}(10.24-7.84)+3.6^{2}(12.96-10.24)+4^{2}(16-12.96))=}
=
π
(
0.4
2
⋅
0.16
+
0.8
2
⋅
0.48
+
1.2
2
⋅
0.8
+
1.6
2
⋅
1.12
+
2
2
⋅
1.44
+
2.4
2
⋅
1.76
+
2.8
2
⋅
2.08
+
3.2
2
⋅
2.4
+
3.6
2
⋅
2.72
+
4
2
⋅
3.04
)
=
{\displaystyle =\pi (0.4^{2}\cdot 0.16+0.8^{2}\cdot 0.48+1.2^{2}\cdot 0.8+1.6^{2}\cdot 1.12+2^{2}\cdot 1.44+2.4^{2}\cdot 1.76+2.8^{2}\cdot 2.08+3.2^{2}\cdot 2.4+3.6^{2}\cdot 2.72+4^{2}\cdot 3.04)=}
=
π
(
0.16
⋅
0.16
+
0.64
⋅
0.48
+
1.44
⋅
0.8
+
2.56
⋅
1.12
+
4
⋅
1.44
+
5.76
⋅
1.76
+
7.84
⋅
2.08
+
10.24
⋅
2.4
+
12.96
⋅
2.72
+
16
⋅
3.04
)
=
{\displaystyle =\pi (0.16\cdot 0.16+0.64\cdot 0.48+1.44\cdot 0.8+2.56\cdot 1.12+4\cdot 1.44+5.76\cdot 1.76+7.84\cdot 2.08+10.24\cdot 2.4+12.96\cdot 2.72+16\cdot 3.04)=}
=
π
(
0.0256
+
0.3072
+
1.152
+
2.8672
+
5.76
+
10.1376
+
16.3072
+
24.576
+
35.2512
+
48.64
)
=
145.024
π
=
455.606333.
{\displaystyle =\pi (0.0256+0.3072+1.152+2.8672+5.76+10.1376+16.3072+24.576+35.2512+48.64)=145.024\pi =455.606333.}
Na, gavosi daugiau nei integravimo budu
V
=
402.1238597
,
{\displaystyle V=402.1238597,}
Palyginimui cilindro tūris lygus dviems paraboloido turiams
V
c
i
l
=
π
r
2
h
=
π
⋅
4
2
⋅
16
=
256
π
=
804.2477193.
{\displaystyle V_{cil}=\pi r^{2}h=\pi \cdot 4^{2}\cdot 16=256\pi =804.2477193.}
1. Figūrą riboja kreivės
x
2
+
y
2
=
4
x
,
{\displaystyle x^{2}+y^{2}=4x,}
x
2
+
y
2
=
8
x
,
{\displaystyle x^{2}+y^{2}=8x,}
y
=
x
,
{\displaystyle y=x,}
y
=
x
3
.
{\displaystyle y=x{\sqrt {3}}.}
x
2
+
y
2
=
4
x
,
{\displaystyle x^{2}+y^{2}=4x,}
x
2
+
y
2
=
8
x
.
{\displaystyle x^{2}+y^{2}=8x.}
x
2
+
y
2
=
4
x
{\displaystyle x^{2}+y^{2}=4x}
(
x
−
2
)
2
+
y
2
=
4
,
{\displaystyle (x-2)^{2}+y^{2}=4,}
x
2
+
y
2
=
8
x
{\displaystyle x^{2}+y^{2}=8x}
(
x
−
4
)
2
+
y
2
=
16.
{\displaystyle (x-4)^{2}+y^{2}=16.}
Tai apskritimo lygtys. Pirmojo apskritimo centras yra taškas (2; 0), o spindulys lygus 2, antrojo centras - taškas (4; 0), o spindulys lygus 4. Parašykime tų apskritimų lygtis polinėje koordinačių sistemoje:
ρ
=
4
cos
ϕ
{\displaystyle \rho =4\cos \phi }
ρ
=
8
cos
ϕ
.
{\displaystyle \rho =8\cos \phi .}
y
=
x
{\displaystyle y=x}
Ox sudaro kampą
π
4
,
{\displaystyle {\pi \over 4},}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
x
=
3
;
tan
ϕ
=
3
;
ϕ
=
arctan
3
=
π
3
.
{\displaystyle {y \over x}={\sqrt {3}};\;\tan \phi ={\sqrt {3}};\;\phi =\arctan {\sqrt {3}}={\pi \over 3}.}
D gauname, kai
ϕ
{\displaystyle \phi }
π
4
{\displaystyle {\pi \over 4}}
π
3
,
{\displaystyle {\pi \over 3},}
ρ
{\displaystyle \rho }
4
cos
ϕ
{\displaystyle 4\cos \phi }
8
cos
ϕ
.
{\displaystyle 8\cos \phi .}
S
=
∬
D
d
x
d
y
=
∬
D
ρ
d
ρ
d
ϕ
,
{\displaystyle S=\iint _{D}dxdy=\iint _{D}\rho d\rho d\phi ,}
S
=
∫
π
4
π
3
d
ϕ
∫
4
cos
ϕ
8
cos
ϕ
ρ
d
ρ
=
1
2
∫
π
4
π
3
ρ
2
|
4
cos
ϕ
8
cos
ϕ
d
ϕ
=
1
2
∫
π
4
π
3
(
64
cos
2
ϕ
−
16
cos
2
ϕ
)
d
ϕ
=
24
∫
π
4
π
3
cos
2
ϕ
d
ϕ
=
{\displaystyle S=\int _{\pi \over 4}^{\pi \over 3}d\phi \int _{4\cos \phi }^{8\cos \phi }\rho d\rho ={1 \over 2}\int _{\pi \over 4}^{\pi \over 3}\rho ^{2}|_{4\cos \phi }^{8\cos \phi }d\phi ={1 \over 2}\int _{\pi \over 4}^{\pi \over 3}(64\cos ^{2}\phi -16\cos ^{2}\phi )d\phi =24\int _{\pi \over 4}^{\pi \over 3}\cos ^{2}\phi d\phi =}
=
12
∫
π
4
π
3
(
1
+
cos
(
2
ϕ
)
)
d
ϕ
=
12
(
ϕ
+
1
2
sin
(
2
ϕ
)
)
|
π
4
π
3
=
12
(
π
3
−
π
4
+
1
2
sin
2
π
3
−
1
2
sin
π
2
)
=
{\displaystyle =12\int _{\pi \over 4}^{\pi \over 3}(1+\cos(2\phi ))d\phi =12(\phi +{1 \over 2}\sin(2\phi ))|_{\pi \over 4}^{\pi \over 3}=12({\pi \over 3}-{\pi \over 4}+{1 \over 2}\sin {2\pi \over 3}-{1 \over 2}\sin {\pi \over 2})=}
=
12
(
π
12
+
1
2
⋅
3
2
−
1
2
⋅
1
)
=
π
+
3
3
−
6
≈
2
,
337745.
{\displaystyle =12({\pi \over 12}+{1 \over 2}\cdot {{\sqrt {3}} \over 2}-{1 \over 2}\cdot 1)=\pi +3{\sqrt {3}}-6\approx 2,337745.}
Figūrą riboja kreivės
x
2
+
y
2
=
4
,
{\displaystyle x^{2}+y^{2}=4,}
x
2
+
y
2
=
8
x
,
{\displaystyle x^{2}+y^{2}=8x,}
y
=
x
,
{\displaystyle y=x,}
y
=
x
3
.
{\displaystyle y=x{\sqrt {3}}.}
x
2
+
y
2
=
4
,
{\displaystyle x^{2}+y^{2}=4,}
x
2
+
y
2
=
8
x
.
{\displaystyle x^{2}+y^{2}=8x.}
x
2
+
y
2
=
4
{\displaystyle x^{2}+y^{2}=4}
(
x
−
0
)
2
+
y
2
=
4
,
{\displaystyle (x-0)^{2}+y^{2}=4,}
x
2
+
y
2
=
8
x
{\displaystyle x^{2}+y^{2}=8x}
(
x
−
4
)
2
+
y
2
=
16.
{\displaystyle (x-4)^{2}+y^{2}=16.}
Tai apskritimo lygtys. Pirmojo apskritimo centras yra taškas (0; 0), o spindulys lygus 2, antrojo centras - taškas (4; 0), o spindulys lygus 4. Parašykime tų apskritimų lygtis polinėje koordinačių sistemoje:
ρ
2
=
4
{\displaystyle \rho ^{2}=4}
ρ
=
2
{\displaystyle \rho =2}
ρ
=
8
cos
ϕ
.
{\displaystyle \rho =8\cos \phi .}
y
=
x
{\displaystyle y=x}
Ox sudaro kampą
π
4
,
{\displaystyle {\pi \over 4},}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
x
=
3
;
tan
ϕ
=
3
;
ϕ
=
arctan
3
=
π
3
.
{\displaystyle {y \over x}={\sqrt {3}};\;\tan \phi ={\sqrt {3}};\;\phi =\arctan {\sqrt {3}}={\pi \over 3}.}
D gauname, kai
ϕ
{\displaystyle \phi }
π
4
{\displaystyle {\pi \over 4}}
π
3
,
{\displaystyle {\pi \over 3},}
ρ
{\displaystyle \rho }
2
{\displaystyle 2}
8
cos
ϕ
.
{\displaystyle 8\cos \phi .}
S
=
∬
D
d
x
d
y
=
∬
D
ρ
d
ρ
d
ϕ
,
{\displaystyle S=\iint _{D}dxdy=\iint _{D}\rho d\rho d\phi ,}
S
=
∫
π
4
π
3
d
ϕ
∫
2
8
cos
ϕ
ρ
d
ρ
=
1
2
∫
π
4
π
3
ρ
2
|
2
8
cos
ϕ
d
ϕ
=
1
2
∫
π
4
π
3
(
64
cos
2
ϕ
−
2
2
)
d
ϕ
=
∫
π
4
π
3
(
32
cos
2
ϕ
−
2
)
d
ϕ
=
{\displaystyle S=\int _{\pi \over 4}^{\pi \over 3}d\phi \int _{2}^{8\cos \phi }\rho d\rho ={1 \over 2}\int _{\pi \over 4}^{\pi \over 3}\rho ^{2}|_{2}^{8\cos \phi }d\phi ={1 \over 2}\int _{\pi \over 4}^{\pi \over 3}(64\cos ^{2}\phi -2^{2})d\phi =\int _{\pi \over 4}^{\pi \over 3}(32\cos ^{2}\phi -2)d\phi =}
=
∫
π
4
π
3
(
16
(
1
+
cos
(
2
ϕ
)
)
−
2
)
d
ϕ
=
14
ϕ
+
16
2
sin
(
2
ϕ
)
|
π
4
π
3
=
14
π
3
−
14
π
4
+
8
sin
2
π
3
−
8
sin
π
2
=
{\displaystyle =\int _{\pi \over 4}^{\pi \over 3}(16(1+\cos(2\phi ))-2)d\phi =14\phi +{16 \over 2}\sin(2\phi )|_{\pi \over 4}^{\pi \over 3}={14\pi \over 3}-{14\pi \over 4}+8\sin {2\pi \over 3}-8\sin {\pi \over 2}=}
=
14
π
12
+
8
3
2
−
8
≈
2
,
593395.
{\displaystyle ={14\pi \over 12}+{8{\sqrt {3}} \over 2}-8\approx 2,593395.}
Figūrą riboja kreivės
x
2
+
y
2
=
8
x
,
{\displaystyle x^{2}+y^{2}=8x,}
y
=
x
,
{\displaystyle y=x,}
y
=
x
3
.
{\displaystyle y=x{\sqrt {3}}.}
x
2
+
y
2
=
8
x
.
{\displaystyle x^{2}+y^{2}=8x.}
x
2
+
y
2
=
8
x
{\displaystyle x^{2}+y^{2}=8x}
(
x
−
4
)
2
+
y
2
=
16.
{\displaystyle (x-4)^{2}+y^{2}=16.}
Tai apskritimo lygtis. Apskritimo centras yra taškas (4; 0), o spindulys lygus 4. Parašykime to apskritimo lygtį polinėje koordinačių sistemoje:
ρ
2
=
8
ρ
cos
ϕ
{\displaystyle \rho ^{2}=8\rho \cos \phi }
ρ
=
8
cos
ϕ
.
{\displaystyle \rho =8\cos \phi .}
y
=
x
{\displaystyle y=x}
Ox sudaro kampą
π
4
,
{\displaystyle {\pi \over 4},}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
x
=
3
;
tan
ϕ
=
3
;
ϕ
=
arctan
3
=
π
3
.
{\displaystyle {y \over x}={\sqrt {3}};\;\tan \phi ={\sqrt {3}};\;\phi =\arctan {\sqrt {3}}={\pi \over 3}.}
D gauname, kai
ϕ
{\displaystyle \phi }
π
4
{\displaystyle {\pi \over 4}}
π
3
,
{\displaystyle {\pi \over 3},}
ρ
{\displaystyle \rho }
0
{\displaystyle 0}
8
cos
ϕ
.
{\displaystyle 8\cos \phi .}
S
=
∬
D
d
x
d
y
=
∬
D
ρ
d
ρ
d
ϕ
,
{\displaystyle S=\iint _{D}dxdy=\iint _{D}\rho d\rho d\phi ,}
S
=
∫
π
4
π
3
d
ϕ
∫
0
8
cos
ϕ
ρ
d
ρ
=
1
2
∫
π
4
π
3
ρ
2
|
0
8
cos
ϕ
d
ϕ
=
1
2
∫
π
4
π
3
(
64
cos
2
ϕ
−
0
2
)
d
ϕ
=
∫
π
4
π
3
32
cos
2
ϕ
d
ϕ
=
{\displaystyle S=\int _{\pi \over 4}^{\pi \over 3}d\phi \int _{0}^{8\cos \phi }\rho d\rho ={1 \over 2}\int _{\pi \over 4}^{\pi \over 3}\rho ^{2}|_{0}^{8\cos \phi }d\phi ={1 \over 2}\int _{\pi \over 4}^{\pi \over 3}(64\cos ^{2}\phi -0^{2})d\phi =\int _{\pi \over 4}^{\pi \over 3}32\cos ^{2}\phi d\phi =}
=
32
∫
π
4
π
3
1
2
(
1
+
cos
(
2
ϕ
)
)
d
ϕ
=
16
ϕ
+
16
2
sin
(
2
ϕ
)
|
π
4
π
3
=
16
π
3
−
16
π
4
+
8
sin
2
π
3
−
8
sin
π
2
=
{\displaystyle =32\int _{\pi \over 4}^{\pi \over 3}{1 \over 2}(1+\cos(2\phi ))d\phi =16\phi +{16 \over 2}\sin(2\phi )|_{\pi \over 4}^{\pi \over 3}={16\pi \over 3}-{16\pi \over 4}+8\sin {2\pi \over 3}-8\sin {\pi \over 2}=}
=
4
π
3
+
8
3
2
−
8
⋅
1
=
3
,
116993435.
{\displaystyle ={4\pi \over 3}+{8{\sqrt {3}} \over 2}-8\cdot 1=3,116993435.}
Apskritimas su spinduliu
r
=
4
{\displaystyle r=4}
Dabar rasime šį plotą be integralų. Yra 2 trikampiai, kuriuos sudaro 3 tiesės:
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
=
x
{\displaystyle y=x}
y
=
4
3
−
x
3
{\displaystyle y=4{\sqrt {3}}-x{\sqrt {3}}}
y
=
x
{\displaystyle y=x}
Ox sudaro kampą 45 laipsniu arba
π
/
4
{\displaystyle \pi /4}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
x
=
3
{\displaystyle {y \over x}={\sqrt {3}}}
sin
α
cos
α
=
3
{\displaystyle {\sin \alpha \over \cos \alpha }={\sqrt {3}}}
tan
α
=
3
{\displaystyle \tan \alpha ={\sqrt {3}}}
α
=
arctan
3
{\displaystyle \alpha =\arctan {\sqrt {3}}}
α
=
π
3
{\displaystyle \alpha ={\pi \over 3}}
y
=
4
3
−
x
3
{\displaystyle y=4{\sqrt {3}}-x{\sqrt {3}}}
y
=
x
{\displaystyle y=x}
x
=
2
,
535898385
{\displaystyle x=2,535898385}
y
=
2
,
535898385
{\displaystyle y=2,535898385}
{
y
=
x
,
y
=
4
3
−
x
3
.
{\displaystyle {\begin{cases}y=x,&\\y=4{\sqrt {3}}-x{\sqrt {3}}.&\end{cases}}}
Keitimo butu gauname,
x
=
4
3
−
x
3
{\displaystyle x=4{\sqrt {3}}-x{\sqrt {3}}}
x
2
=
(
3
(
4
−
x
)
)
2
{\displaystyle x^{2}=({\sqrt {3}}(4-x))^{2}}
x
2
=
3
(
16
−
8
x
+
x
2
)
{\displaystyle x^{2}=3(16-8x+x^{2})}
x
2
−
3
x
2
+
24
x
−
48
=
0
{\displaystyle x^{2}-3x^{2}+24x-48=0}
−
2
x
2
+
24
x
−
48
=
0
{\displaystyle -2x^{2}+24x-48=0}
D
=
b
2
−
4
a
c
=
24
2
−
4
⋅
(
−
2
)
⋅
(
−
48
)
=
576
−
384
=
192
{\displaystyle D=b^{2}-4ac=24^{2}-4\cdot (-2)\cdot (-48)=576-384=192}
x
1
=
−
b
+
D
2
a
=
−
24
+
192
2
⋅
(
−
2
)
=
−
24
+
8
3
−
4
=
6
−
2
3
≈
2
,
535898385
{\displaystyle x_{1}={-b+{\sqrt {D}} \over 2a}={-24+{\sqrt {192}} \over 2\cdot (-2)}={-24+8{\sqrt {3}} \over -4}=6-2{\sqrt {3}}\approx 2,535898385}
x
2
=
−
24
−
192
−
4
=
6
+
2
3
≈
9
,
464101615
{\displaystyle x_{2}={-24-{\sqrt {192}} \over -4}=6+2{\sqrt {3}}\approx 9,464101615}
x
2
{\displaystyle x_{2}}
y
=
x
{\displaystyle y=x}
y , kai kertasi šios dvi tiesės; y=x, taigi y=2,535898385, tą patį gauname ir įstačius x į lygtį
y
=
3
(
4
−
x
)
=
3
(
4
−
(
6
−
2
3
)
)
=
3
(
4
−
6
+
2
3
)
=
−
2
3
+
2
⋅
3
=
2
,
535898385
{\displaystyle y={\sqrt {3}}(4-x)={\sqrt {3}}(4-(6-2{\sqrt {3}}))={\sqrt {3}}(4-6+2{\sqrt {3}})=-2{\sqrt {3}}+2\cdot 3=2,535898385}
Dabar galime rasti ilgį tiesės
y
=
x
{\displaystyle y=x}
y
=
3
(
4
−
x
)
{\displaystyle y={\sqrt {3}}(4-x)}
c
=
a
2
+
b
2
=
2
,
535898385
2
⋅
2
≈
3
,
586301889.
{\displaystyle c={\sqrt {a^{2}+b^{2}}}={\sqrt {2,535898385^{2}\cdot 2}}\approx 3,586301889.}
y
=
3
(
4
−
x
)
{\displaystyle y={\sqrt {3}}(4-x)}
y
=
x
{\displaystyle y=x}
c
=
a
2
+
b
2
=
(
4
−
2
,
535898385
)
2
+
2
,
535898385
2
=
1
,
464101615
2
+
2
,
535898385
2
=
2
,
92820323.
{\displaystyle c={\sqrt {a^{2}+b^{2}}}={\sqrt {(4-2,535898385)^{2}+2,535898385^{2}}}={\sqrt {1,464101615^{2}+2,535898385^{2}}}=2,92820323.}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
=
4
3
−
x
3
{\displaystyle y=4{\sqrt {3}}-x{\sqrt {3}}}
x
3
=
4
3
−
x
3
{\displaystyle x{\sqrt {3}}=4{\sqrt {3}}-x{\sqrt {3}}}
2
x
3
=
4
3
{\displaystyle 2x{\sqrt {3}}=4{\sqrt {3}}}
2
x
=
4
{\displaystyle 2x=4}
x
=
2
{\displaystyle x=2}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
=
2
3
{\displaystyle y=2{\sqrt {3}}}
x
=
2
{\displaystyle x=2}
y
=
2
3
.
{\displaystyle y=2{\sqrt {3}}.}
Pagal pitagoro teorema surandame vienu metu ir tiesės
y
=
3
(
4
−
x
)
{\displaystyle y={\sqrt {3}}(4-x)}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
c
=
x
2
+
y
2
=
2
2
+
(
2
3
)
2
=
4
+
4
⋅
3
=
16
=
4.
{\displaystyle c={\sqrt {x^{2}+y^{2}}}={\sqrt {2^{2}+(2{\sqrt {3}})^{2}}}={\sqrt {4+4\cdot 3}}={\sqrt {16}}=4.}
Dabar galime rasti iškirptą tiesės
y
=
3
(
4
−
x
)
{\displaystyle y={\sqrt {3}}(4-x)}
y
=
x
{\displaystyle y=x}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
e
=
4
−
2
,
92820323
=
1
,
07179677
{\displaystyle e=4-2,92820323=1,07179677}
Žinodami kampą tarp tiesės
y
=
x
{\displaystyle y=x}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
α
=
60
−
45
=
15
{\displaystyle \alpha =60-45=15}
α
=
π
3
−
π
4
=
π
12
{\displaystyle \alpha ={\pi \over 3}-{\pi \over 4}={\pi \over 12}}
S
=
1
2
a
b
sin
α
=
4
⋅
3
,
586301889
2
sin
π
12
=
7
,
172603778
⋅
0
,
258819045
=
1
,
856406461
{\displaystyle S={\frac {1}{2}}ab\sin \alpha ={4\cdot 3,586301889 \over 2}\sin {\pi \over 12}=7,172603778\cdot 0,258819045=1,856406461}
y
=
x
{\displaystyle y=x}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
=
4
3
−
x
3
{\displaystyle y=4{\sqrt {3}}-x{\sqrt {3}}}
p
=
(
a
+
b
+
c
)
/
2
=
(
4
+
3
,
586301889
+
1
,
07179677
)
/
2
=
4
,
32904933
{\displaystyle p=(a+b+c)/2=(4+3,586301889+1,07179677)/2=4,32904933}
S
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
=
{\displaystyle S={\sqrt {p(p-a)(p-b)(p-c)}}=}
=
4
,
32904933
(
4
,
32904933
−
4
)
(
4
,
32904933
−
3
,
586301889
)
(
4
,
32904933
−
1
,
07179677
)
=
{\displaystyle ={\sqrt {4,32904933(4,32904933-4)(4,32904933-3,586301889)(4,32904933-1,07179677)}}=}
=
4
,
32904933
⋅
0
,
329049329
⋅
0
,
74274744
⋅
3
,
257252559
=
3
,
446244942
≈
1
,
856406459.
{\displaystyle ={\sqrt {4,32904933\cdot 0,329049329\cdot 0,74274744\cdot 3,257252559}}={\sqrt {3,446244942}}\approx 1,856406459.}
Tiesė
y
=
x
{\displaystyle y=x}
x
2
+
y
2
=
8
x
{\displaystyle x^{2}+y^{2}=8x}
r
=
4
{\displaystyle r=4}
d
=
4
2
+
4
2
=
4
2
≈
5
,
656854249
{\displaystyle d={\sqrt {4^{2}+4^{2}}}=4{\sqrt {2}}\approx 5,656854249}
y
=
4
3
−
x
3
{\displaystyle y=4{\sqrt {3}}-x{\sqrt {3}}}
g
=
5
,
656854249
−
2
,
535898385
2
+
2
,
535898385
2
=
5
,
656854249
−
3
,
586301889
=
2
,
070552361.
{\displaystyle g=5,656854249-{\sqrt {2,535898385^{2}+2,535898385^{2}}}=5,656854249-3,586301889=2,070552361.}
y
=
x
{\displaystyle y=x}
y
=
4
3
−
x
3
{\displaystyle y=4{\sqrt {3}}-x{\sqrt {3}}}
p
=
(
a
+
b
+
c
)
/
2
=
(
4
+
2
,
070552361
+
2
,
92820323
)
/
2
=
4
,
499377796
{\displaystyle p=(a+b+c)/2=(4+2,070552361+2,92820323)/2=4,499377796}
S
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
=
{\displaystyle S={\sqrt {p(p-a)(p-b)(p-c)}}=}
=
4
,
499377796
(
4
,
499377796
−
4
)
(
4
,
499377796
−
2
,
070552361
)
(
4
,
499377796
−
2
,
92820323
)
=
{\displaystyle ={\sqrt {4,499377796(4,499377796-4)(4,499377796-2,070552361)(4,499377796-2,92820323)}}=}
=
4
,
499377796
⋅
0
,
499377795
⋅
2
,
428825435
⋅
1
,
571174566
≈
8
,
57437415
≈
2
,
928203229.
{\displaystyle ={\sqrt {4,499377796\cdot 0,499377795\cdot 2,428825435\cdot 1,571174566}}\approx {\sqrt {8,57437415}}\approx 2,928203229.}
x
=
4
{\displaystyle x=4}
y
=
3
(
4
−
x
)
{\displaystyle y={\sqrt {3}}(4-x)}
β
=
90
−
60
=
30
{\displaystyle \beta =90-60=30}
π
6
{\displaystyle {\pi \over 6}}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
Ox ašimi. Šios dalies plotą galime apskaičiuoti taip:
1
2
⋅
π
6
r
2
=
1
2
⋅
π
6
⋅
4
2
=
8
π
6
≈
4
,
188790205.
{\displaystyle {1 \over 2}\cdot {\pi \over 6}r^{2}={1 \over 2}\cdot {\pi \over 6}\cdot 4^{2}={8\pi \over 6}\approx 4,188790205.}
S
1
=
4
,
188790205
−
2
,
928203229
=
1
,
260586976
{\displaystyle S_{1}=4,188790205-2,928203229=1,260586976}
S
v
=
S
1
+
S
t
r
i
k
a
m
=
1
,
260586976
+
1
,
856406461
≈
3
,
116993437.
{\displaystyle S_{v}=S_{1}+S_{trikam}=1,260586976+1,856406461\approx 3,116993437.}
Figūrą riboja kreivės
x
2
+
y
2
=
8
x
,
{\displaystyle x^{2}+y^{2}=8x,}
0
=
x
{\displaystyle 0=x}
x reikšme imtum, y visada yra lygus 0),
y
=
0
{\displaystyle y=0}
x visada yra lygus nuliui). Apskaičiuokime tos figuros plotą (tai yra plotas pusė apskritimo, kurio spindulys lygus 4, o greitai skaičiuojant jo plotas yra
S
=
π
⋅
r
2
2
=
π
⋅
4
2
2
=
8
π
=
25.13274123
{\displaystyle S={\frac {\pi \cdot r^{2}}{2}}={\frac {\pi \cdot 4^{2}}{2}}=8\pi =25.13274123}
x
2
+
y
2
=
8
x
.
{\displaystyle x^{2}+y^{2}=8x.}
x
2
+
y
2
=
8
x
{\displaystyle x^{2}+y^{2}=8x}
(
x
−
4
)
2
+
y
2
=
16.
{\displaystyle (x-4)^{2}+y^{2}=16.}
Tai apskritimo lygtis. Apskritimo centras yra taškas (4; 0), o spindulys lygus 4. Parašykime to apskritimo lygtį polinėje koordinačių sistemoje:
ρ
2
=
8
ρ
cos
ϕ
{\displaystyle \rho ^{2}=8\rho \cos \phi }
ρ
=
8
cos
ϕ
.
{\displaystyle \rho =8\cos \phi .}
0
=
x
{\displaystyle 0=x}
Ox sudaro kampą
ϕ
=
0
{\displaystyle \phi =0}
y
=
0
{\displaystyle y=0}
ϕ
=
π
2
.
{\displaystyle \phi ={\pi \over 2}.}
D gauname, kai
ϕ
{\displaystyle \phi }
0
{\displaystyle 0}
π
2
,
{\displaystyle {\pi \over 2},}
ρ
{\displaystyle \rho }
0
{\displaystyle 0}
8
cos
ϕ
.
{\displaystyle 8\cos \phi .}
S
=
∬
D
d
x
d
y
=
∬
D
ρ
d
ρ
d
ϕ
,
{\displaystyle S=\iint _{D}dxdy=\iint _{D}\rho d\rho d\phi ,}
S
=
∫
0
π
2
d
ϕ
∫
0
8
cos
ϕ
ρ
d
ρ
=
1
2
∫
0
π
2
ρ
2
|
0
8
cos
ϕ
d
ϕ
=
1
2
∫
0
π
2
(
64
cos
2
ϕ
−
0
2
)
d
ϕ
=
∫
0
π
2
32
cos
2
ϕ
d
ϕ
=
{\displaystyle S=\int _{0}^{\pi \over 2}d\phi \int _{0}^{8\cos \phi }\rho d\rho ={1 \over 2}\int _{0}^{\pi \over 2}\rho ^{2}|_{0}^{8\cos \phi }d\phi ={1 \over 2}\int _{0}^{\pi \over 2}(64\cos ^{2}\phi -0^{2})d\phi =\int _{0}^{\pi \over 2}32\cos ^{2}\phi d\phi =}
=
32
∫
0
π
2
1
2
(
1
+
cos
(
2
ϕ
)
)
d
ϕ
=
16
ϕ
+
16
2
sin
(
2
ϕ
)
|
0
π
2
=
16
π
2
−
16
⋅
0
+
8
sin
2
π
2
−
8
sin
(
2
⋅
0
)
=
{\displaystyle =32\int _{0}^{\pi \over 2}{1 \over 2}(1+\cos(2\phi ))d\phi =16\phi +{16 \over 2}\sin(2\phi )|_{0}^{\pi \over 2}={16\pi \over 2}-16\cdot 0+8\sin {2\pi \over 2}-8\sin(2\cdot 0)=}
=
8
π
+
8
⋅
sin
π
−
8
sin
0
=
8
π
+
8
⋅
0
−
8
⋅
0
=
8
π
=
25.13274123.
{\displaystyle =8\pi +8\cdot \sin \pi -8\sin 0=8\pi +8\cdot 0-8\cdot 0=8\pi =25.13274123.}
Figūrą riboja kreivės
x
2
+
y
2
=
8
x
,
{\displaystyle x^{2}+y^{2}=8x,}
0
=
x
{\displaystyle 0=x}
x reikšme imtum, y visada yra lygus 0),
y
=
0
{\displaystyle y=0}
x visada yra lygus nuliui). Apskaičiuokime tos figuros plotą (tai yra plotas pusė apskritimo, kurio spindulys lygus 4, o greitai skaičiuojant jo plotas yra
S
=
π
⋅
r
2
2
=
π
⋅
4
2
2
=
8
π
=
25.13274123
{\displaystyle S={\frac {\pi \cdot r^{2}}{2}}={\frac {\pi \cdot 4^{2}}{2}}=8\pi =25.13274123}
x
2
+
y
2
=
8
x
.
{\displaystyle x^{2}+y^{2}=8x.}
x
2
+
y
2
=
8
x
{\displaystyle x^{2}+y^{2}=8x}
(
x
−
4
)
2
+
y
2
=
16.
{\displaystyle (x-4)^{2}+y^{2}=16.}
Tai apskritimo lygtis. Apskritimo centras yra taškas (4; 0), o spindulys lygus 4. Parašykime to apskritimo lygtį polinėje koordinačių sistemoje:
ρ
2
=
8
ρ
cos
ϕ
{\displaystyle \rho ^{2}=8\rho \cos \phi }
ρ
=
8
cos
ϕ
.
{\displaystyle \rho =8\cos \phi .}
0
=
x
{\displaystyle 0=x}
Ox sudaro kampą
ϕ
=
0
{\displaystyle \phi =0}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
x
=
3
;
tan
ϕ
=
3
;
ϕ
=
arctan
3
=
π
3
{\displaystyle {y \over x}={\sqrt {3}};\;\tan \phi ={\sqrt {3}};\;\phi =\arctan {\sqrt {3}}={\pi \over 3}}
D gauname, kai
ϕ
{\displaystyle \phi }
0
{\displaystyle 0}
π
3
,
{\displaystyle {\pi \over 3},}
ρ
{\displaystyle \rho }
0
{\displaystyle 0}
8
cos
ϕ
.
{\displaystyle 8\cos \phi .}
S
=
∬
D
d
x
d
y
=
∬
D
ρ
d
ρ
d
ϕ
,
{\displaystyle S=\iint _{D}dxdy=\iint _{D}\rho d\rho d\phi ,}
S
=
∫
0
π
3
d
ϕ
∫
0
8
cos
ϕ
ρ
d
ρ
=
1
2
∫
0
π
3
ρ
2
|
0
8
cos
ϕ
d
ϕ
=
1
2
∫
0
π
3
(
64
cos
2
ϕ
−
0
2
)
d
ϕ
=
∫
0
π
3
32
cos
2
ϕ
d
ϕ
=
{\displaystyle S=\int _{0}^{\pi \over 3}d\phi \int _{0}^{8\cos \phi }\rho d\rho ={1 \over 2}\int _{0}^{\pi \over 3}\rho ^{2}|_{0}^{8\cos \phi }d\phi ={1 \over 2}\int _{0}^{\pi \over 3}(64\cos ^{2}\phi -0^{2})d\phi =\int _{0}^{\pi \over 3}32\cos ^{2}\phi d\phi =}
=
32
∫
0
π
3
1
2
(
1
+
cos
(
2
ϕ
)
)
d
ϕ
=
16
ϕ
+
16
2
sin
(
2
ϕ
)
|
0
π
3
=
16
π
3
−
16
⋅
0
+
8
sin
2
π
3
−
8
sin
(
2
⋅
0
)
=
{\displaystyle =32\int _{0}^{\pi \over 3}{1 \over 2}(1+\cos(2\phi ))d\phi =16\phi +{16 \over 2}\sin(2\phi )|_{0}^{\pi \over 3}={16\pi \over 3}-16\cdot 0+8\sin {2\pi \over 3}-8\sin(2\cdot 0)=}
=
16
π
3
+
8
⋅
sin
(
2
,
094395102
)
−
8
sin
0
=
16
π
3
+
8
⋅
3
2
−
8
⋅
0
=
16
,
75516082
+
6.92820323
=
23.68336405.
{\displaystyle ={\frac {16\pi }{3}}+8\cdot \sin(2,094395102)-8\sin 0={\frac {16\pi }{3}}+8\cdot {\frac {\sqrt {3}}{2}}-8\cdot 0=16,75516082+6.92820323=23.68336405.}
Sprendžiant tokio tipo uždavinius (kitokio tipo ir negali būti, turiu galvoje, ne su apskritimais, o su parabolėmis ir kad būtų polinėse koordinatėse), formulę galima supaprastint iki tokio lygio:
S
=
r
2
⋅
ϕ
+
r
2
2
⋅
sin
(
2
ϕ
)
,
{\displaystyle S=r^{2}\cdot \phi +{\frac {r^{2}}{2}}\cdot \sin(2\phi ),}
kur apskritimo kairys šonas visada liečiasi su koordinačių sistemos O tašku ir ašis Ox visada apskrimą (arba skritulį, jei manyti, kad ten yra ieškomas plotas) dalina į dvi dalis. Visada! Kitaip neįmanoma užrašyti apskritimo lygties, kuri tiktų polinėms koordinatėms, dėl to polinių koordinačių pritaikymas dvilypiams integralams yra labai ribotas. Taigi, apskritimas visada būna koordinačių sistemos pirmame ketvirtyje, kai kampas
ϕ
{\displaystyle \phi }
ϕ
{\displaystyle \phi }
ϕ
{\displaystyle \phi }
Figura riboja kreivės
x
2
+
y
2
=
8
x
,
{\displaystyle x^{2}+y^{2}=8x,}
y
=
x
3
{\displaystyle y={\frac {x}{\sqrt {3}}}}
Ox ašis. Tiesė
y
=
x
3
{\displaystyle y={\frac {x}{\sqrt {3}}}}
Ox sudaro 30 laipsnių kampą,
ϕ
=
π
6
,
{\displaystyle \phi ={\frac {\pi }{6}},}
tan
ϕ
=
y
x
=
1
3
;
ϕ
=
arctan
1
3
=
π
6
=
0.523598775.
{\displaystyle \tan \phi ={\frac {y}{x}}={\frac {1}{\sqrt {3}}};\;\phi =\arctan {\frac {1}{\sqrt {3}}}={\frac {\pi }{6}}=0.523598775.}
Kreivė
y
=
8
x
−
x
2
{\displaystyle y={\sqrt {8x-x^{2}}}}
Ox . Apskritimo centro koordinatės yra (x; y)=(4; 0). Pereidami į polines koordinates apskritimo lygtį perašome
ρ
2
=
8
ρ
cos
ϕ
;
ρ
=
8
cos
ϕ
.
{\displaystyle \rho ^{2}=8\rho \cos \phi ;\;\rho =8\cos \phi .}
ρ
{\displaystyle \rho }
8
cos
ϕ
,
{\displaystyle 8\cos \phi ,}
ϕ
{\displaystyle \phi }
π
6
,
{\displaystyle {\frac {\pi }{6}},}
S
=
∫
0
π
6
d
ϕ
∫
0
8
cos
ϕ
ρ
d
ρ
=
1
2
∫
0
π
6
ρ
2
|
0
8
cos
ϕ
d
ϕ
=
1
2
∫
0
π
6
(
64
cos
2
ϕ
−
0
2
)
d
ϕ
=
∫
0
π
6
32
cos
2
ϕ
d
ϕ
=
{\displaystyle S=\int _{0}^{\pi \over 6}d\phi \int _{0}^{8\cos \phi }\rho d\rho ={1 \over 2}\int _{0}^{\pi \over 6}\rho ^{2}|_{0}^{8\cos \phi }d\phi ={1 \over 2}\int _{0}^{\pi \over 6}(64\cos ^{2}\phi -0^{2})d\phi =\int _{0}^{\pi \over 6}32\cos ^{2}\phi d\phi =}
=
32
∫
0
π
6
1
2
(
1
+
cos
(
2
ϕ
)
)
d
ϕ
=
16
ϕ
+
16
2
sin
(
2
ϕ
)
|
0
π
6
=
16
π
6
−
16
⋅
0
+
8
sin
2
π
6
−
8
sin
(
2
⋅
0
)
=
{\displaystyle =32\int _{0}^{\pi \over 6}{1 \over 2}(1+\cos(2\phi ))d\phi =16\phi +{16 \over 2}\sin(2\phi )|_{0}^{\pi \over 6}={16\pi \over 6}-16\cdot 0+8\sin {2\pi \over 6}-8\sin(2\cdot 0)=}
=
8
π
3
+
8
⋅
sin
π
3
−
8
sin
0
=
8
π
3
+
8
⋅
3
2
−
8
⋅
0
=
8.37758041
+
6.92820323
=
15.30578364.
{\displaystyle ={\frac {8\pi }{3}}+8\cdot \sin {\pi \over 3}-8\sin 0={\frac {8\pi }{3}}+8\cdot {\frac {\sqrt {3}}{2}}-8\cdot 0=8.37758041+6.92820323=15.30578364.}
Patikrinimui, rasime šį plotą elementariosios matematikos metodais. Visu pirma reikia surasti kuriame taške kertasi tiesė
y
=
x
3
{\displaystyle y={\frac {x}{\sqrt {3}}}}
x
2
+
y
2
=
8
x
{\displaystyle x^{2}+y^{2}=8x}
{
y
=
x
3
;
x
2
+
y
2
=
8
x
.
<=>
(
x
−
4
)
2
+
y
2
=
16.
{\displaystyle {\begin{cases}y={\frac {x}{\sqrt {3}}};&\\x^{2}+y^{2}=8x.\;<=>\;(x-4)^{2}+y^{2}=16.&\end{cases}}}
Tada:
{
y
3
=
x
;
(
x
−
4
)
2
+
y
2
=
16.
{\displaystyle {\begin{cases}y{\sqrt {3}}=x;&\\(x-4)^{2}+y^{2}=16.&\end{cases}}}
Toliau:
(
y
3
−
4
)
2
+
y
2
=
16
;
{\displaystyle (y{\sqrt {3}}-4)^{2}+y^{2}=16;}
3
y
2
+
8
y
3
−
16
+
y
2
=
16
;
{\displaystyle 3y^{2}+8y{\sqrt {3}}-16+y^{2}=16;}
4
y
2
+
8
y
3
=
32
;
{\displaystyle 4y^{2}+8y{\sqrt {3}}=32;}
y
2
+
2
y
3
=
8
;
{\displaystyle y^{2}+2y{\sqrt {3}}=8;}
y
2
+
2
y
3
−
8
=
0
;
{\displaystyle y^{2}+2y{\sqrt {3}}-8=0;}
D
=
b
2
−
4
a
c
=
(
2
3
)
2
−
4
⋅
1
⋅
(
−
8
)
=
12
+
32
=
44
;
{\displaystyle D=b^{2}-4ac=(2{\sqrt {3}})^{2}-4\cdot 1\cdot (-8)=12+32=44;}
y
1
=
−
b
+
D
2
a
=
−
2
3
+
44
2
=
−
3.464101615
+
6.633249581
2
=
1.584573983
;
{\displaystyle y_{1}={\frac {-b+{\sqrt {D}}}{2a}}={\frac {-2{\sqrt {3}}+{\sqrt {44}}}{2}}={\frac {-3.464101615+6.633249581}{2}}=1.584573983;}
y
2
=
−
b
−
D
2
a
=
−
2
3
−
44
2
=
−
3.464101615
−
6.633249581
2
=
−
5.048675598.
{\displaystyle y_{2}={\frac {-b-{\sqrt {D}}}{2a}}={\frac {-2{\sqrt {3}}-{\sqrt {44}}}{2}}={\frac {-3.464101615-6.633249581}{2}}=-5.048675598.}
Reikšmė
y
2
{\displaystyle y_{2}}
Įstatę
y
1
{\displaystyle y_{1}}
x
1
=
y
1
3
=
1.584573983
⋅
3
=
2.744562647.
{\displaystyle x_{1}=y_{1}{\sqrt {3}}=1.584573983\cdot {\sqrt {3}}=2.744562647.}
Taigi, turime tiesės
y
=
x
3
{\displaystyle y={\frac {x}{\sqrt {3}}}}
A
=
(
x
1
;
y
1
)
=
(
2.744562647
;
1.584573983
)
{\displaystyle A=(x_{1};y_{1})=(2.744562647;1.584573983)}
Toliau iš taško
A
=
(
x
1
;
y
1
)
=
(
2.744562647
;
1.584573983
)
{\displaystyle A=(x_{1};y_{1})=(2.744562647;1.584573983)}
Ox ašiai, kuri susikerta taške B=(2.744562647; 0). Tiesės AB ilgis yra
a
=
y
1
=
1.584573983.
{\displaystyle a=y_{1}=1.584573983.}
OB ilgis yra
b
=
2.744562647
{\displaystyle b=2.744562647}
OA ilgį:
c
=
a
2
+
b
2
=
1.584573983
2
+
2.744562647
2
=
10.04349883
=
3.169147966.
{\displaystyle c={\sqrt {a^{2}+b^{2}}}={\sqrt {1.584573983^{2}+2.744562647^{2}}}={\sqrt {10.04349883}}=3.169147966.}
Žinodami visų trikampio OAB kraštinių ilgius, randame pagal Herono formulę trikampio OAB plotą:
S
O
A
B
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
=
3.749142298
(
3.749142298
−
1.584573983
)
(
3.749142298
−
2.744562647
)
(
3.749142298
−
3.169147966
)
=
{\displaystyle S_{OAB}={\sqrt {p(p-a)(p-b)(p-c)}}={\sqrt {3.749142298(3.749142298-1.584573983)(3.749142298-2.744562647)(3.749142298-3.169147966)}}=}
=
3.749142298
⋅
2.164568315
⋅
1.004579651
⋅
0.579994332
=
4.728368849
=
2.174481283.
{\displaystyle ={\sqrt {3.749142298\cdot 2.164568315\cdot 1.004579651\cdot 0.579994332}}={\sqrt {4.728368849}}=2.174481283.}
kur
p
=
a
+
b
+
c
2
=
1.584573983
+
2.744562647
+
3.169147966
2
=
3.749142298.
{\displaystyle p={\frac {a+b+c}{2}}={\frac {1.584573983+2.744562647+3.169147966}{2}}=3.749142298.}
S
O
A
B
=
a
⋅
b
2
=
1.584573983
⋅
2.744562647
2
=
2.174481283.
{\displaystyle S_{OAB}={\frac {a\cdot b}{2}}={\frac {1.584573983\cdot 2.744562647}{2}}=2.174481283.}
Toliau sujungiame tašką A su apskritimo centro tašku C=(4; 0). Gauname tiesės atkarpą AC , kurią pavadiname f . Iš atkarpos OB =b atimame atimame tiesę OC =d, kurios ilgis yra OC=d=4 ir gauname
e
=
b
−
d
=
2.744562647
−
4
=
−
1.255437353.
{\displaystyle e=b-d=2.744562647-4=-1.255437353.}
A nuleista tiesė statmena ašiai Ox ir susikirtusi su ašim Ox taške B turėjo sudaryti tiesę OB , kurios ilgis būtų daugiau nei 4. Vadinasi lygčių sistema išspręsta neteisingai. Ši anomalija gali padėti nekartoti klaidų, kai atrodo viskas padaryta teisingai. Todėl nebus trinamas blogas lygčių sistemos išsprendimas ir diskriminanto radimas. Taigi, iš naujo sprendžiame lygčių sistemą:
{
y
=
x
3
;
x
2
+
y
2
=
8
x
.
<=>
(
x
−
4
)
2
+
y
2
=
16.
{\displaystyle {\begin{cases}y={\frac {x}{\sqrt {3}}};&\\x^{2}+y^{2}=8x.\;<=>\;(x-4)^{2}+y^{2}=16.&\end{cases}}}
x
2
+
(
x
3
)
2
=
8
x
;
{\displaystyle x^{2}+({\frac {x}{\sqrt {3}}})^{2}=8x;}
x
2
+
3
x
2
=
8
x
;
{\displaystyle x^{2}+3x^{2}=8x;}
4
x
2
−
8
x
=
0
;
{\displaystyle 4x^{2}-8x=0;}
D
=
b
2
−
4
a
c
=
(
−
8
)
2
−
4
⋅
1
⋅
0
=
64.
{\displaystyle D=b^{2}-4ac=(-8)^{2}-4\cdot 1\cdot 0=64.}
x
1
=
−
b
+
D
2
⋅
4
=
−
(
−
8
)
+
64
2
⋅
4
=
8
+
8
8
=
2
;
{\displaystyle x_{1}={\frac {-b+{\sqrt {D}}}{2\cdot 4}}={\frac {-(-8)+{\sqrt {64}}}{2\cdot 4}}={\frac {8+8}{8}}=2;}
x
2
=
−
b
+
D
2
⋅
4
=
−
(
−
8
)
−
64
2
⋅
4
=
8
−
8
8
=
0.
{\displaystyle x_{2}={\frac {-b+{\sqrt {D}}}{2\cdot 4}}={\frac {-(-8)-{\sqrt {64}}}{2\cdot 4}}={\frac {8-8}{8}}=0.}
y
=
x
3
=
2
3
=
3.464101615.
{\displaystyle y={\frac {x}{\sqrt {3}}}={\frac {2}{\sqrt {3}}}=3.464101615.}
2
2
+
y
2
=
8
⋅
2
,
{\displaystyle 2^{2}+y^{2}=8\cdot 2,}
y
2
=
16
−
4
=
12
,
{\displaystyle y^{2}=16-4=12,}
y
=
12
=
2
3
=
3.464101615.
{\displaystyle y={\sqrt {12}}=2{\sqrt {3}}=3.464101615.}
(
x
−
4
)
2
+
y
2
=
16
,
{\displaystyle (x-4)^{2}+y^{2}=16,}
(
2
−
4
)
2
+
y
2
=
16
,
{\displaystyle (2-4)^{2}+y^{2}=16,}
4
+
y
2
=
16
,
{\displaystyle 4+y^{2}=16,}
y
=
16
−
4
=
12
=
2
3
=
3.464101615.
{\displaystyle y={\sqrt {16-4}}={\sqrt {12}}=2{\sqrt {3}}=3.464101615.}
Matyt, negalima patikrinti integravimo budu gauto ploto, nes nepavyksta rasti susikritimo taško tiesės
y
=
x
3
{\displaystyle y={\frac {x}{\sqrt {3}}}}
x
2
+
y
2
=
8
x
.
{\displaystyle x^{2}+y^{2}=8x.}
y surastas teisingai, o prie x , dėl kažkokių priežasčių, dar reikia pridėti spindulį r=4. Todėl galima tęsti ieškoti ploto, kurį riboja apskritimas
x
2
+
y
2
=
8
x
{\displaystyle x^{2}+y^{2}=8x}
y
=
x
3
{\displaystyle y={\frac {x}{\sqrt {3}}}}
Ox . Iš naujo nuleidžiamę statmenį iš taško
A
=
(
6
;
2
3
)
{\displaystyle A=(6;\;2{\sqrt {3}})}
A
B
=
a
=
2
3
=
3.464101615.
{\displaystyle AB=a=2{\sqrt {3}}=3.464101615.}
O
B
=
b
=
6
{\displaystyle OB=b=6}
OAB plotas yra
S
Δ
O
A
B
=
a
⋅
b
2
=
1
2
⋅
2
3
⋅
6
=
6
3
=
10.39230485.
{\displaystyle S_{\Delta OAB}={\frac {a\cdot b}{2}}={\frac {1}{2}}\cdot 2{\sqrt {3}}\cdot 6=6{\sqrt {3}}=10.39230485.}
CB ilgis yra d=b-r=6-4=2. Trikampio CBA plotas yra
S
Δ
C
B
A
=
a
⋅
d
2
=
1
2
⋅
2
3
⋅
2
=
2
3
=
3.464101615.
{\displaystyle S_{\Delta CBA}={\frac {a\cdot d}{2}}={\frac {1}{2}}\cdot 2{\sqrt {3}}\cdot 2=2{\sqrt {3}}=3.464101615.}
Sukūriame naują tašką D=(8; 0). Atkarpos CD ilgis yra 4. Kampas ACD turi būti surastas. Žinome, kad tiesė CA yra spindulys r=4. Todėl
4
cos
(
α
)
=
d
=
2
;
{\displaystyle 4\cos(\alpha )=d=2;}
cos
α
=
1
2
;
α
=
arccos
1
2
=
1.047197551
{\displaystyle \cos \alpha ={\frac {1}{2}};\;\alpha =\arccos {\frac {1}{2}}=1.047197551}
α
=
60
{\displaystyle \alpha =60}
π
r
2
,
{\displaystyle \pi r^{2},}
ACD plotas yra
S
A
C
D
=
60
360
⋅
π
⋅
r
2
=
1
6
⋅
π
⋅
4
2
=
8
π
3
=
8.37758041.
{\displaystyle S_{ACD}={\frac {60}{360}}\cdot \pi \cdot r^{2}={\frac {1}{6}}\cdot \pi \cdot 4^{2}={\frac {8\pi }{3}}=8.37758041.}
Visas ieškomas plotas yra lygus:
S
=
S
Δ
O
A
B
−
S
Δ
C
B
A
+
S
A
C
D
=
6
3
−
2
3
+
8
π
3
=
4
3
+
8
π
3
=
6.92820323
+
8.37758041
=
15.30578364.
{\displaystyle S=S_{\Delta OAB}-S_{\Delta CBA}+S_{ACD}=6{\sqrt {3}}-2{\sqrt {3}}+{\frac {8\pi }{3}}=4{\sqrt {3}}+{\frac {8\pi }{3}}=6.92820323+8.37758041=15.30578364.}
Figūrą riboja kreivės
x
2
+
y
2
=
4
x
,
{\displaystyle x^{2}+y^{2}=4x,}
y
=
x
,
{\displaystyle y=x,}
y
=
x
3
.
{\displaystyle y=x{\sqrt {3}}.}
x
2
+
y
2
=
4
x
.
{\displaystyle x^{2}+y^{2}=4x.}
x
2
+
y
2
=
4
x
{\displaystyle x^{2}+y^{2}=4x}
(
x
−
2
)
2
+
y
2
=
4.
{\displaystyle (x-2)^{2}+y^{2}=4.}
Tai apskritimo lygtis. Apskritimo centras yra taškas (2; 0), o spindulys lygus 2. Parašykime to apskritimo lygtį polinėje koordinačių sistemoje:
ρ
2
=
4
ρ
cos
ϕ
{\displaystyle \rho ^{2}=4\rho \cos \phi }
ρ
=
4
cos
ϕ
.
{\displaystyle \rho =4\cos \phi .}
y
=
x
{\displaystyle y=x}
Ox sudaro kampą
π
4
,
{\displaystyle {\pi \over 4},}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
y
x
=
3
;
tan
ϕ
=
3
;
ϕ
=
arctan
3
=
π
3
.
{\displaystyle {y \over x}={\sqrt {3}};\;\tan \phi ={\sqrt {3}};\;\phi =\arctan {\sqrt {3}}={\pi \over 3}.}
D gauname, kai
ϕ
{\displaystyle \phi }
π
4
{\displaystyle {\pi \over 4}}
π
3
,
{\displaystyle {\pi \over 3},}
ρ
{\displaystyle \rho }
0
{\displaystyle 0}
8
cos
ϕ
.
{\displaystyle 8\cos \phi .}
S
=
∬
D
d
x
d
y
=
∬
D
ρ
d
ρ
d
ϕ
,
{\displaystyle S=\iint _{D}dxdy=\iint _{D}\rho d\rho d\phi ,}
S
=
∫
π
4
π
3
d
ϕ
∫
0
4
cos
ϕ
ρ
d
ρ
=
1
2
∫
π
4
π
3
ρ
2
|
0
4
cos
ϕ
d
ϕ
=
1
2
∫
π
4
π
3
(
16
cos
2
ϕ
−
0
2
)
d
ϕ
=
∫
π
4
π
3
8
cos
2
ϕ
d
ϕ
=
{\displaystyle S=\int _{\pi \over 4}^{\pi \over 3}d\phi \int _{0}^{4\cos \phi }\rho d\rho ={1 \over 2}\int _{\pi \over 4}^{\pi \over 3}\rho ^{2}|_{0}^{4\cos \phi }d\phi ={1 \over 2}\int _{\pi \over 4}^{\pi \over 3}(16\cos ^{2}\phi -0^{2})d\phi =\int _{\pi \over 4}^{\pi \over 3}8\cos ^{2}\phi d\phi =}
=
8
∫
π
4
π
3
1
2
(
1
+
cos
(
2
ϕ
)
)
d
ϕ
=
4
ϕ
+
4
2
sin
(
2
ϕ
)
|
π
4
π
3
=
4
π
3
−
4
π
4
+
2
sin
2
π
3
−
2
sin
π
2
=
{\displaystyle =8\int _{\pi \over 4}^{\pi \over 3}{1 \over 2}(1+\cos(2\phi ))d\phi =4\phi +{4 \over 2}\sin(2\phi )|_{\pi \over 4}^{\pi \over 3}={4\pi \over 3}-{4\pi \over 4}+2\sin {2\pi \over 3}-2\sin {\pi \over 2}=}
=
π
3
+
2
3
2
−
2
=
0
,
779248358.
{\displaystyle ={\pi \over 3}+{2{\sqrt {3}} \over 2}-2=0,779248358.}
Be kita ko,
0
,
779248358
⋅
4
=
3
,
116993432.
{\displaystyle 0,779248358\cdot 4=3,116993432.}
0
,
779248358
⋅
3
=
2
,
337745074.
{\displaystyle 0,779248358\cdot 3=2,337745074.}
(
R
/
r
)
2
{\displaystyle (R/r)^{2}}
3
2
=
9
{\displaystyle 3^{2}=9}
Figūrą riboja kreivės
x
2
+
y
2
=
4
,
{\displaystyle x^{2}+y^{2}=4,}
y
=
x
,
{\displaystyle y=x,}
y
=
x
3
.
{\displaystyle y=x{\sqrt {3}}.}
x
2
+
y
2
=
4
{\displaystyle x^{2}+y^{2}=4}
ρ
2
=
4
{\displaystyle \rho ^{2}=4}
ρ
=
r
=
2.
{\displaystyle \rho =r=2.}
y
=
x
{\displaystyle y=x}
Ox sudaro kampą
π
4
,
{\displaystyle {\pi \over 4},}
y
=
x
3
{\displaystyle y=x{\sqrt {3}}}
π
3
.
{\displaystyle {\pi \over 3}.}
Taigi sritį D gauname, kai
ϕ
{\displaystyle \phi }
π
4
{\displaystyle {\pi \over 4}}
π
3
.
{\displaystyle {\pi \over 3}.}
S
=
π
r
2
,
{\displaystyle S=\pi r^{2},}
2
π
{\displaystyle 2\pi }
S
=
1
2
(
π
3
−
π
4
)
⋅
r
2
=
1
2
⋅
π
12
⋅
2
2
=
π
6
=
0
,
523598775.
{\displaystyle S={1 \over 2}({\pi \over 3}-{\pi \over 4})\cdot r^{2}={1 \over 2}\cdot {\pi \over 12}\cdot 2^{2}={\pi \over 6}=0,523598775.}
3
,
116993432
−
0
,
523598775
=
2
,
593394656.
{\displaystyle 3,116993432-0,523598775=2,593394656.}
2. Apskaičiuosime tūrį kūno, apriboto paviršiais
x
2
+
y
2
=
2
y
,
{\displaystyle x^{2}+y^{2}=2y,}
z
=
0
,
{\displaystyle z=0,}
x
+
y
+
z
=
4.
{\displaystyle x+y+z=4.}
z
=
4
−
x
−
y
{\displaystyle z=4-x-y}
V
=
∬
D
(
4
−
x
−
y
)
d
x
d
y
,
{\displaystyle V=\iint _{D}(4-x-y)dxdy,}
D - kūno pagrindas - apskritimas
x
2
+
(
y
−
1
)
2
≤
1
{\displaystyle x^{2}+(y-1)^{2}\leq 1}
XOY . Perėję į poliarines koordinates gauname:
V
=
∬
P
(
4
−
ρ
cos
ϕ
−
ρ
sin
ϕ
)
ρ
d
ρ
d
ϕ
=
{\displaystyle V=\iint _{P}(4-\rho \cos \phi -\rho \sin \phi )\rho d\rho d\phi =}
=
∫
0
π
d
ϕ
∫
0
2
sin
ϕ
(
4
−
ρ
cos
ϕ
−
ρ
sin
ϕ
)
ρ
d
ρ
=
3
π
.
{\displaystyle =\int _{0}^{\pi }d\phi \int _{0}^{2\sin \phi }(4-\rho \cos \phi -\rho \sin \phi )\rho d\rho =3\pi .}
Šio kūno tūrį galima rasti gerokai greičiau. Cilindro pjuvis tesiasi nuo
z
=
4
{\displaystyle z=4}
z
=
2
{\displaystyle z=2}
z
=
4
−
x
−
y
{\displaystyle z=4-x-y}
z yra žemiausiame taške 2, kai
x
=
0
{\displaystyle x=0}
y
=
2
{\displaystyle y=2}
z
=
4
−
0
−
2
=
2
{\displaystyle z=4-0-2=2}
r
=
2
{\displaystyle r=2}
h
1
=
2
{\displaystyle h_{1}=2}
V
1
=
h
1
π
r
2
=
2
π
⋅
1
2
=
2
π
.
{\displaystyle V_{1}=h_{1}\pi r^{2}=2\pi \cdot 1^{2}=2\pi .}
Aukščiau esančią cilindro dalį, kurios aukštis yra
h
2
=
2
{\displaystyle h_{2}=2}
zOy ir matome, kad gaunasi 2 vienodi trikampiai (vienas iš jų nepriklauso tai daliai). Taigi tiesiog aukštesnės dalies tūrį gauname padalinę cilindrą pusiau:
V
2
=
h
2
π
r
2
2
=
2
π
⋅
1
2
2
=
π
.
{\displaystyle V_{2}={h_{2}\pi r^{2} \over 2}={2\pi \cdot 1^{2} \over 2}=\pi .}
V
=
V
1
+
V
2
=
2
π
+
π
=
3
π
{\displaystyle V=V_{1}+V_{2}=2\pi +\pi =3\pi }
3. Apskaičiuosime paviršiaus dalį paraboloido
x
2
+
y
2
=
z
,
{\displaystyle x^{2}+y^{2}=z,}
x
2
+
y
2
=
1.
{\displaystyle x^{2}+y^{2}=1.}
1
+
f
x
′
2
(
x
;
y
)
+
f
y
′
2
(
x
;
y
)
.
{\displaystyle {\sqrt {1+f_{x}'^{2}(x;y)+f_{y}'^{2}(x;y)}}.}
∂
z
∂
x
=
2
x
,
∂
z
∂
y
=
2
y
,
{\displaystyle {\partial z \over \partial x}=2x,\;{\partial z \over \partial y}=2y,}
S
=
∬
D
1
+
4
x
2
+
4
y
2
d
x
d
y
,
{\displaystyle S=\iint _{D}{\sqrt {1+4x^{2}+4y^{2}}}\;dx\;dy,}
kur D - apskritimas
x
2
+
y
2
≤
1
{\displaystyle x^{2}+y^{2}\leq 1}
XOY . Pereidami į poliarines koordinates gauname:
S
=
∬
P
1
+
4
ρ
2
ρ
d
ρ
d
ϕ
=
∫
0
2
π
d
ϕ
∫
0
1
ρ
1
+
4
ρ
2
d
ρ
=
{\displaystyle S=\iint _{P}{\sqrt {1+4\rho ^{2}}}\rho d\rho d\phi =\int _{0}^{2\pi }d\phi \int _{0}^{1}\rho {\sqrt {1+4\rho ^{2}}}d\rho =}
=
∫
0
2
π
d
ϕ
∫
0
1
1
+
4
ρ
2
d
(
1
+
4
ρ
2
)
8
=
1
8
∫
0
2
π
2
3
(
1
+
4
ρ
2
)
3
2
|
0
1
d
ϕ
=
1
12
∫
0
2
π
(
5
3
2
−
1
)
d
ϕ
=
{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{1}{\sqrt {1+4\rho ^{2}}}{d(1+4\rho ^{2}) \over 8}={1 \over 8}\int _{0}^{2\pi }{2 \over 3}(1+4\rho ^{2})^{3 \over 2}|_{0}^{1}d\phi ={1 \over 12}\int _{0}^{2\pi }(5^{3 \over 2}-1)d\phi =}
=
1
12
(
5
5
−
1
)
ϕ
|
0
2
π
=
π
6
(
5
5
−
1
)
=
5.3304135
,
{\displaystyle ={1 \over 12}(5{\sqrt {5}}-1)\phi |_{0}^{2\pi }={\pi \over 6}(5{\sqrt {5}}-1)=5.3304135,}
d
(
1
+
4
ρ
2
)
=
8
ρ
d
ρ
;
d
ρ
=
d
(
1
+
4
ρ
2
)
8
ρ
.
{\displaystyle d(1+4\rho ^{2})=8\rho d\rho ;\;d\rho ={d(1+4\rho ^{2}) \over 8\rho }.}
Palyginimui paviršiaus plotas cilindro be dviejų pagrindų, kurio spindulys r=1 ir aukštis h=1, yra lygus:
S
1
=
c
⋅
h
=
2
π
r
⋅
h
=
2
⋅
π
⋅
1
⋅
1
=
2
π
=
6.283185307.
{\displaystyle S_{1}=c\cdot h=2\pi r\cdot h=2\cdot \pi \cdot 1\cdot 1=2\pi =6.283185307.}
Vaizdas:1314abpav.jpg 13.14. Ritinys
x
2
+
y
2
=
a
x
{\displaystyle x^{2}+y^{2}=ax}
x
2
+
y
2
+
z
2
≤
a
2
{\displaystyle x^{2}+y^{2}+z^{2}\leq a^{2}}
V . Apskaičiuokime 1/4 ieškomo tūrio, nes kūnas simetriškas plokštumų xOz ir xOy atžvilgiu. Integravimo sritis D yra duotojo ritinio pagrindas. Kūną iš viršaus riboja paviršius
z
=
a
2
−
x
2
−
y
2
,
{\displaystyle z={\sqrt {a^{2}-x^{2}-y^{2}}},}
V
=
4
∬
D
a
2
−
x
2
−
y
2
d
x
d
y
.
{\displaystyle V=4\iint _{D}{\sqrt {a^{2}-x^{2}-y^{2}}}dxdy.}
V
=
4
∬
D
a
2
−
ρ
2
ρ
d
ρ
d
ϕ
.
{\displaystyle V=4\iint _{D}{\sqrt {a^{2}-\rho ^{2}}}\rho d\rho d\phi .}
ϕ
{\displaystyle \phi }
ρ
{\displaystyle \rho }
x
2
+
y
2
=
a
x
{\displaystyle x^{2}+y^{2}=ax}
ρ
2
=
a
ρ
cos
ϕ
,
ρ
=
a
cos
ϕ
.
{\displaystyle \rho ^{2}=a\rho \cos \phi ,\;\rho =a\cos \phi .}
D gauname, kai
ϕ
{\displaystyle \phi }
π
2
,
{\displaystyle {\pi \over 2},}
ρ
{\displaystyle \rho }
a
cos
ϕ
.
{\displaystyle a\cos \phi .}
V
=
4
∫
0
π
2
d
ϕ
∫
0
a
cos
ϕ
a
2
−
ρ
2
ρ
d
ρ
=
−
4
2
∫
0
π
2
d
ϕ
∫
0
a
cos
ϕ
a
2
−
ρ
2
d
(
a
2
−
ρ
2
)
=
{\displaystyle V=4\int _{0}^{\pi \over 2}d\phi \int _{0}^{a\cos \phi }{\sqrt {a^{2}-\rho ^{2}}}\rho d\rho =-{4 \over 2}\int _{0}^{\pi \over 2}d\phi \int _{0}^{a\cos \phi }{\sqrt {a^{2}-\rho ^{2}}}d(a^{2}-\rho ^{2})=}
=
−
4
3
∫
0
π
2
(
a
2
−
ρ
2
)
3
|
0
a
cos
ϕ
d
ϕ
=
−
4
3
∫
0
π
2
(
(
a
2
−
(
a
cos
ϕ
)
2
)
3
−
(
a
2
−
0
2
)
3
)
d
ϕ
=
{\displaystyle =-{4 \over 3}\int _{0}^{\pi \over 2}{\sqrt {(a^{2}-\rho ^{2})^{3}}}|_{0}^{a\cos \phi }d\phi =-{4 \over 3}\int _{0}^{\pi \over 2}({\sqrt {(a^{2}-(a\cos \phi )^{2})^{3}}}-{\sqrt {(a^{2}-0^{2})^{3}}})d\phi =}
=
−
4
3
∫
0
π
2
(
a
6
(
1
−
cos
2
ϕ
)
3
−
a
3
)
d
ϕ
=
−
4
3
∫
0
π
2
(
a
6
(
sin
2
ϕ
)
3
−
a
3
)
d
ϕ
=
−
4
3
∫
0
π
2
(
a
3
sin
3
ϕ
−
a
3
)
d
ϕ
=
{\displaystyle =-{4 \over 3}\int _{0}^{\pi \over 2}({\sqrt {a^{6}(1-\cos ^{2}\phi )^{3}}}-a^{3})d\phi =-{4 \over 3}\int _{0}^{\pi \over 2}({\sqrt {a^{6}(\sin ^{2}\phi )^{3}}}-a^{3})d\phi =-{4 \over 3}\int _{0}^{\pi \over 2}(a^{3}\sin ^{3}\phi -a^{3})d\phi =}
=
−
4
3
a
3
∫
0
π
2
(
sin
3
ϕ
−
1
)
d
ϕ
=
−
4
3
a
3
(
2
!
!
3
!
!
−
π
2
)
=
4
3
a
3
(
π
2
−
2
3
)
=
2
a
3
9
(
3
π
−
4
)
,
{\displaystyle =-{4 \over 3}a^{3}\int _{0}^{\pi \over 2}(\sin ^{3}\phi -1)d\phi =-{4 \over 3}a^{3}({2!! \over 3!!}-{\pi \over 2})={4 \over 3}a^{3}({\pi \over 2}-{2 \over 3})={2a^{3} \over 9}(3\pi -4),}
pasinaudojom dvigubu faktorialu trigonometrijoje .
Pasinaudodami trigonometrijos formule
sin
3
A
=
3
sin
A
−
4
sin
3
A
,
{\displaystyle \sin 3A=3\sin A-4\sin ^{3}A,}
4
sin
3
A
=
3
sin
A
−
sin
3
A
;
sin
3
A
=
1
4
⋅
(
3
sin
A
−
sin
3
A
)
,
{\displaystyle 4\sin ^{3}A=3\sin A-\sin 3A;\;\sin ^{3}A={\frac {1}{4}}\cdot (3\sin A-\sin 3A),}
V
=
−
4
3
a
3
∫
0
π
2
(
sin
3
ϕ
−
1
)
d
ϕ
=
−
4
3
a
3
∫
0
π
2
(
1
4
⋅
(
3
sin
ϕ
−
sin
(
3
ϕ
)
)
−
1
)
d
ϕ
=
−
4
3
⋅
a
3
⋅
(
1
4
⋅
(
−
3
cos
ϕ
+
1
3
⋅
cos
(
3
ϕ
)
)
−
ϕ
)
|
0
π
2
=
{\displaystyle V=-{4 \over 3}a^{3}\int _{0}^{\pi \over 2}(\sin ^{3}\phi -1)d\phi =-{4 \over 3}a^{3}\int _{0}^{\pi \over 2}({\frac {1}{4}}\cdot (3\sin \phi -\sin(3\phi ))-1)d\phi =-{4 \over 3}\cdot a^{3}\cdot ({\frac {1}{4}}\cdot (-3\cos \phi +{\frac {1}{3}}\cdot \cos(3\phi ))-\phi )|_{0}^{\pi \over 2}=}
=
a
3
3
⋅
(
3
cos
ϕ
−
1
3
⋅
cos
(
3
ϕ
)
+
4
ϕ
)
|
0
π
2
=
a
3
3
⋅
(
3
cos
π
2
−
1
3
⋅
cos
(
3
⋅
π
2
)
+
4
⋅
π
2
)
−
a
3
3
⋅
(
3
cos
(
0
)
−
1
3
⋅
cos
(
3
⋅
0
)
+
4
⋅
0
)
=
{\displaystyle ={\frac {a^{3}}{3}}\cdot (3\cos \phi -{\frac {1}{3}}\cdot \cos(3\phi )+4\phi )|_{0}^{\pi \over 2}={\frac {a^{3}}{3}}\cdot (3\cos {\pi \over 2}-{\frac {1}{3}}\cdot \cos(3\cdot {\pi \over 2})+4\cdot {\frac {\pi }{2}})-{\frac {a^{3}}{3}}\cdot (3\cos(0)-{\frac {1}{3}}\cdot \cos(3\cdot 0)+4\cdot 0)=}
=
a
3
3
⋅
(
3
⋅
0
−
1
3
⋅
0
+
2
π
)
−
a
3
3
⋅
(
3
⋅
1
−
1
3
⋅
1
+
0
)
=
a
3
3
⋅
2
π
−
a
3
3
⋅
(
3
⋅
−
1
3
)
=
2
⋅
π
⋅
a
3
3
−
a
3
3
⋅
8
3
=
2
π
a
3
3
−
8
a
3
9
=
6
π
a
3
−
8
a
3
9
.
{\displaystyle ={\frac {a^{3}}{3}}\cdot (3\cdot 0-{\frac {1}{3}}\cdot 0+2\pi )-{\frac {a^{3}}{3}}\cdot (3\cdot 1-{\frac {1}{3}}\cdot 1+0)={\frac {a^{3}}{3}}\cdot 2\pi -{\frac {a^{3}}{3}}\cdot (3\cdot -{\frac {1}{3}})={\frac {2\cdot \pi \cdot a^{3}}{3}}-{\frac {a^{3}}{3}}\cdot {\frac {8}{3}}={\frac {2\pi a^{3}}{3}}-{\frac {8a^{3}}{9}}={\frac {6\pi a^{3}-8a^{3}}{9}}.}
Čia a yra rutulio spindulys ir ritinio pagrindo skersmuo. Kai a=1, tai cilindru iš rutulio išpjautas tūris yra lygus:
V
=
6
π
a
3
−
8
a
3
9
=
6
π
−
8
9
=
1.205506214.
{\displaystyle V={\frac {6\pi a^{3}-8a^{3}}{9}}={\frac {6\pi -8}{9}}=1.205506214.}
V
r
i
t
.
=
S
p
a
g
r
.
⋅
h
=
π
⋅
r
2
⋅
h
=
π
⋅
(
a
2
)
2
⋅
2
a
=
π
⋅
a
2
4
⋅
2
a
=
π
a
3
2
=
π
⋅
1
3
2
=
1.570796327.
{\displaystyle V_{rit.}=S_{pagr.}\cdot h=\pi \cdot r^{2}\cdot h=\pi \cdot ({\frac {a}{2}})^{2}\cdot 2a=\pi \cdot {\frac {a^{2}}{4}}\cdot 2a={\frac {\pi a^{3}}{2}}={\frac {\pi \cdot 1^{3}}{2}}=1.570796327.}
Rasime tūrį V kūno, gauto iš rutulio išpjovus du cilindrus
x
2
+
y
2
≤
a
x
2
{\displaystyle x^{2}+y^{2}\leq ax^{2}}
x
2
+
y
2
≤
−
a
x
2
.
{\displaystyle x^{2}+y^{2}\leq -ax^{2}.}
V
0
{\displaystyle V_{0}}
z
2
+
x
2
+
y
2
≤
a
2
.
{\displaystyle z^{2}+x^{2}+y^{2}\leq a^{2}.}
V
0
=
8
∬
D
a
2
−
x
2
−
y
2
d
x
d
y
=
8
∬
D
a
2
−
ρ
2
ρ
d
ρ
d
ϕ
=
8
∫
0
π
2
d
ϕ
∫
0
a
cos
ϕ
ρ
a
2
−
ρ
2
d
ρ
=
{\displaystyle V_{0}=8\iint _{D}{\sqrt {a^{2}-x^{2}-y^{2}}}dxdy=8\iint _{D}{\sqrt {a^{2}-\rho ^{2}}}\rho d\rho d\phi =8\int _{0}^{\pi \over 2}d\phi \int _{0}^{a\cos \phi }\rho {\sqrt {a^{2}-\rho ^{2}}}d\rho =}
=
−
4
∫
0
π
2
d
ϕ
∫
0
a
cos
ϕ
a
2
−
ρ
2
d
(
a
2
−
ρ
2
)
=
8
3
a
3
∫
0
π
2
(
1
−
sin
3
ϕ
)
d
ϕ
=
8
3
(
π
2
−
2
!
!
3
!
!
)
a
3
=
4
a
3
9
(
3
π
−
4
)
.
{\displaystyle =-4\int _{0}^{\pi \over 2}d\phi \int _{0}^{a\cos \phi }{\sqrt {a^{2}-\rho ^{2}}}d(a^{2}-\rho ^{2})={8 \over 3}a^{3}\int _{0}^{\pi \over 2}(1-\sin ^{3}\phi )d\phi ={8 \over 3}({\pi \over 2}-{2!! \over 3!!})a^{3}={4a^{3} \over 9}(3\pi -4).}
V tūris yra lygus rutulio ir išpjautojo kūno
V
0
{\displaystyle V_{0}}
|
V
|
=
4
3
π
a
3
−
|
V
0
|
=
4
π
a
3
3
−
4
a
3
(
3
π
−
4
)
9
=
12
π
a
3
−
12
a
3
π
−
16
a
3
9
=
|
−
16
a
3
9
|
=
16
a
3
9
.
{\displaystyle |V|={4 \over 3}\pi a^{3}-|V_{0}|={4\pi a^{3} \over 3}-{4a^{3}(3\pi -4) \over 9}={12\pi a^{3}-12a^{3}\pi -16a^{3} \over 9}=|-{16a^{3} \over 9}|={16a^{3} \over 9}.}
Rasime kūno tūrį V , išpjauto iš cilindro, kurio pagrindas yra apskritimas
x
2
+
y
2
=
1
{\displaystyle x^{2}+y^{2}=1}
x
2
+
y
2
=
4
−
z
;
{\displaystyle x^{2}+y^{2}=4-z;}
x
2
+
y
2
=
4
(
z
+
2
)
.
{\displaystyle x^{2}+y^{2}=4(z+2).}
V randame kaip sumą tūrių
V
1
{\displaystyle V_{1}}
V
2
{\displaystyle V_{2}}
XOY . Tokiu budu pereinant į polinę koordinačių sistemą apskritimo lygtis tampa
ρ
2
=
1
{\displaystyle \rho ^{2}=1}
ρ
2
=
4
−
z
{\displaystyle \rho ^{2}=4-z}
z
=
4
−
ρ
2
{\displaystyle z=4-\rho ^{2}}
ρ
2
=
4
(
z
+
2
)
{\displaystyle \rho ^{2}=4(z+2)}
z
=
ρ
2
4
−
2.
{\displaystyle z={\frac {\rho ^{2}}{4}}-2.}
V
1
=
4
∫
0
π
2
d
ϕ
∫
0
1
(
4
−
ρ
2
)
ρ
d
ρ
=
4
∫
0
π
2
d
ϕ
∫
0
1
(
4
ρ
−
ρ
3
)
d
ρ
=
4
∫
0
π
2
d
ϕ
(
2
ρ
2
−
ρ
4
4
)
|
0
1
=
{\displaystyle V_{1}=4\int _{0}^{\pi \over 2}{\mathsf {d}}\phi \int _{0}^{1}(4-\rho ^{2})\rho \;{\mathsf {d}}\rho =4\int _{0}^{\pi \over 2}{\mathsf {d}}\phi \int _{0}^{1}(4\rho -\rho ^{3}){\mathsf {d}}\rho =4\int _{0}^{\pi \over 2}{\mathsf {d}}\phi (2\rho ^{2}-{\frac {\rho ^{4}}{4}})|_{0}^{1}=}
=
4
∫
0
π
2
(
2
⋅
1
2
−
1
4
4
−
(
2
⋅
0
2
−
0
4
4
)
)
d
ϕ
=
4
∫
0
π
2
(
2
−
1
4
)
d
ϕ
=
4
∫
0
π
2
8
−
1
4
d
ϕ
=
{\displaystyle =4\int _{0}^{\pi \over 2}\left(2\cdot 1^{2}-{\frac {1^{4}}{4}}-(2\cdot 0^{2}-{\frac {0^{4}}{4}})\right){\mathsf {d}}\phi =4\int _{0}^{\pi \over 2}\left(2-{\frac {1}{4}}\right){\mathsf {d}}\phi =4\int _{0}^{\pi \over 2}{\frac {8-1}{4}}{\mathsf {d}}\phi =}
=
4
⋅
7
4
∫
0
π
2
d
ϕ
=
7
ϕ
|
0
π
2
=
7
π
2
=
10.99557429.
{\displaystyle =4\cdot {\frac {7}{4}}\int _{0}^{\pi \over 2}{\mathsf {d}}\phi =7\phi |_{0}^{\pi \over 2}={\frac {7\pi }{2}}=10.99557429.}
V
2
=
4
∫
0
π
2
d
ϕ
∫
0
1
(
ρ
2
4
−
2
)
ρ
d
ρ
=
4
∫
0
π
2
d
ϕ
∫
0
1
(
ρ
3
4
−
2
ρ
)
d
ρ
=
4
∫
0
π
2
d
ϕ
(
ρ
4
16
−
ρ
2
)
|
0
1
=
4
∫
0
π
2
(
1
4
16
−
1
2
)
d
ϕ
=
{\displaystyle V_{2}=4\int _{0}^{\pi \over 2}{\mathsf {d}}\phi \int _{0}^{1}({\frac {\rho ^{2}}{4}}-2)\rho \;{\mathsf {d}}\rho =4\int _{0}^{\pi \over 2}{\mathsf {d}}\phi \int _{0}^{1}({\frac {\rho ^{3}}{4}}-2\rho )\;{\mathsf {d}}\rho =4\int _{0}^{\pi \over 2}{\mathsf {d}}\phi ({\frac {\rho ^{4}}{16}}-\rho ^{2})|_{0}^{1}=4\int _{0}^{\pi \over 2}({\frac {1^{4}}{16}}-1^{2}){\mathsf {d}}\phi =}
=
4
∫
0
π
2
(
1
16
−
1
)
d
ϕ
=
4
16
∫
0
π
2
(
1
−
16
)
d
ϕ
=
1
4
∫
0
π
2
(
−
15
)
d
ϕ
=
−
15
4
⋅
ϕ
|
0
π
2
=
−
15
π
8
=
−
5.890486225.
{\displaystyle =4\int _{0}^{\pi \over 2}({\frac {1}{16}}-1){\mathsf {d}}\phi ={\frac {4}{16}}\int _{0}^{\pi \over 2}(1-16){\mathsf {d}}\phi ={\frac {1}{4}}\int _{0}^{\pi \over 2}(-15){\mathsf {d}}\phi =-{\frac {15}{4}}\cdot \phi |_{0}^{\pi \over 2}=-{\frac {15\pi }{8}}=-5.890486225.}
V
=
V
1
+
|
V
2
|
=
7
π
2
+
|
−
15
π
8
|
=
7
π
2
+
15
π
8
=
π
(
7
⋅
4
+
15
)
8
=
43
π
8
=
16.88606051.
{\displaystyle V=V_{1}+|V_{2}|={\frac {7\pi }{2}}+|-{\frac {15\pi }{8}}|={\frac {7\pi }{2}}+{\frac {15\pi }{8}}={\frac {\pi (7\cdot 4+15)}{8}}={43\pi \over 8}=16.88606051.}
Dvilypio integralo taikymas mechanikoje [ keisti ] [ keisti ] Dvilypiu integralu masė apskaičiuojama pagal formulę:
m
=
∬
D
γ
(
x
,
y
)
d
x
d
y
.
{\displaystyle m=\iint _{D}\gamma (x,y)dxdy.}
Pavyzdžiai
Skritulinės plokštelės spindulys R , o jos plokštuminis tankis tiesiog proporcingas atstumo nuo taško iki plokštelės centro kvadratui. Plokštelės kontūro taškuose tankis lygus a . Apskaičiuokime tos plokštelės masę. Pagal sąlyga, tankis taške (x; y) lygus atstumo nuo to taško iki taško (0; 0) kvadratui:
γ
(
x
,
y
)
=
k
(
x
2
+
y
2
)
;
{\displaystyle \gamma (x,y)=k(x^{2}+y^{2});}
x
2
+
y
2
=
R
2
,
{\displaystyle x^{2}+y^{2}=R^{2},}
γ
(
x
,
y
)
=
a
.
{\displaystyle \gamma (x,y)=a.}
a
=
k
R
2
.
{\displaystyle a=kR^{2}.}
k
=
a
R
2
.
{\displaystyle k={a \over R^{2}}.}
γ
(
x
,
y
)
=
a
R
2
(
x
2
+
y
2
)
.
{\displaystyle \gamma (x,y)={a \over R^{2}}(x^{2}+y^{2}).}
m
=
a
R
2
∬
D
(
x
2
+
y
2
)
d
x
d
y
.
{\displaystyle m={a \over R^{2}}\iint _{D}(x^{2}+y^{2})dxdy.}
Šį integralą apskaičiuosime pakeisdami jį kartotiniu integralu, užrašytu polinėje koordinačių sistemoje. Taigi
m
=
a
R
2
∬
D
ρ
3
d
ρ
d
ϕ
=
a
R
2
∫
0
2
π
d
ϕ
∫
0
R
ρ
3
d
ρ
=
a
R
2
∫
0
2
π
R
4
4
d
ϕ
=
a
R
2
4
ϕ
|
0
2
π
=
1
2
π
a
R
2
.
{\displaystyle m={a \over R^{2}}\iint _{D}\rho ^{3}d\rho d\phi ={a \over R^{2}}\int _{0}^{2\pi }d\phi \int _{0}^{R}\rho ^{3}d\rho ={a \over R^{2}}\int _{0}^{2\pi }{R^{4} \over 4}d\phi ={aR^{2} \over 4}\phi |_{0}^{2\pi }={1 \over 2}\pi aR^{2}.}
kvadratinė plokštelė Rasime kvadratinės plokštelės masę su kraštine 2a , jeigu tankis
γ
(
x
;
y
)
{\displaystyle \gamma (x;y)}
M
(
x
;
y
)
{\displaystyle M(x;y)}
M iki įžambinių susikirtimo (iki centro), ir proporcingumo koeficientas lygus k . Parinksime koordinačių sistemą kaip parodyta paveiksliuke. Po šito galima rasti funkciją
γ
(
x
;
y
)
{\displaystyle \gamma (x;y)}
M iki taško suskirtimo įstrižainių lygus
x
2
+
y
2
.
{\displaystyle x^{2}+y^{2}.}
M :
γ
(
M
)
=
γ
(
x
;
y
)
=
k
(
x
2
+
y
2
)
.
{\displaystyle \gamma (M)=\gamma (x;y)=k(x^{2}+y^{2}).}
m
=
∬
D
k
(
x
2
+
y
2
)
d
x
d
y
.
{\displaystyle m=\iint _{D}k(x^{2}+y^{2})dxdy.}
x ir y , o integravimo sritis simteriška koordinačių ašių atžvilgiu, galima apsiriboti apskaičiavimu integralo toje dalyje srities D , kuri yra I ketvirtyje, t. y.
m
=
4
k
∫
0
a
d
x
∫
0
a
(
x
2
+
y
2
)
d
y
=
4
k
∫
0
a
(
x
2
y
+
y
3
3
)
|
0
a
d
x
=
{\displaystyle m=4k\int _{0}^{a}dx\int _{0}^{a}(x^{2}+y^{2})dy=4k\int _{0}^{a}(x^{2}y+{y^{3} \over 3})|_{0}^{a}dx=}
=
4
k
∫
0
a
(
a
x
2
+
a
3
3
)
d
x
=
4
k
(
a
x
3
3
+
a
3
x
3
)
|
0
a
=
4
k
2
a
4
3
=
8
3
k
a
4
.
{\displaystyle =4k\int _{0}^{a}(ax^{2}+{a^{3} \over 3})dx=4k({ax^{3} \over 3}+{a^{3}x \over 3})|_{0}^{a}=4k{2a^{4} \over 3}={8 \over 3}ka^{4}.}
1. [ keisti ] Masės centro koordinatės randamos pagal formules:
x
c
=
∬
D
x
γ
(
x
,
y
)
d
x
d
y
∬
D
γ
(
x
,
y
)
d
x
d
y
,
y
c
=
∬
D
y
γ
(
x
,
y
)
d
x
d
y
∬
D
γ
(
x
,
y
)
d
x
d
y
.
{\displaystyle x_{c}={\iint _{D}x\gamma (x,y)dxdy \over \iint _{D}\gamma (x,y)dxdy},\;y_{c}={\iint _{D}y\gamma (x,y)dxdy \over \iint _{D}\gamma (x,y)dxdy}.}
Pavyzdžiai
Homogeninę plokštelę
(
γ
(
x
,
y
)
=
c
o
n
s
t
)
{\displaystyle (\gamma (x,y)=const)}
y
=
x
2
{\displaystyle y=x^{2}}
y
−
x
=
2.
{\displaystyle y-x=2.}
Kai
γ
(
x
,
y
)
=
c
o
n
s
t
,
{\displaystyle \gamma (x,y)=const,}
x
c
=
∬
D
x
d
x
d
y
∬
D
d
x
d
y
,
y
c
=
∬
D
y
d
x
d
y
∬
D
d
x
d
y
.
{\displaystyle x_{c}={\iint _{D}xdxdy \over \iint _{D}dxdy},\;y_{c}={\iint _{D}ydxdy \over \iint _{D}dxdy}.}
∬
D
d
x
d
y
=
∫
−
1
2
d
x
∫
x
2
2
+
x
d
y
=
∫
−
1
2
(
2
+
x
−
x
2
)
d
x
=
(
2
x
+
x
2
2
−
x
3
3
)
|
−
1
2
=
9
2
;
{\displaystyle \iint _{D}dxdy=\int _{-1}^{2}dx\int _{x^{2}}^{2+x}dy=\int _{-1}^{2}(2+x-x^{2})dx=(2x+{x^{2} \over 2}-{x^{3} \over 3})|_{-1}^{2}={9 \over 2};}
∬
D
x
d
x
d
y
=
∫
−
1
2
x
d
x
∫
x
2
2
+
x
d
y
=
∫
−
1
2
(
2
x
+
x
2
−
x
3
)
d
x
=
(
x
2
+
x
3
3
−
x
4
4
)
|
−
1
2
=
9
4
;
{\displaystyle \iint _{D}xdxdy=\int _{-1}^{2}xdx\int _{x^{2}}^{2+x}dy=\int _{-1}^{2}(2x+x^{2}-x^{3})dx=(x^{2}+{x^{3} \over 3}-{x^{4} \over 4})|_{-1}^{2}={9 \over 4};}
∬
D
y
d
x
d
y
=
∫
−
1
2
d
x
∫
x
2
2
+
x
y
d
y
=
1
2
∫
−
1
2
(
4
+
4
x
+
x
2
−
x
4
)
d
x
=
1
2
(
4
x
+
2
x
2
+
x
3
3
−
x
5
5
)
|
−
1
2
=
36
5
.
{\displaystyle \iint _{D}ydxdy=\int _{-1}^{2}dx\int _{x^{2}}^{2+x}ydy={1 \over 2}\int _{-1}^{2}(4+4x+x^{2}-x^{4})dx={1 \over 2}(4x+2x^{2}+{x^{3} \over 3}-{x^{5} \over 5})|_{-1}^{2}={36 \over 5}.}
Vadinasi
x
c
=
9
4
9
2
=
9
4
⋅
2
9
=
1
2
,
y
c
=
36
5
:
9
2
=
36
5
⋅
2
9
=
8
5
.
{\displaystyle x_{c}={{9 \over 4} \over {9 \over 2}}={9 \over 4}\cdot {2 \over 9}={1 \over 2},\;y_{c}={36 \over 5}:{9 \over 2}={36 \over 5}\cdot {2 \over 9}={8 \over 5}.}
Homogeninę plokštelę
(
γ
(
x
,
y
)
=
c
o
n
s
t
)
{\displaystyle (\gamma (x,y)=const)}
y
=
x
{\displaystyle y=x}
y
=
2
x
{\displaystyle y=2x}
x
=
2
{\displaystyle x=2}
Oy ašiai. Apskaičiuokime tos plokštelės masės centro koordinates. Kai
γ
(
x
,
y
)
=
c
o
n
s
t
,
{\displaystyle \gamma (x,y)=const,}
x
c
=
∬
D
x
d
x
d
y
∬
D
d
x
d
y
,
y
c
=
∬
D
y
d
x
d
y
∬
D
d
x
d
y
.
{\displaystyle x_{c}={\iint _{D}xdxdy \over \iint _{D}dxdy},\;y_{c}={\iint _{D}ydxdy \over \iint _{D}dxdy}.}
∬
D
d
x
d
y
=
∫
0
2
d
x
∫
x
2
x
d
y
=
∫
0
2
(
2
x
−
x
)
d
x
=
x
2
2
|
0
2
=
2
;
{\displaystyle \iint _{D}dxdy=\int _{0}^{2}dx\int _{x}^{2x}dy=\int _{0}^{2}(2x-x)dx={x^{2} \over 2}|_{0}^{2}=2;}
∬
D
x
d
x
d
y
=
∫
0
2
x
d
x
∫
x
2
x
d
y
=
∫
0
2
x
(
2
x
−
x
)
d
x
=
x
3
3
|
0
2
=
8
3
;
{\displaystyle \iint _{D}xdxdy=\int _{0}^{2}xdx\int _{x}^{2x}dy=\int _{0}^{2}x(2x-x)dx={x^{3} \over 3}|_{0}^{2}={8 \over 3};}
∬
D
y
d
x
d
y
=
∫
0
2
d
x
∫
x
2
x
y
d
y
=
∫
0
2
(
2
x
)
2
−
x
2
2
d
x
=
∫
0
2
3
x
2
2
d
x
=
3
2
x
3
3
|
0
2
=
2
3
2
−
0
3
2
=
4.
{\displaystyle \iint _{D}ydxdy=\int _{0}^{2}dx\int _{x}^{2x}ydy=\int _{0}^{2}{(2x)^{2}-x^{2} \over 2}dx=\int _{0}^{2}{3x^{2} \over 2}dx={3 \over 2}{x^{3} \over 3}|_{0}^{2}={2^{3} \over 2}-{0^{3} \over 2}=4.}
x
c
=
8
3
2
=
4
3
=
1
,
(
3
)
;
y
c
=
4
2
=
2.
{\displaystyle x_{c}={{8 \over 3} \over 2}={4 \over 3}=1,(3);\;y_{c}={4 \over 2}=2.}
plokštelė. Rasime centro koordinates homogeninės plokštelės, apribotos dvejomis parabolėmis
y
2
=
x
{\displaystyle y^{2}=x}
x
2
=
y
.
{\displaystyle x^{2}=y.}
m
=
∬
D
d
x
d
y
=
∫
0
1
d
x
∫
x
2
x
d
y
=
∫
0
1
(
x
0.5
−
x
2
)
d
x
=
(
2
3
x
x
−
x
3
3
)
|
0
1
=
{\displaystyle m=\iint _{D}dxdy=\int _{0}^{1}dx\int _{x^{2}}^{\sqrt {x}}dy=\int _{0}^{1}(x^{0.5}-x^{2})dx=({2 \over 3}x{\sqrt {x}}-{x^{3} \over 3})|_{0}^{1}=}
=
2
3
−
1
3
=
1
3
.
{\displaystyle ={2 \over 3}-{1 \over 3}={1 \over 3}.}
M
y
=
∬
D
x
d
x
d
y
=
∫
0
1
x
d
x
∫
x
2
x
d
y
=
∫
0
1
(
x
1.5
−
x
3
)
d
x
=
{\displaystyle M_{y}=\iint _{D}xdxdy=\int _{0}^{1}xdx\int _{x^{2}}^{\sqrt {x}}dy=\int _{0}^{1}(x^{1.5}-x^{3})dx=}
=
(
2
5
x
2
x
−
x
4
4
)
|
0
1
=
2
5
−
1
4
=
3
20
;
{\displaystyle =({2 \over 5}x^{2}{\sqrt {x}}-{x^{4} \over 4})|_{0}^{1}={2 \over 5}-{1 \over 4}={3 \over 20};}
M
x
=
∬
D
y
d
x
d
y
=
∫
0
1
d
x
∫
x
2
x
y
d
y
=
1
2
∫
0
1
(
x
−
x
4
)
d
x
=
1
2
(
x
2
2
−
x
5
5
)
|
0
1
=
1
2
(
1
2
−
1
5
)
=
3
20
.
{\displaystyle M_{x}=\iint _{D}ydxdy=\int _{0}^{1}dx\int _{x^{2}}^{\sqrt {x}}y\;dy={1 \over 2}\int _{0}^{1}(x-x^{4})dx={1 \over 2}({x^{2} \over 2}-{x^{5} \over 5})|_{0}^{1}={1 \over 2}({1 \over 2}-{1 \over 5})={3 \over 20}.}
x
c
=
M
y
m
=
3
20
:
1
3
=
9
20
;
x
c
=
M
x
m
=
3
20
:
1
3
=
9
20
.
{\displaystyle x_{c}={M_{y} \over m}={3 \over 20}:{1 \over 3}={9 \over 20};\;x_{c}={M_{x} \over m}={3 \over 20}:{1 \over 3}={9 \over 20}.}
[ keisti ] Inercijos momentai ašių Ox , Oy ir koordinačių pradžios atžvilgiu lygūs
I
x
=
∬
D
y
2
γ
(
x
,
y
)
d
x
d
y
,
I
y
=
∬
D
x
2
γ
(
x
,
y
)
d
x
d
y
,
I
0
=
∬
D
(
x
2
+
y
2
)
γ
(
x
,
y
)
d
x
d
y
.
{\displaystyle I_{x}=\iint _{D}y^{2}\gamma (x,y)dxdy,\;I_{y}=\iint _{D}x^{2}\gamma (x,y)dxdy,\;I_{0}=\iint _{D}(x^{2}+y^{2})\gamma (x,y)dxdy.}
Pavyzdžiai
2. Kardioidė Homogeninę figūrą
(
γ
(
x
,
y
)
=
1
)
{\displaystyle (\gamma (x,y)=1)}
ρ
=
a
(
1
+
cos
ϕ
)
.
{\displaystyle \rho =a(1+\cos \phi ).}
Ox , Oy ir poliaus O atžvilgiu. Kai
γ
(
x
,
y
)
=
1
,
{\displaystyle \gamma (x,y)=1,}
I
x
=
∬
D
y
2
d
x
d
y
,
I
y
=
∬
D
x
2
d
x
d
y
,
I
0
=
∬
D
(
x
2
+
y
2
)
d
x
d
y
.
{\displaystyle I_{x}=\iint _{D}y^{2}dxdy,\;I_{y}=\iint _{D}x^{2}dxdy,\;I_{0}=\iint _{D}(x^{2}+y^{2})dxdy.}
I
x
{\displaystyle I_{x}}
I
0
,
{\displaystyle I_{0},}
I
y
{\displaystyle I_{y}}
I
y
=
I
0
−
I
x
.
{\displaystyle I_{y}=I_{0}-I_{x}.}
I
x
=
∬
D
ρ
2
sin
2
ϕ
ρ
d
ρ
d
ϕ
=
∬
ρ
3
sin
2
ϕ
d
ρ
d
ϕ
.
I
0
=
∬
D
=
ρ
3
d
ρ
d
ϕ
.
{\displaystyle I_{x}=\iint _{D}\rho ^{2}\sin ^{2}\phi \rho d\rho d\phi =\iint \rho ^{3}\sin ^{2}\phi d\rho d\phi .\;I_{0}=\iint _{D}=\rho ^{3}d\rho d\phi .}
D dalimi. Taigi
I
x
=
2
∫
0
π
sin
2
ϕ
d
ϕ
∫
0
a
(
1
+
cos
ϕ
)
ρ
3
d
ρ
=
a
4
2
∫
0
π
sin
2
ϕ
(
1
+
cos
ϕ
)
4
d
ϕ
=
{\displaystyle I_{x}=2\int _{0}^{\pi }\sin ^{2}\phi d\phi \int _{0}^{a(1+\cos \phi )}\rho ^{3}d\rho ={a^{4} \over 2}\int _{0}^{\pi }\sin ^{2}\phi (1+\cos \phi )^{4}d\phi =}
=
a
4
2
∫
0
π
(
2
sin
ϕ
2
cos
ϕ
2
)
2
(
2
cos
2
ϕ
2
)
4
d
ϕ
=
32
a
4
∫
0
π
sin
2
ϕ
2
cos
10
ϕ
2
d
ϕ
.
{\displaystyle ={a^{4} \over 2}\int _{0}^{\pi }(2\sin {\phi \over 2}\cos {\phi \over 2})^{2}(2\cos ^{2}{\phi \over 2})^{4}d\phi =32a^{4}\int _{0}^{\pi }\sin ^{2}{\phi \over 2}\cos ^{10}{\phi \over 2}d\phi .}
t
=
ϕ
2
,
{\displaystyle t={\phi \over 2},}
d
ϕ
=
2
d
t
,
{\displaystyle d\phi =2dt,}
I
x
=
64
a
4
∫
0
π
2
sin
2
t
cos
10
t
d
t
=
64
a
4
∫
0
π
2
(
cos
10
t
−
cos
12
t
)
d
t
=
64
a
4
(
9
!
!
10
!
!
−
11
!
!
12
!
!
)
⋅
π
2
=
21
32
π
a
4
;
{\displaystyle I_{x}=64a^{4}\int _{0}^{\pi \over 2}\sin ^{2}t\cos ^{10}t\;dt=64a^{4}\int _{0}^{\pi \over 2}(\cos ^{10}t-\cos ^{12}t)dt=64a^{4}({9!! \over 10!!}-{11!! \over 12!!})\cdot {\pi \over 2}={21 \over 32}\pi a^{4};}
I
0
=
2
∫
0
π
d
ϕ
∫
0
a
(
1
+
cos
ϕ
)
ρ
3
d
ρ
=
a
4
2
∫
0
π
(
1
+
cos
ϕ
)
4
d
ϕ
=
8
a
4
∫
0
π
cos
8
ϕ
2
d
ϕ
=
16
a
4
∫
0
π
2
cos
8
t
d
t
=
{\displaystyle I_{0}=2\int _{0}^{\pi }d\phi \int _{0}^{a(1+\cos \phi )}\rho ^{3}d\rho ={a^{4} \over 2}\int _{0}^{\pi }(1+\cos \phi )^{4}d\phi =8a^{4}\int _{0}^{\pi }\cos ^{8}{\phi \over 2}d\phi =16a^{4}\int _{0}^{\pi \over 2}\cos ^{8}t\;dt=}
=
16
a
4
⋅
7
!
!
8
!
!
⋅
π
2
=
35
16
π
a
4
.
{\displaystyle =16a^{4}\cdot {7!! \over 8!!}\cdot {\pi \over 2}={35 \over 16}\pi a^{4}.}
I
y
=
I
0
−
I
x
=
35
16
π
a
4
−
21
32
π
a
4
=
49
32
π
a
4
.
{\displaystyle I_{y}=I_{0}-I_{x}={35 \over 16}\pi a^{4}-{21 \over 32}\pi a^{4}={49 \over 32}\pi a^{4}.}
Rasime inercijos momentą skritulio su spinduliu R su vienodu tankiu
γ
(
x
,
y
)
=
1
{\displaystyle \gamma (x,y)=1}
I
0
=
∬
D
(
x
2
+
y
2
)
d
x
d
y
.
{\displaystyle I_{0}=\iint _{D}(x^{2}+y^{2})dxdy.}
ρ
=
R
.
{\displaystyle \rho =R.}
I
0
=
∫
0
2
π
d
ϕ
∫
0
R
ρ
2
ρ
d
ρ
=
∫
0
2
π
ρ
4
4
|
0
R
d
ϕ
=
1
4
R
4
ϕ
|
0
2
π
=
π
R
4
2
.
{\displaystyle I_{0}=\int _{0}^{2\pi }d\phi \int _{0}^{R}\rho ^{2}\rho \;d\rho =\int _{0}^{2\pi }{\rho ^{4} \over 4}|_{0}^{R}d\phi ={1 \over 4}R^{4}\phi |_{0}^{2\pi }={\pi R^{4} \over 2}.}
Dvilypis integralas ir Jakobiano determinantas [ keisti ] Apskaičiuoti integralą
I
=
∬
R
(
x
2
+
y
2
)
d
x
d
y
,
{\displaystyle I=\iint _{R}(x^{2}+y^{2})dxdy,}
R yra plotas kurį iš kairio šono riboja kreivė
u
=
x
2
−
y
2
=
1
{\displaystyle u=x^{2}-y^{2}=1}
u
=
x
2
−
y
2
=
9
{\displaystyle u=x^{2}-y^{2}=9}
v
=
2
x
y
=
8
{\displaystyle v=2xy=8}
v
=
2
x
y
=
4
{\displaystyle v=2xy=4}
Sprendimas . Taikydami Jakobiano determinantą randame:
∂
u
∂
x
=
2
x
,
∂
u
∂
y
=
−
2
y
;
{\displaystyle {\frac {\partial u}{\partial x}}=2x,\quad {\frac {\partial u}{\partial y}}=-2y;}
∂
v
∂
x
=
2
y
,
∂
v
∂
y
=
2
x
;
{\displaystyle {\frac {\partial v}{\partial x}}=2y,\quad {\frac {\partial v}{\partial y}}=2x;}
∂
(
x
,
y
)
∂
(
u
,
v
)
=
[
∂
(
u
,
v
)
∂
(
x
,
y
)
]
−
1
=
|
∂
u
∂
x
∂
u
∂
y
∂
v
∂
x
∂
v
∂
y
|
−
1
=
|
2
x
−
2
y
2
y
2
x
|
−
1
=
1
2
x
⋅
2
−
(
−
2
y
)
⋅
2
y
=
1
4
x
2
+
4
y
2
=
1
4
(
x
2
+
y
2
)
.
{\displaystyle {\frac {\partial (x,\;y)}{\partial (u,\;v)}}=\left[{\frac {\partial (u,\;v)}{\partial (x,\;y)}}\right]^{-1}={\begin{vmatrix}{\frac {\partial u}{\partial x}}&{\frac {\partial u}{\partial y}}\\{\frac {\partial v}{\partial x}}&{\frac {\partial v}{\partial y}}\end{vmatrix}}^{-1}={\begin{vmatrix}2x&-2y\\2y&2x\end{vmatrix}}^{-1}={\frac {1}{2x\cdot 2-(-2y)\cdot 2y}}={\frac {1}{4x^{2}+4y^{2}}}={\frac {1}{4(x^{2}+y^{2})}}.}
Bet
u
2
+
v
2
=
(
x
2
−
y
2
)
2
+
(
2
x
y
)
2
=
x
4
−
2
x
2
y
2
+
y
4
+
4
x
2
y
2
=
x
4
+
2
x
2
y
2
+
y
4
=
(
x
2
+
y
2
)
2
.
{\displaystyle u^{2}+v^{2}=(x^{2}-y^{2})^{2}+(2xy)^{2}=x^{4}-2x^{2}y^{2}+y^{4}+4x^{2}y^{2}=x^{4}+2x^{2}y^{2}+y^{4}=(x^{2}+y^{2})^{2}.}
Todėl
∂
(
x
,
y
)
∂
(
u
,
v
)
=
1
4
u
2
+
v
2
.
{\displaystyle {\frac {\partial (x,\;y)}{\partial (u,\;v)}}={\frac {1}{4{\sqrt {u^{2}+v^{2}}}}}.}
Nustatę ribas randame integralą:
I
=
∬
R
(
x
2
+
y
2
)
d
x
d
y
=
∫
4
8
(
∫
1
9
(
x
2
+
y
2
)
4
u
2
+
v
2
d
u
)
d
v
=
∫
4
8
(
∫
1
9
u
2
+
v
2
4
u
2
+
v
2
d
u
)
d
v
=
1
4
∫
4
8
(
∫
1
9
d
u
)
d
v
=
{\displaystyle I=\iint _{R}(x^{2}+y^{2})dxdy=\int _{4}^{8}(\int _{1}^{9}{\frac {(x^{2}+y^{2})}{4{\sqrt {u^{2}+v^{2}}}}}du)dv=\int _{4}^{8}(\int _{1}^{9}{\frac {\sqrt {u^{2}+v^{2}}}{4{\sqrt {u^{2}+v^{2}}}}}du)dv={\frac {1}{4}}\int _{4}^{8}(\int _{1}^{9}du)dv=}
=
1
4
∫
4
8
(
u
|
1
9
)
d
v
=
1
4
∫
4
8
(
9
−
1
)
d
v
=
8
4
v
|
4
8
=
2
(
8
−
4
)
=
8.
{\displaystyle ={\frac {1}{4}}\int _{4}^{8}(u|_{1}^{9})dv={\frac {1}{4}}\int _{4}^{8}(9-1)dv={\frac {8}{4}}v|_{4}^{8}=2(8-4)=8.}
Pastaba . Mes radome ne plokščios figūros plotą R , o tiesiog apskaičiavome integralą, kuris neturi gilesnės prasmės (nebent susiję su kokiais nors inercijos momentais arba tankiu ir/ar jo kitimu pagal tam tikrą funkciją).
![{\displaystyle V={\frac {3}{30}}\cdot [3\cdot 9+(3-1\cdot 0.1)(9-(1\cdot 0.1)^{2})+(3-2\cdot 0.1)(9-(2\cdot 0.1)^{2})+(3-3\cdot 0.1)(9-(3\cdot 0.1)^{2})+...+(3-30\cdot 0.1)(9-(30\cdot 0.1)^{2})]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c909f0965e83b5f9ff4c7f78d6953ea58a396124)
![{\displaystyle {\frac {\partial (x,\;y)}{\partial (u,\;v)}}=\left[{\frac {\partial (u,\;v)}{\partial (x,\;y)}}\right]^{-1}={\begin{vmatrix}{\frac {\partial u}{\partial x}}&{\frac {\partial u}{\partial y}}\\{\frac {\partial v}{\partial x}}&{\frac {\partial v}{\partial y}}\end{vmatrix}}^{-1}={\begin{vmatrix}2x&-2y\\2y&2x\end{vmatrix}}^{-1}={\frac {1}{2x\cdot 2-(-2y)\cdot 2y}}={\frac {1}{4x^{2}+4y^{2}}}={\frac {1}{4(x^{2}+y^{2})}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/022d42e795c74e4aeb379d217bb227187af5f8ee)
