Trilypis integralas naudojamas tūriui apskaičiuoti ir mechanikoje – tose vietose, kur dvilypio integralo savybių neužtenka greitesniam apskaičiavimui.
Trilypio integralo apskaičiavimas[ keisti ]
∬
V
f
(
x
,
y
,
z
)
d
x
d
y
d
z
=
∫
a
b
d
x
∫
y
1
(
x
)
y
2
(
x
)
d
y
∫
z
1
(
x
,
y
)
z
2
(
x
,
y
)
f
(
x
,
y
,
z
)
d
z
.
{\displaystyle \iint _{V}f(x,y,z)dxdydz=\int _{a}^{b}dx\int _{y_{1}(x)}^{y_{2}(x)}dy\int _{z_{1}(x,y)}^{z_{2}(x,y)}f(x,y,z)dz.}
Apskaičiuosime tūrį V tetraedro , apriboto plokštumų
x
=
0
,
{\displaystyle x=0,}
y
=
0
,
{\displaystyle y=0,}
z
=
0
,
{\displaystyle z=0,}
x
+
y
+
z
=
2.
{\displaystyle x+y+z=2.}
Integravimo sritis D projektuojama į plokštumą xOy . Tūrį V iš apačios riboja plokštuma
z
=
0
,
{\displaystyle z=0,}
iš viršaus - plokštuma
z
=
2
−
x
−
y
.
{\displaystyle z=2-x-y.}
Trilypį integralą pakeičiame kartotiniu:
V
=
∭
V
d
x
d
y
d
z
=
∫
0
2
d
x
∫
0
2
−
x
d
y
∫
0
2
−
x
−
y
d
z
=
∫
0
2
d
x
∫
0
2
−
x
z
|
0
2
−
x
−
y
d
y
=
{\displaystyle V=\iiint _{V}dxdydz=\int _{0}^{2}dx\int _{0}^{2-x}dy\int _{0}^{2-x-y}dz=\int _{0}^{2}dx\int _{0}^{2-x}z|_{0}^{2-x-y}dy=}
=
∫
0
2
d
x
∫
0
2
−
x
(
2
−
x
−
y
)
d
y
=
∫
0
2
(
2
y
−
x
y
−
y
2
2
)
|
0
2
−
x
d
x
=
∫
0
2
(
4
−
2
x
−
2
x
+
x
2
−
1
2
(
4
−
4
x
+
x
2
)
)
d
x
=
{\displaystyle =\int _{0}^{2}dx\int _{0}^{2-x}(2-x-y)dy=\int _{0}^{2}(2y-xy-{y^{2} \over 2})|_{0}^{2-x}dx=\int _{0}^{2}(4-2x-2x+x^{2}-{1 \over 2}(4-4x+x^{2}))dx=}
=
∫
0
2
(
2
−
2
x
+
x
2
2
)
d
x
=
(
2
x
−
x
2
+
x
3
6
)
|
0
2
=
4
−
4
+
8
6
=
8
6
.
{\displaystyle =\int _{0}^{2}(2-2x+{x^{2} \over 2})dx=(2x-x^{2}+{x^{3} \over 6})|_{0}^{2}=4-4+{8 \over 6}={8 \over 6}.}
Šį atsakymą galima buvo gauti naudojantis mišriąja vektorių sandauga.
V
=
(
a
×
b
)
⋅
c
=
|
a
x
a
y
a
z
b
z
b
y
b
z
c
x
c
y
c
z
|
=
|
2
0
0
0
2
0
0
0
2
|
=
8.
{\displaystyle V=(a\times b)\cdot c={\begin{vmatrix}a_{x}&a_{y}&a_{z}\\b_{z}&b_{y}&b_{z}\\c_{x}&c_{y}&c_{z}\end{vmatrix}}={\begin{vmatrix}2&0&0\\0&2&0\\0&0&2\end{vmatrix}}=8.}
Gretasienio tūris yra 8. Rasime piramidės (t. y. netaisyklingo tetraedro) su 4 viršūnėmis, kurios pagrindas yra trikampis, tūrį:
V
=
1
6
|
(
a
×
b
)
⋅
c
|
=
8
6
.
{\displaystyle V={\frac {1}{6}}|(a\times b)\cdot c|={\frac {8}{6}}.}
∭
V
d
x
d
y
d
z
=
∫
−
1
1
d
x
∫
0
1
d
y
∫
0
2
d
z
=
∫
−
1
1
d
x
∫
0
1
2
d
y
=
∫
−
1
1
2
d
x
=
2
x
|
−
1
1
=
2
(
1
−
(
−
1
)
)
=
4.
{\displaystyle \iiint _{V}dxdydz=\int _{-1}^{1}dx\int _{0}^{1}dy\int _{0}^{2}dz=\int _{-1}^{1}dx\int _{0}^{1}2dy=\int _{-1}^{1}2dx=2x|_{-1}^{1}=2(1-(-1))=4.}
Šį tūrį galima buvo gauti nustačius kiekvienos kraštinės ilgį palei koordinačių ašis. M (1-(-1); 1-0; 2-0)=M (2; 1; 2). Sudauginus kraštinių ilgius gauname stačiakampio gretasienio tūrį
V
=
2
⋅
1
⋅
2
=
4.
{\displaystyle V=2\cdot 1\cdot 2=4.}
Arba per vektorius
V
=
(
a
×
b
)
⋅
c
=
|
2
0
0
0
1
0
0
0
2
|
=
4.
{\displaystyle V=(a\times b)\cdot c={\begin{vmatrix}2&0&0\\0&1&0\\0&0&2\end{vmatrix}}=4.}
Apskaičiuosime tetraedro tūrį V , apriboto plokštumomis x+y+z=2, z=1, x=0, y=0. tetraedro trys kraštinės a=b=c=1 ir lygiagrečios atitinkamai x , y ir z ašims, o kitos trys kraštinės
d
=
e
=
f
=
1
2
+
1
2
=
2
.
{\displaystyle d=e=f={\sqrt {1^{2}+1^{2}}}={\sqrt {2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∬
D
d
x
d
y
∫
1
2
−
x
−
y
d
z
=
∫
0
1
d
x
∫
0
1
−
x
d
y
∫
1
2
−
x
−
y
d
z
=
{\displaystyle V=\iiint _{V}dxdydz=\iint _{D}dxdy\int _{1}^{2-x-y}dz=\int _{0}^{1}dx\int _{0}^{1-x}dy\int _{1}^{2-x-y}dz=}
=
∫
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 =\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}-{1-2x+x^{2} \over 2})dx=\int _{0}^{1}({1 \over 2}-x+{x^{2} \over 2})dx=}
=
(
1
2
x
−
x
2
2
+
x
3
6
)
|
0
1
=
1
2
−
1
2
+
1
6
=
1
6
.
{\displaystyle =({1 \over 2}x-{x^{2} \over 2}+{x^{3} \over 6})|_{0}^{1}={1 \over 2}-{1 \over 2}+{1 \over 6}={1 \over 6}.}
Tą patį atsakymą galėjome gauti pasinaudodami piramidės tūrio skaičiavimu per vektorius M (1-0; 1-0; 2-1)=M (1; 1; 1):
V
=
1
6
|
(
a
×
b
)
⋅
c
|
=
1
6
|
1
0
0
0
1
0
0
0
1
|
=
1
6
.
{\displaystyle V={1 \over 6}|(a\times b)\cdot c|={1 \over 6}{\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}}={1 \over 6}.}
Vaizdas:Trilypis1321.jpg 13.21.
Pirmajame oktante esantį kūną riboja paviršiai
z
=
4
−
y
2
,
{\displaystyle z=4-y^{2},}
x
+
y
=
2
,
{\displaystyle x+y=2,}
2
x
+
y
=
2
,
{\displaystyle 2x+y=2,}
y
=
0
,
{\displaystyle y=0,}
z
=
0
{\displaystyle z=0}
(pav. 13.21). Apskaičiuokime to kūno tūrį. Kūno tūrį apskaičiuosime pagal formulę
V
=
∭
V
d
x
d
y
d
z
=
∬
D
d
x
d
y
∫
0
4
−
y
2
.
{\displaystyle V=\iiint _{V}dxdydz=\iint _{D}dxdy\int _{0}^{4-y^{2}}.}
Integravimo sritits D yra kūno projekcija plokštumoje xOy . Parinkus vienokią integravimo tvarką, dvilypis integralas šioje srityje išreiškiamas vienu kartotiniu integralu, o pakeitus tą tvarką dviem kartotiniais integralais:
∬
D
f
(
x
,
y
)
d
x
d
y
=
∫
0
2
d
y
∫
2
−
y
2
2
−
y
f
(
x
,
y
)
d
x
{\displaystyle \iint _{D}f(x,y)dxdy=\int _{0}^{2}dy\int _{2-y \over 2}^{2-y}f(x,y)dx}
arba
∬
D
f
(
x
,
y
)
d
x
d
y
=
∫
0
1
d
x
∫
2
−
2
x
2
−
x
f
(
x
,
y
)
d
y
+
∫
1
2
d
x
∫
0
2
−
x
f
(
x
,
y
)
d
y
.
{\displaystyle \iint _{D}f(x,y)dxdy=\int _{0}^{1}dx\int _{2-2x}^{2-x}f(x,y)dy+\int _{1}^{2}dx\int _{0}^{2-x}f(x,y)dy.}
Todėl trilypį integralą keisdami kartotiniu, remkimės trumpesne formule:
V
=
∫
0
2
d
y
∫
2
−
y
2
2
−
y
d
x
∫
0
4
−
y
2
d
z
=
∫
0
2
d
y
∫
2
−
y
2
2
−
y
z
|
0
4
−
y
2
d
x
=
∫
0
2
d
y
∫
2
−
y
2
2
−
y
(
4
−
y
2
)
d
x
=
{\displaystyle V=\int _{0}^{2}dy\int _{2-y \over 2}^{2-y}dx\int _{0}^{4-y^{2}}dz=\int _{0}^{2}dy\int _{2-y \over 2}^{2-y}z|_{0}^{4-y^{2}}dx=\int _{0}^{2}dy\int _{2-y \over 2}^{2-y}(4-y^{2})dx=}
=
∫
0
2
(
4
−
y
2
)
x
|
2
−
y
2
2
−
y
d
y
=
∫
0
2
(
4
−
2
y
−
y
2
+
y
3
2
)
d
y
=
(
4
y
−
y
2
−
y
3
3
+
y
4
8
)
|
0
2
=
10
3
.
{\displaystyle =\int _{0}^{2}(4-y^{2})x|_{2-y \over 2}^{2-y}dy=\int _{0}^{2}(4-2y-y^{2}+{y^{3} \over 2})dy=(4y-y^{2}-{y^{3} \over 3}+{y^{4} \over 8})|_{0}^{2}={10 \over 3}.}
Apskaičiuosime tūrį kūno apriboto šiais paviršiais:
y
=
x
,
y
=
2
x
,
{\displaystyle y={\sqrt {x}},\;y=2{\sqrt {x}},}
z
=
0
{\displaystyle z=0}
ir
x
+
z
=
6.
{\displaystyle x+z=6.}
Iš lygties
x
+
z
=
6
,
{\displaystyle x+z=6,}
kai z lygi nuliui
x
=
6.
{\displaystyle x=6.}
Kai
x
=
6
{\displaystyle x=6}
parabolės įgija reikšmes
y
=
6
{\displaystyle y={\sqrt {6}}}
ir
y
=
2
6
.
{\displaystyle y=2{\sqrt {6}}.}
Todėl tūris lygus
V
=
∫
0
6
d
x
∫
x
2
x
d
y
∫
0
6
−
x
d
z
=
∫
0
6
d
x
∫
x
2
x
(
6
−
x
)
d
y
=
{\displaystyle V=\int _{0}^{6}dx\int _{\sqrt {x}}^{2{\sqrt {x}}}dy\int _{0}^{6-x}dz=\int _{0}^{6}dx\int _{\sqrt {x}}^{2{\sqrt {x}}}(6-x)dy=}
=
∫
0
6
(
6
−
x
)
x
d
x
=
(
6
⋅
x
3
2
3
2
−
2
5
⋅
x
5
2
)
|
0
6
=
4
⋅
6
6
−
2
5
⋅
6
2
6
=
24
6
−
72
5
6
=
48
6
5
.
{\displaystyle =\int _{0}^{6}(6-x){\sqrt {x}}dx=(6\cdot {x^{3 \over 2} \over {3 \over 2}}-{2 \over 5}\cdot x^{5 \over 2})|_{0}^{6}=4\cdot 6{\sqrt {6}}-{2 \over 5}\cdot 6^{2}{\sqrt {6}}=24{\sqrt {6}}-{72 \over 5}{\sqrt {6}}={48{\sqrt {6}} \over 5}.}
Vaizdas:Integral379380.jpg 379.
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
y
=
x
2
{\displaystyle y=x^{2}}
(parabolė ant plokštumos xOy ),
y
=
1
,
{\displaystyle y=1,}
z
=
0
{\displaystyle z=0}
(plokštuma ant plokštumos xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
(paraboloidas) (pav. 379).
V
=
∫
0
1
d
x
∫
x
2
1
d
y
∫
0
x
2
+
y
2
d
z
=
∫
0
1
d
x
∫
x
2
1
d
y
z
|
0
x
2
+
y
2
=
∫
0
1
d
x
∫
x
2
1
(
(
x
2
+
y
2
)
−
0
)
d
y
=
{\displaystyle V=\int _{0}^{1}{\mathsf {d}}x\int _{x^{2}}^{1}{\mathsf {d}}y\int _{0}^{x^{2}+y^{2}}{\mathsf {d}}z=\int _{0}^{1}{\mathsf {d}}x\int _{x^{2}}^{1}{\mathsf {d}}y\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{1}{\mathsf {d}}x\int _{x^{2}}^{1}((x^{2}+y^{2})-0){\mathsf {d}}y=}
=
∫
0
1
d
x
∫
x
2
1
(
x
2
+
y
2
)
d
y
=
∫
0
1
d
x
(
x
2
y
+
y
3
3
)
|
x
2
1
=
∫
0
1
d
x
[
(
x
2
⋅
1
+
1
3
3
)
−
(
x
2
⋅
x
2
+
(
x
2
)
3
3
)
]
=
{\displaystyle =\int _{0}^{1}{\mathsf {d}}x\int _{x^{2}}^{1}(x^{2}+y^{2}){\mathsf {d}}y=\int _{0}^{1}{\mathsf {d}}x\;(x^{2}y+{\frac {y^{3}}{3}})|_{x^{2}}^{1}=\int _{0}^{1}{\mathsf {d}}x\;[(x^{2}\cdot 1+{\frac {1^{3}}{3}})-(x^{2}\cdot x^{2}+{\frac {(x^{2})^{3}}{3}})]=}
=
∫
0
1
[
x
2
+
1
3
−
x
4
−
x
6
3
]
d
x
=
(
x
3
3
+
x
3
−
x
5
5
−
x
7
3
⋅
7
)
|
0
1
=
(
1
3
3
+
1
3
−
1
5
5
−
1
7
21
)
−
(
0
3
3
+
0
3
−
0
5
5
−
0
7
21
)
=
{\displaystyle =\int _{0}^{1}[x^{2}+{\frac {1}{3}}-x^{4}-{\frac {x^{6}}{3}}]{\mathsf {d}}x=\left({\frac {x^{3}}{3}}+{\frac {x}{3}}-{\frac {x^{5}}{5}}-{\frac {x^{7}}{3\cdot 7}}\right)|_{0}^{1}=\left({\frac {1^{3}}{3}}+{\frac {1}{3}}-{\frac {1^{5}}{5}}-{\frac {1^{7}}{21}}\right)-\left({\frac {0^{3}}{3}}+{\frac {0}{3}}-{\frac {0^{5}}{5}}-{\frac {0^{7}}{21}}\right)=}
=
(
1
3
+
1
3
−
1
5
−
1
21
)
=
(
2
3
−
1
5
−
1
21
)
=
(
2
⋅
7
−
1
21
−
1
5
)
=
(
13
21
−
1
5
)
=
13
⋅
5
−
21
21
⋅
5
=
65
−
21
105
=
44
105
=
0.419047619.
{\displaystyle =\left({\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{5}}-{\frac {1}{21}}\right)=\left({\frac {2}{3}}-{\frac {1}{5}}-{\frac {1}{21}}\right)=\left({\frac {2\cdot 7-1}{21}}-{\frac {1}{5}}\right)=\left({\frac {13}{21}}-{\frac {1}{5}}\right)={\frac {13\cdot 5-21}{21\cdot 5}}={\frac {65-21}{105}}={\frac {44}{105}}=0.419047619.}
Kad gauti tūrį dviejuose oktantuose, reikia padauginti iš 2.
Autoriaus manymu, tikrasis tūris gali buti apskaičiuotas (o kad geriau suprasti kaip apskaičiuoti, reikėtų įsigilinti į sukimo tūrio radimą) taip:
V
=
∫
0
1
y
⋅
y
2
d
y
=
y
1
2
+
2
d
y
=
∫
0
1
y
5
2
d
y
=
y
5
2
+
1
5
2
+
1
|
0
1
=
y
7
2
7
2
|
0
1
=
2
⋅
y
7
2
7
|
0
1
=
2
⋅
1
7
2
7
=
2
7
=
0.285714285
,
{\displaystyle V=\int _{0}^{1}{\sqrt {y}}\cdot y^{2}dy=y^{{\frac {1}{2}}+2}dy=\int _{0}^{1}y^{5 \over 2}dy={\frac {y^{{\frac {5}{2}}+1}}{{\frac {5}{2}}+1}}|_{0}^{1}={\frac {y^{\frac {7}{2}}}{\frac {7}{2}}}|_{0}^{1}={\frac {2\cdot y^{\frac {7}{2}}}{7}}|_{0}^{1}={\frac {2\cdot 1^{\frac {7}{2}}}{7}}={\frac {2}{7}}=0.285714285,}
arba galbūt net taip:
V
=
∫
0
1
y
⋅
y
2
⋅
y
d
y
,
{\displaystyle V=\int _{0}^{1}{\sqrt {y}}\cdot y^{2}\cdot y\;dy,}
arba taip:
V
=
∫
0
1
y
⋅
y
2
⋅
y
d
y
.
{\displaystyle V=\int _{0}^{1}{\sqrt {y}}\cdot y^{2}\cdot {\sqrt {y}}\;dy.}
Bent jau elipsinio paraboloido, tokio kaip
x
2
10000
+
y
2
=
z
{\displaystyle {\frac {x^{2}}{10000}}+y^{2}=z}
(x turi būti 100, kai y =0, kad z būtų lygus 1), pakeitimu, šiame uždavinyje, tūris turėtų būti
V
=
∫
0
1
y
⋅
y
2
d
y
.
{\displaystyle V=\int _{0}^{1}{\sqrt {y}}\cdot y^{2}\;dy.}
Pavyzdis . Rasti kūno tūrį 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}}
(pav. 379).
V
=
∫
0
1
d
y
∫
0
y
d
x
∫
0
x
2
+
y
2
d
z
=
∫
0
1
d
y
∫
0
y
d
x
z
|
0
x
2
+
y
2
=
∫
0
1
d
y
∫
0
y
(
(
x
2
+
y
2
)
−
0
)
d
x
=
{\displaystyle V=\int _{0}^{1}{\mathsf {d}}y\int _{0}^{\sqrt {y}}{\mathsf {d}}x\int _{0}^{x^{2}+y^{2}}{\mathsf {d}}z=\int _{0}^{1}{\mathsf {d}}y\int _{0}^{\sqrt {y}}{\mathsf {d}}x\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{1}{\mathsf {d}}y\int _{0}^{\sqrt {y}}((x^{2}+y^{2})-0){\mathsf {d}}x=}
=
∫
0
1
d
y
∫
0
y
(
x
2
+
y
2
)
d
x
=
∫
0
1
d
y
(
x
3
3
+
y
2
x
)
|
0
y
=
∫
0
1
d
y
[
(
(
y
)
3
3
+
y
2
y
)
−
(
0
3
3
+
y
2
⋅
0
)
]
=
{\displaystyle =\int _{0}^{1}{\mathsf {d}}y\int _{0}^{\sqrt {y}}(x^{2}+y^{2}){\mathsf {d}}x=\int _{0}^{1}{\mathsf {d}}y\;({\frac {x^{3}}{3}}+y^{2}x)|_{0}^{\sqrt {y}}=\int _{0}^{1}{\mathsf {d}}y\;[({\frac {({\sqrt {y}})^{3}}{3}}+y^{2}{\sqrt {y}})-({\frac {0^{3}}{3}}+y^{2}\cdot 0)]=}
=
∫
0
1
[
y
3
2
3
+
y
5
2
]
d
y
=
(
y
3
2
+
1
3
(
3
2
+
1
)
+
y
5
2
+
1
5
2
+
1
)
|
0
1
=
(
y
5
2
3
⋅
5
2
+
y
7
2
5
+
2
2
)
|
0
1
=
(
2
y
5
2
15
+
2
y
7
2
7
)
|
0
1
=
{\displaystyle =\int _{0}^{1}[{\frac {y^{3 \over 2}}{3}}+y^{5 \over 2}]{\mathsf {d}}y=\left({\frac {y^{{\frac {3}{2}}+1}}{3({\frac {3}{2}}+1)}}+{\frac {y^{{\frac {5}{2}}+1}}{{\frac {5}{2}}+1}}\right)|_{0}^{1}=\left({\frac {y^{\frac {5}{2}}}{3\cdot {\frac {5}{2}}}}+{\frac {y^{\frac {7}{2}}}{\frac {5+2}{2}}}\right)|_{0}^{1}=\left({\frac {2y^{\frac {5}{2}}}{15}}+{\frac {2y^{\frac {7}{2}}}{7}}\right)|_{0}^{1}=}
=
(
2
⋅
1
5
2
15
+
2
⋅
1
7
2
7
)
−
(
2
⋅
0
5
2
15
+
2
⋅
0
7
2
7
)
=
2
15
+
2
7
=
2
⋅
7
+
2
⋅
15
15
⋅
7
=
14
+
30
105
=
44
105
=
0.419047619.
{\displaystyle =\left({\frac {2\cdot 1^{\frac {5}{2}}}{15}}+{\frac {2\cdot 1^{\frac {7}{2}}}{7}}\right)-\left({\frac {2\cdot 0^{\frac {5}{2}}}{15}}+{\frac {2\cdot 0^{\frac {7}{2}}}{7}}\right)={\frac {2}{15}}+{\frac {2}{7}}={\frac {2\cdot 7+2\cdot 15}{15\cdot 7}}={\frac {14+30}{105}}={\frac {44}{105}}=0.419047619.}
Kad gauti tūrį dviejuose oktantuose, reikia padauginti iš 2.
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
y
=
x
2
{\displaystyle y=x^{2}}
(parabolė ant plokštumos xOy ),
y
=
9
,
{\displaystyle y=9,}
z
=
0
{\displaystyle z=0}
(plokštuma ant plokštumos xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
(paraboloidas).
V
=
∫
0
3
d
x
∫
x
2
9
d
y
∫
0
x
2
+
y
2
d
z
=
∫
0
3
d
x
∫
x
2
9
d
y
z
|
0
x
2
+
y
2
=
∫
0
3
d
x
∫
x
2
9
(
(
x
2
+
y
2
)
−
0
)
d
y
=
{\displaystyle V=\int _{0}^{3}{\mathsf {d}}x\int _{x^{2}}^{9}{\mathsf {d}}y\int _{0}^{x^{2}+y^{2}}{\mathsf {d}}z=\int _{0}^{3}{\mathsf {d}}x\int _{x^{2}}^{9}{\mathsf {d}}y\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{3}{\mathsf {d}}x\int _{x^{2}}^{9}((x^{2}+y^{2})-0){\mathsf {d}}y=}
=
∫
0
3
d
x
∫
x
2
9
(
x
2
+
y
2
)
d
y
=
∫
0
3
d
x
(
x
2
y
+
y
3
3
)
|
x
2
9
=
∫
0
1
d
x
[
(
x
2
⋅
9
+
9
3
3
)
−
(
x
2
⋅
x
2
+
(
x
2
)
3
3
)
]
=
{\displaystyle =\int _{0}^{3}{\mathsf {d}}x\int _{x^{2}}^{9}(x^{2}+y^{2}){\mathsf {d}}y=\int _{0}^{3}{\mathsf {d}}x\;(x^{2}y+{\frac {y^{3}}{3}})|_{x^{2}}^{9}=\int _{0}^{1}{\mathsf {d}}x\;[(x^{2}\cdot 9+{\frac {9^{3}}{3}})-(x^{2}\cdot x^{2}+{\frac {(x^{2})^{3}}{3}})]=}
=
∫
0
3
[
9
x
2
+
729
3
−
x
4
−
x
6
3
]
d
x
=
(
9
x
3
3
+
729
x
3
−
x
5
5
−
x
7
3
⋅
7
)
|
0
3
=
(
9
⋅
3
3
3
+
729
⋅
3
3
−
3
5
5
−
3
7
21
)
−
(
9
⋅
0
3
3
+
729
⋅
0
3
−
0
5
5
−
0
7
21
)
=
{\displaystyle =\int _{0}^{3}[9x^{2}+{\frac {729}{3}}-x^{4}-{\frac {x^{6}}{3}}]{\mathsf {d}}x=\left({\frac {9x^{3}}{3}}+{\frac {729x}{3}}-{\frac {x^{5}}{5}}-{\frac {x^{7}}{3\cdot 7}}\right)|_{0}^{3}=\left({\frac {9\cdot 3^{3}}{3}}+{\frac {729\cdot 3}{3}}-{\frac {3^{5}}{5}}-{\frac {3^{7}}{21}}\right)-\left({\frac {9\cdot 0^{3}}{3}}+{\frac {729\cdot 0}{3}}-{\frac {0^{5}}{5}}-{\frac {0^{7}}{21}}\right)=}
=
(
9
⋅
9
+
729
−
243
5
−
2187
21
)
=
81
+
729
−
243
5
−
729
7
=
810
−
243
5
−
729
7
=
810
⋅
5
⋅
7
−
243
⋅
7
−
729
⋅
5
5
⋅
7
=
{\displaystyle =\left(9\cdot 9+729-{\frac {243}{5}}-{\frac {2187}{21}}\right)=81+729-{\frac {243}{5}}-{\frac {729}{7}}=810-{\frac {243}{5}}-{\frac {729}{7}}={\frac {810\cdot 5\cdot 7-243\cdot 7-729\cdot 5}{5\cdot 7}}=}
=
28350
−
1701
−
3645
35
=
23004
35
=
657.2571429.
{\displaystyle ={\frac {28350-1701-3645}{35}}={\frac {23004}{35}}=657.2571429.}
Kad gauti tūrį dviejuose oktantuose, reikia padauginti iš 2.
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
y
=
x
2
{\displaystyle y=x^{2}}
,
y
=
9
,
{\displaystyle y=9,}
z
=
0
{\displaystyle z=0}
,
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
.
V
=
∫
0
9
d
y
∫
0
y
d
x
∫
0
x
2
+
y
2
d
z
=
∫
0
9
d
y
∫
0
y
d
x
z
|
0
x
2
+
y
2
=
∫
0
9
d
y
∫
0
y
(
(
x
2
+
y
2
)
−
0
)
d
x
=
{\displaystyle V=\int _{0}^{9}{\mathsf {d}}y\int _{0}^{\sqrt {y}}{\mathsf {d}}x\int _{0}^{x^{2}+y^{2}}{\mathsf {d}}z=\int _{0}^{9}{\mathsf {d}}y\int _{0}^{\sqrt {y}}{\mathsf {d}}x\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{9}{\mathsf {d}}y\int _{0}^{\sqrt {y}}((x^{2}+y^{2})-0){\mathsf {d}}x=}
=
∫
0
9
d
y
∫
0
y
(
x
2
+
y
2
)
d
x
=
∫
0
9
d
y
(
x
3
3
+
y
2
x
)
|
0
y
=
∫
0
9
d
y
[
(
(
y
)
3
3
+
y
2
y
)
−
(
0
3
3
+
y
2
⋅
0
)
]
=
{\displaystyle =\int _{0}^{9}{\mathsf {d}}y\int _{0}^{\sqrt {y}}(x^{2}+y^{2}){\mathsf {d}}x=\int _{0}^{9}{\mathsf {d}}y\;({\frac {x^{3}}{3}}+y^{2}x)|_{0}^{\sqrt {y}}=\int _{0}^{9}{\mathsf {d}}y\;[({\frac {({\sqrt {y}})^{3}}{3}}+y^{2}{\sqrt {y}})-({\frac {0^{3}}{3}}+y^{2}\cdot 0)]=}
=
∫
0
9
[
y
3
2
3
+
y
5
2
]
d
y
=
(
y
3
2
+
1
3
(
3
2
+
1
)
+
y
5
2
+
1
5
2
+
1
)
|
0
9
=
(
y
5
2
3
⋅
5
2
+
y
7
2
5
+
2
2
)
|
0
9
=
(
2
y
5
2
15
+
2
y
7
2
7
)
|
0
9
=
{\displaystyle =\int _{0}^{9}[{\frac {y^{3 \over 2}}{3}}+y^{5 \over 2}]{\mathsf {d}}y=\left({\frac {y^{{\frac {3}{2}}+1}}{3({\frac {3}{2}}+1)}}+{\frac {y^{{\frac {5}{2}}+1}}{{\frac {5}{2}}+1}}\right)|_{0}^{9}=\left({\frac {y^{\frac {5}{2}}}{3\cdot {\frac {5}{2}}}}+{\frac {y^{\frac {7}{2}}}{\frac {5+2}{2}}}\right)|_{0}^{9}=\left({\frac {2y^{\frac {5}{2}}}{15}}+{\frac {2y^{\frac {7}{2}}}{7}}\right)|_{0}^{9}=}
=
(
2
⋅
9
5
2
15
+
2
⋅
9
7
2
7
)
−
(
2
⋅
0
5
2
15
+
2
⋅
0
7
2
7
)
=
2
59049
15
+
2
4782969
7
=
2
⋅
243
15
+
2
⋅
2187
7
=
486
15
+
4374
7
=
{\displaystyle =\left({\frac {2\cdot 9^{\frac {5}{2}}}{15}}+{\frac {2\cdot 9^{\frac {7}{2}}}{7}}\right)-\left({\frac {2\cdot 0^{\frac {5}{2}}}{15}}+{\frac {2\cdot 0^{\frac {7}{2}}}{7}}\right)={\frac {2{\sqrt {59049}}}{15}}+{\frac {2{\sqrt {4782969}}}{7}}={\frac {2\cdot 243}{15}}+{\frac {2\cdot 2187}{7}}={\frac {486}{15}}+{\frac {4374}{7}}=}
=
486
⋅
7
+
4374
⋅
15
15
⋅
7
=
3402
+
65610
105
=
69012
105
=
23004
35
=
657.2571429.
{\displaystyle ={\frac {486\cdot 7+4374\cdot 15}{15\cdot 7}}={\frac {3402+65610}{105}}={\frac {69012}{105}}={\frac {23004}{35}}=657.2571429.}
Kad gauti tūrį dviejuose oktantuose, reikia padauginti iš 2.
Vaizdas:Integral379380.jpg 380.
Pavyzdis . Rasti kūno tūrį V , išpjaunamą iš begalinės prizmės su kraštais
x
=
±
1
,
y
=
±
1
{\displaystyle x=\pm 1,\;y=\pm 1}
paraboloidais
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)}
(pav. 380).
z
1
=
4
−
x
2
−
y
2
,
{\displaystyle z_{1}=4-x^{2}-y^{2},}
z
2
=
x
2
+
y
2
4
−
2.
{\displaystyle z_{2}={\frac {x^{2}+y^{2}}{4}}-2.}
Kai reikšmės x ir y yra 0, tai
z
1
=
4
{\displaystyle z_{1}=4}
,
z
2
=
−
2
,
{\displaystyle z_{2}=-2,}
šie taškai ir yra aukščiausias ir žemiausias taškai.
V
=
∫
0
1
d
x
∫
0
1
d
y
∫
x
2
+
y
2
4
−
2
4
−
x
2
−
y
2
d
z
=
∫
0
1
d
x
∫
0
1
d
y
z
|
x
2
+
y
2
4
−
2
4
−
x
2
−
y
2
=
∫
0
1
d
x
∫
0
1
[
(
4
−
x
2
−
y
2
)
−
(
x
2
+
y
2
4
−
2
)
]
d
y
=
{\displaystyle V=\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}{\mathsf {d}}y\int _{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}{\mathsf {d}}z=\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}{\mathsf {d}}y\;z|_{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}=\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}[(4-x^{2}-y^{2})-({\frac {x^{2}+y^{2}}{4}}-2)]{\mathsf {d}}y=}
=
∫
0
1
d
x
∫
0
1
(
6
−
x
2
−
y
2
−
x
2
+
y
2
4
)
d
y
=
∫
0
1
d
x
(
6
y
−
x
2
y
−
y
3
3
−
x
2
y
4
−
y
3
4
⋅
3
)
|
0
1
=
∫
0
1
(
6
⋅
1
−
x
2
⋅
1
−
1
3
3
−
x
2
⋅
1
4
−
1
3
12
)
d
x
=
{\displaystyle =\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}(6-x^{2}-y^{2}-{\frac {x^{2}+y^{2}}{4}}){\mathsf {d}}y=\int _{0}^{1}{\mathsf {d}}x(6y-x^{2}y-{\frac {y^{3}}{3}}-{\frac {x^{2}y}{4}}-{\frac {y^{3}}{4\cdot 3}})|_{0}^{1}=\int _{0}^{1}(6\cdot 1-x^{2}\cdot 1-{\frac {1^{3}}{3}}-{\frac {x^{2}\cdot 1}{4}}-{\frac {1^{3}}{12}}){\mathsf {d}}x=}
=
∫
0
1
(
6
−
x
2
−
1
3
−
x
2
4
−
1
12
)
d
x
=
∫
0
1
(
6
⋅
12
−
1
⋅
4
−
1
12
−
x
2
−
x
2
4
)
d
x
=
∫
0
1
(
72
−
4
−
1
12
−
x
2
−
x
2
4
)
d
x
=
{\displaystyle =\int _{0}^{1}(6-x^{2}-{\frac {1}{3}}-{\frac {x^{2}}{4}}-{\frac {1}{12}}){\mathsf {d}}x=\int _{0}^{1}({\frac {6\cdot 12-1\cdot 4-1}{12}}-x^{2}-{\frac {x^{2}}{4}}){\mathsf {d}}x=\int _{0}^{1}({\frac {72-4-1}{12}}-x^{2}-{\frac {x^{2}}{4}}){\mathsf {d}}x=}
=
∫
0
1
(
67
12
−
x
2
−
x
2
4
)
d
x
=
(
67
x
12
−
x
3
3
−
x
3
4
⋅
3
)
|
0
1
=
67
⋅
1
12
−
1
3
3
−
1
3
12
=
67
−
4
−
1
12
=
67
−
5
12
=
62
12
=
31
6
=
5.166666667.
{\displaystyle =\int _{0}^{1}({\frac {67}{12}}-x^{2}-{\frac {x^{2}}{4}}){\mathsf {d}}x=({\frac {67x}{12}}-{\frac {x^{3}}{3}}-{\frac {x^{3}}{4\cdot 3}})|_{0}^{1}={\frac {67\cdot 1}{12}}-{\frac {1^{3}}{3}}-{\frac {1^{3}}{12}}={\frac {67-4-1}{12}}={\frac {67-5}{12}}={\frac {62}{12}}={\frac {31}{6}}=5.166666667.}
Kad gauti tūrį visuose 8-iuose oktantuose, reikia
31
6
{\displaystyle {\frac {31}{6}}}
padauginti iš 4.
Palyginimui, stačiakampio gretasienio tūris, kurio kraštinės a=1, b=1, c=6 yra lygus
V
b
i
g
=
1
⋅
1
⋅
6
=
6.
{\displaystyle V_{big}=1\cdot 1\cdot 6=6.}
Paraboloidas.
Pavyzdis . Kūną riboja plokštuma xOy , cilindrinis paviršius
x
2
+
y
2
=
1
{\displaystyle x^{2}+y^{2}=1}
ir paraboloidas
z
=
x
2
+
y
2
.
{\displaystyle z=x^{2}+y^{2}.}
Praboloidas su cilindriniu paviršiumi susikerta, kai
z
=
1
{\displaystyle z=1}
Apskaičiuosime to kūno tūrį.
Sprendimas . Kadangi kūnas yra simetriškas koordinačių plokštumų xOz ir yOz atžvilgiu, tai apskaičiuosime tik jo ketvirtadalio, esančio pirmajame oktante tūrį. Taigi
V
=
∫
0
1
d
x
∫
0
1
−
x
2
d
y
∫
0
x
2
+
y
2
d
z
=
∫
0
1
d
x
∫
0
1
−
x
2
(
x
2
+
y
2
)
d
y
=
∫
0
1
d
x
(
x
2
y
+
y
3
3
)
|
0
1
−
x
2
=
{\displaystyle V=\int _{0}^{1}dx\int _{0}^{\sqrt {1-x^{2}}}dy\int _{0}^{x^{2}+y^{2}}dz=\int _{0}^{1}dx\int _{0}^{\sqrt {1-x^{2}}}(x^{2}+y^{2})dy=\int _{0}^{1}dx(x^{2}y+{\frac {y^{3}}{3}})|_{0}^{\sqrt {1-x^{2}}}=}
=
∫
0
1
(
x
2
1
−
x
2
+
(
1
−
x
2
)
3
3
)
d
x
=
∫
0
π
2
(
sin
2
(
t
)
1
−
sin
2
t
+
(
1
−
sin
2
t
)
3
3
)
cos
(
t
)
d
t
=
{\displaystyle =\int _{0}^{1}(x^{2}{\sqrt {1-x^{2}}}+{\frac {({\sqrt {1-x^{2}}})^{3}}{3}})dx=\int _{0}^{\pi \over 2}(\sin ^{2}(t){\sqrt {1-\sin ^{2}t}}+{\frac {({\sqrt {1-\sin ^{2}t}})^{3}}{3}})\cos(t)\;dt=}
=
∫
0
π
2
(
sin
2
(
t
)
cos
2
t
+
(
cos
2
t
)
3
3
)
cos
(
t
)
d
t
=
∫
0
π
2
(
sin
2
(
t
)
cos
t
+
1
3
cos
3
t
)
cos
(
t
)
d
t
=
∫
0
π
2
(
sin
2
(
t
)
cos
2
t
+
1
3
cos
4
t
)
d
t
=
{\displaystyle =\int _{0}^{\pi \over 2}(\sin ^{2}(t){\sqrt {\cos ^{2}t}}+{\frac {({\sqrt {\cos ^{2}t}})^{3}}{3}})\cos(t)\;dt=\int _{0}^{\pi \over 2}(\sin ^{2}(t)\cos t+{\frac {1}{3}}\cos ^{3}t)\cos(t)\;dt=\int _{0}^{\pi \over 2}(\sin ^{2}(t)\cos ^{2}t+{\frac {1}{3}}\cos ^{4}t)\;dt=}
=
∫
0
π
2
(
(
1
−
cos
2
t
)
cos
2
t
+
1
3
cos
4
t
)
d
t
=
∫
0
π
2
(
cos
2
t
−
cos
4
t
+
1
3
cos
4
t
)
d
t
=
∫
0
π
2
(
cos
2
t
−
2
3
cos
4
t
)
d
t
=
{\displaystyle =\int _{0}^{\pi \over 2}((1-\cos ^{2}t)\cos ^{2}t+{\frac {1}{3}}\cos ^{4}t)\;dt=\int _{0}^{\pi \over 2}(\cos ^{2}t-\cos ^{4}t+{\frac {1}{3}}\cos ^{4}t)\;dt=\int _{0}^{\pi \over 2}(\cos ^{2}t-{\frac {2}{3}}\cos ^{4}t)\;dt=}
=
∫
0
π
2
(
cos
(
2
t
)
+
1
2
−
2
3
⋅
cos
(
4
t
)
+
4
cos
(
2
t
)
+
3
8
)
d
t
=
(
1
2
sin
(
2
t
)
+
t
2
−
2
3
⋅
1
4
sin
(
4
t
)
+
2
sin
(
2
t
)
+
3
t
8
)
|
0
π
2
=
{\displaystyle =\int _{0}^{\pi \over 2}({\frac {\cos(2t)+1}{2}}-{\frac {2}{3}}\cdot {\cos(4t)+4\cos(2t)+3 \over 8})\;dt=({\frac {{\frac {1}{2}}\sin(2t)+t}{2}}-{\frac {2}{3}}\cdot {{\frac {1}{4}}\sin(4t)+2\sin(2t)+3t \over 8})|_{0}^{\pi \over 2}=}
=
(
1
2
sin
(
2
⋅
π
2
)
+
π
2
2
−
2
3
⋅
1
4
sin
(
4
⋅
π
2
)
+
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
8
)
−
(
1
2
sin
(
2
⋅
0
)
+
0
2
−
2
3
⋅
1
4
sin
(
4
⋅
0
)
+
2
sin
(
2
⋅
0
)
+
3
⋅
0
8
)
=
{\displaystyle =({\frac {{\frac {1}{2}}\sin(2\cdot {\frac {\pi }{2}})+{\frac {\pi }{2}}}{2}}-{\frac {2}{3}}\cdot {{\frac {1}{4}}\sin(4\cdot {\frac {\pi }{2}})+2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}} \over 8})-({\frac {{\frac {1}{2}}\sin(2\cdot 0)+0}{2}}-{\frac {2}{3}}\cdot {{\frac {1}{4}}\sin(4\cdot 0)+2\sin(2\cdot 0)+3\cdot 0 \over 8})=}
=
(
sin
π
+
π
4
−
2
⋅
1
4
sin
(
2
π
)
+
2
sin
(
π
)
+
3
π
2
24
)
−
(
1
2
sin
(
0
)
2
−
2
24
⋅
(
1
4
sin
(
0
)
+
2
sin
(
0
)
+
0
)
)
=
{\displaystyle =({\frac {\sin \pi +\pi }{4}}-2\cdot {{\frac {1}{4}}\sin(2\pi )+2\sin(\pi )+{\frac {3\pi }{2}} \over 24})-({\frac {{\frac {1}{2}}\sin(0)}{2}}-{\frac {2}{24}}\cdot ({\frac {1}{4}}\sin(0)+2\sin(0)+0))=}
=
(
0
+
π
4
−
2
⋅
1
4
⋅
0
+
2
⋅
0
+
3
π
2
24
)
−
0
=
π
4
−
3
π
2
12
=
π
4
−
3
π
24
=
π
4
−
π
8
=
π
8
=
0.392699081.
{\displaystyle =({\frac {0+\pi }{4}}-2\cdot {{\frac {1}{4}}\cdot 0+2\cdot 0+{\frac {3\pi }{2}} \over 24})-0={\frac {\pi }{4}}-{{\frac {3\pi }{2}} \over 12}={\frac {\pi }{4}}-{\frac {3\pi }{24}}={\frac {\pi }{4}}-{\frac {\pi }{8}}={\frac {\pi }{8}}=0.392699081.}
kur
x
=
sin
t
,
{\displaystyle x=\sin t,}
kai
x
=
0
{\displaystyle x=0}
, tada
t
=
0
{\displaystyle t=0}
ir kai
x
=
1
{\displaystyle x=1}
, tada
t
=
π
2
,
{\displaystyle t={\frac {\pi }{2}},}
d
x
=
cos
(
t
)
d
t
;
{\displaystyle dx=\cos(t)dt;}
cos
2
A
=
cos
(
2
A
)
+
1
2
,
cos
4
A
=
cos
(
4
A
)
+
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \cos ^{2}A={\frac {\cos(2A)+1}{2}},\quad \cos ^{4}A={\cos(4A)+4\cos(2A)+3 \over 8}.}
Pasinaudojant dvigubu faktorialu gauname tą patį atsakymą:
V
=
∫
0
π
2
(
cos
2
t
−
2
3
cos
4
t
)
d
t
=
(
2
−
1
)
!
!
2
!
!
⋅
π
2
−
2
3
⋅
(
4
−
1
)
!
!
4
!
!
⋅
π
2
=
1
!
!
2
!
!
⋅
π
2
−
1
3
⋅
3
!
!
4
!
!
⋅
π
=
1
2
⋅
π
2
−
1
3
⋅
3
4
⋅
2
⋅
π
=
π
4
−
1
8
⋅
π
=
π
8
.
{\displaystyle V=\int _{0}^{\pi \over 2}(\cos ^{2}t-{\frac {2}{3}}\cos ^{4}t)\;dt={\frac {(2-1)!!}{2!!}}\cdot {\frac {\pi }{2}}-{\frac {2}{3}}\cdot {\frac {(4-1)!!}{4!!}}\cdot {\frac {\pi }{2}}={\frac {1!!}{2!!}}\cdot {\frac {\pi }{2}}-{\frac {1}{3}}\cdot {\frac {3!!}{4!!}}\cdot \pi ={\frac {1}{2}}\cdot {\frac {\pi }{2}}-{\frac {1}{3}}\cdot {\frac {3}{4\cdot 2}}\cdot \pi ={\frac {\pi }{4}}-{\frac {1}{8}}\cdot \pi ={\frac {\pi }{8}}.}
Kad gauti tūrį keturiuose oktantuose, reikia padauginti iš 4.
Pavyzdis . Rasti kūno tūrį V , išpjaunamą iš begalinės prizmės su kraštais
x
=
±
1
,
y
=
±
1
{\displaystyle x=\pm 1,\;y=\pm 1}
paraboloidu
z
=
2
−
x
2
−
y
2
.
{\displaystyle z=2-x^{2}-y^{2}.}
V
=
∫
0
1
d
x
∫
0
1
d
y
∫
0
2
−
x
2
−
y
2
d
z
=
∫
0
1
d
x
∫
0
1
(
2
−
x
2
−
y
2
)
d
y
=
∫
0
1
d
x
(
2
y
−
x
2
y
−
y
3
3
)
|
0
1
=
{\displaystyle V=\int _{0}^{1}dx\int _{0}^{1}dy\int _{0}^{2-x^{2}-y^{2}}dz=\int _{0}^{1}dx\int _{0}^{1}(2-x^{2}-y^{2})dy=\int _{0}^{1}dx\;(2y-x^{2}y-{\frac {y^{3}}{3}})|_{0}^{1}=}
=
∫
0
1
(
2
−
x
2
−
1
3
)
d
x
=
(
2
x
−
x
3
3
−
x
3
)
|
0
1
=
2
−
1
3
−
1
3
=
2
−
2
3
=
6
−
2
3
=
4
3
=
1.3
(
3
)
.
{\displaystyle =\int _{0}^{1}(2-x^{2}-{\frac {1}{3}})dx=(2x-{\frac {x^{3}}{3}}-{\frac {x}{3}})|_{0}^{1}=2-{\frac {1}{3}}-{\frac {1}{3}}=2-{\frac {2}{3}}={\frac {6-2}{3}}={\frac {4}{3}}=1.3(3).}
Kad gauti tūrį keturiuose oktantuose, reikia padauginti iš keturių, tuomet tūris bus lygus
V
4
=
4
⋅
4
3
=
16
3
=
5.3
(
3
)
.
{\displaystyle V_{4}=4\cdot {\frac {4}{3}}={\frac {16}{3}}=5.3(3).}
Tūris esanti virš tūrio, kurį radome ir apribotas plokštuma
z
=
2
{\displaystyle z=2}
yra
V
v
i
r
=
a
⋅
b
⋅
c
−
V
4
=
2
⋅
2
⋅
2
−
16
3
=
8
−
16
3
=
8
3
=
2.6
(
6
)
.
{\displaystyle V_{vir}=a\cdot b\cdot c-V_{4}=2\cdot 2\cdot 2-{\frac {16}{3}}=8-{\frac {16}{3}}={\frac {8}{3}}=2.6(6).}
Rasime kūno tūrį V , esantį po paraboloidu
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
ir apribotą begalinės prizmės (stačiakampio gretasienio kurio aukšis begalinis) su kraštinėmis
x
=
1
,
y
=
1
{\displaystyle x=1,\;y=1}
.
V
=
∫
0
1
d
x
∫
0
1
d
y
∫
0
x
2
+
y
2
d
z
=
∫
0
1
d
x
∫
0
1
d
y
z
|
0
x
2
+
y
2
=
∫
0
1
d
x
∫
0
1
(
(
x
2
+
y
2
)
−
0
)
d
y
=
{\displaystyle V=\int _{0}^{1}dx\int _{0}^{1}dy\int _{0}^{x^{2}+y^{2}}dz=\int _{0}^{1}dx\int _{0}^{1}dy\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{1}dx\int _{0}^{1}((x^{2}+y^{2})-0)dy=}
=
∫
0
1
d
x
∫
0
1
(
x
2
+
y
2
)
d
y
=
∫
0
1
d
x
(
x
2
y
+
y
3
3
)
|
0
1
=
∫
0
1
(
x
2
⋅
1
+
1
3
3
)
d
x
=
(
x
3
3
+
x
3
)
|
0
1
=
1
−
1
3
3
+
1
3
=
2
3
=
0.6
(
6
)
.
{\displaystyle =\int _{0}^{1}dx\int _{0}^{1}(x^{2}+y^{2})dy=\int _{0}^{1}dx(x^{2}y+{\frac {y^{3}}{3}})|_{0}^{1}=\int _{0}^{1}(x^{2}\cdot 1+{\frac {1^{3}}{3}})dx=({\frac {x^{3}}{3}}+{\frac {x}{3}})|_{0}^{1}=1-{\frac {1^{3}}{3}}+{\frac {1}{3}}={\frac {2}{3}}=0.6(6).}
Kad gauti tūrį keturiuose oktantuose, reikia padauginti iš keturių, tuomet tūris bus lygus
V
4
=
4
⋅
2
3
=
8
3
=
2.66666667.
{\displaystyle V_{4}=4\cdot {\frac {2}{3}}={\frac {8}{3}}=2.66666667.}
Trilypis integralas cilindrinėje koordinačių sistemoje[ keisti ]
Su stačiakampėmis Dekarto koordinatėmis cilindrines koordinates sieja formulės
x
=
ρ
cos
ϕ
,
y
=
ρ
sin
ϕ
,
z
=
z
.
{\displaystyle x=\rho \cos \phi ,\;y=\rho \sin \phi ,\;z=z.}
∭
V
f
(
x
,
y
,
z
)
d
x
d
y
d
z
=
∭
V
f
(
ρ
cos
ϕ
,
ρ
sin
ϕ
,
z
)
ρ
d
ρ
d
ϕ
d
z
.
{\displaystyle \iiint _{V}f(x,y,z)dxdydz=\iiint _{V}f(\rho \cos \phi ,\rho \sin \phi ,z)\rho d\rho d\phi dz.}
Kadangi kūno tūris
V
=
∭
V
d
x
d
y
d
z
,
{\displaystyle V=\iiint _{V}dxdydz,}
tai cilindrinėje koordinačių sistemoje jis išreiškiamas formule
V
=
∭
V
ρ
d
ρ
d
ϕ
d
z
.
{\displaystyle V=\iiint _{V}\rho d\rho d\phi dz.}
Kūną V riboja paviršiai
x
2
+
y
2
=
x
,
{\displaystyle x^{2}+y^{2}=x,}
x
2
+
y
2
=
2
x
,
{\displaystyle x^{2}+y^{2}=2x,}
z
=
4
−
x
2
+
y
2
,
{\displaystyle z=4-{\sqrt {x^{2}+y^{2}}},}
z=0. Apskaičiuokime to kūno tūrį.
Kūnas V iš šonų apribotas dviejų cilindrų, kurių sudaromosios lygiagrečios ašiai Oz , o vedamosios - apskritimai
x
2
+
y
2
=
x
{\displaystyle x^{2}+y^{2}=x}
ir
x
2
+
y
2
=
2
x
.
{\displaystyle x^{2}+y^{2}=2x.}
Iš apačios kūną riboja plokštuma z=0, iš viršaus - kūgis
z
=
4
−
x
2
+
y
2
,
{\displaystyle z=4-{\sqrt {x^{2}+y^{2}}},}
kurio viršūnė yra taške (0; 0; 4) o sudaromosios nukreiptos žemyn. Kadangi kūnas yra simetriškas plokštumos xOy atžvilgiu, tai apskaičiuosime
1
2
{\displaystyle {1 \over 2}}
to kūno tūrio. Integravimo sritis D , t. y. kūno prjokecija plokštumoje xOy .
Cilindrinėje koordinačių sistemoje apskritimų lygtys yra
ρ
=
cos
ϕ
{\displaystyle \rho =\cos \phi }
ir
ρ
=
2
cos
ϕ
,
{\displaystyle \rho =2\cos \phi ,}
o kūgio lygtis yra
z
=
4
−
ρ
.
{\displaystyle z=4-\rho .}
Figūra D gaunama, kai kampas
ϕ
{\displaystyle \phi }
kinta nuo 0 iki
π
2
,
{\displaystyle {\pi \over 2},}
o dydis
ρ
{\displaystyle \rho }
- nuo
cos
ϕ
{\displaystyle \cos \phi }
iki
2
cos
ϕ
.
{\displaystyle 2\cos \phi .}
Todėl, pritaikę formulę, gauname
V
=
2
∫
0
π
2
d
ϕ
∫
cos
ϕ
2
cos
ϕ
ρ
d
ρ
∫
0
4
−
ρ
d
z
=
2
∫
0
π
2
d
ϕ
∫
cos
ϕ
2
cos
ϕ
ρ
z
|
0
4
−
ρ
d
ρ
=
2
∫
0
π
2
d
ϕ
∫
cos
ϕ
2
cos
ϕ
(
4
ρ
−
ρ
2
)
d
ρ
=
{\displaystyle V=2\int _{0}^{\pi \over 2}d\phi \int _{\cos \phi }^{2\cos \phi }\rho d\rho \int _{0}^{4-\rho }dz=2\int _{0}^{\pi \over 2}d\phi \int _{\cos \phi }^{2\cos \phi }\rho z|_{0}^{4-\rho }d\rho =2\int _{0}^{\pi \over 2}d\phi \int _{\cos \phi }^{2\cos \phi }(4\rho -\rho ^{2})d\rho =}
=
2
∫
0
π
2
(
2
ρ
2
−
ρ
3
3
)
|
cos
ϕ
2
cos
ϕ
d
ϕ
=
2
∫
0
π
2
(
6
cos
2
ϕ
−
7
3
cos
3
ϕ
)
d
ϕ
=
12
⋅
1
2
⋅
π
2
−
14
3
⋅
2
!
!
3
!
!
=
3
π
−
28
9
.
{\displaystyle =2\int _{0}^{\pi \over 2}(2\rho ^{2}-{\rho ^{3} \over 3})|_{\cos \phi }^{2\cos \phi }d\phi =2\int _{0}^{\pi \over 2}(6\cos ^{2}\phi -{7 \over 3}\cos ^{3}\phi )d\phi =12\cdot {1 \over 2}\cdot {\pi \over 2}-{14 \over 3}\cdot {2!! \over 3!!}=3\pi -{28 \over 9}.}
Kur du šauktukai dvigubas faktorialas .
Kūną V riboja viršutinė sferos
z
=
6
−
x
2
−
y
2
{\displaystyle z={\sqrt {6-x^{2}-y^{2}}}}
dalis ir paraboloidas
z
=
x
2
+
y
2
.
{\displaystyle z=x^{2}+y^{2}.}
Apskaičiuokime kūno tūrį.
Kadangi kūnas yra simteriškas plokštumų xOz ir yOz atžvilgiu, tai apskaičiuosime
1
4
{\displaystyle {1 \over 4}}
jo tūrio. Norėdami rasti sritį D , turime suprojektuoti į plokštumą xOy sferos paraboloido susikirtimo kreivę, kurios lygtį gausime išsprendę jų lygčių sistemą. Į lygtį
z
=
6
−
x
2
−
y
2
{\displaystyle z={\sqrt {6-x^{2}-y^{2}}}}
vietoje z įrašome reiškinį
x
2
+
y
2
.
{\displaystyle x^{2}+y^{2}.}
Gauname lygtį
x
2
+
y
2
=
6
−
x
2
−
y
2
;
{\displaystyle x^{2}+y^{2}={\sqrt {6-x^{2}-y^{2}}};}
r
=
6
−
r
;
{\displaystyle r={\sqrt {6-r}};}
r
2
+
r
−
6
=
0.
{\displaystyle r^{2}+r-6=0.}
D
=
b
2
−
4
a
c
=
1
−
4
(
−
6
)
=
25
;
{\displaystyle D=b^{2}-4ac=1-4(-6)=25;}
r
1
;
2
=
−
b
±
D
2
a
=
−
1
±
5
2
=
−
3
;
2.
{\displaystyle r_{1;2}={-b\pm {\sqrt {D}} \over 2a}={-1\pm 5 \over 2}=-3;2.}
Iš čia
r
=
ρ
2
=
x
2
+
y
2
=
2.
{\displaystyle r=\rho ^{2}=x^{2}+y^{2}=2.}
Šiuo atveju r yra susikirtimo parabaloido ir pusapskritimo koordinate z , o kadangi parabolės projekcija į plokštumą xOz yra nusakoma formule
z
=
x
2
,
{\displaystyle z=x^{2},}
tai, kai
z
=
r
=
2
=
x
2
{\displaystyle z=r=2=x^{2}}
(arba
z
=
r
=
2
=
y
2
{\displaystyle z=r=2=y^{2}}
), tada
x
=
r
=
2
,
{\displaystyle x={\sqrt {r}}={\sqrt {2}},}
kaip parodyta paveiksliuke.
Taigi viso kūno tūris
V
=
4
∫
0
π
2
d
ϕ
∫
0
2
ρ
d
ρ
∫
ρ
2
6
−
ρ
d
z
=
4
∫
0
π
2
d
ϕ
∫
0
2
(
ρ
6
−
ρ
2
−
ρ
3
)
d
ρ
=
{\displaystyle V=4\int _{0}^{\pi \over 2}d\phi \int _{0}^{\sqrt {2}}\rho d\rho \int _{\rho ^{2}}^{\sqrt {6-\rho }}dz=4\int _{0}^{\pi \over 2}d\phi \int _{0}^{\sqrt {2}}(\rho {\sqrt {6-\rho ^{2}}}-\rho ^{3})d\rho =}
=
4
∫
0
π
2
(
−
(
6
−
ρ
2
)
3
3
−
ρ
4
4
)
|
0
2
d
ϕ
=
−
4
∫
0
π
2
[
(
(
6
−
2
)
3
3
+
4
4
)
−
(
(
6
−
0
)
3
3
+
0
4
)
]
d
ϕ
=
{\displaystyle =4\int _{0}^{\pi \over 2}(-{{\sqrt {(6-\rho ^{2})^{3}}} \over 3}-{\rho ^{4} \over 4})|_{0}^{\sqrt {2}}d\phi =-4\int _{0}^{\pi \over 2}[({{\sqrt {(6-2)^{3}}} \over 3}+{4 \over 4})-({{\sqrt {(6-0)^{3}}} \over 3}+{0 \over 4})]d\phi =}
=
−
4
∫
0
π
2
[
(
64
3
+
1
)
−
216
3
]
d
ϕ
=
−
4
∫
0
π
2
[
(
8
3
+
1
)
−
6
6
3
]
d
ϕ
=
4
3
(
6
6
−
11
)
∫
0
π
2
d
ϕ
=
4
π
6
(
6
6
−
11
)
.
{\displaystyle =-4\int _{0}^{\pi \over 2}[({{\sqrt {64}} \over 3}+1)-{{\sqrt {216}} \over 3}]d\phi =-4\int _{0}^{\pi \over 2}[({8 \over 3}+1)-{6{\sqrt {6}} \over 3}]d\phi ={4 \over 3}(6{\sqrt {6}}-11)\int _{0}^{\pi \over 2}d\phi ={4\pi \over 6}(6{\sqrt {6}}-11).}
Integralas integruojamas taip:
∫
ρ
6
−
ρ
2
d
ρ
=
∫
ρ
6
−
ρ
2
d
(
6
−
ρ
2
)
−
2
ρ
=
−
1
2
∫
6
−
ρ
2
d
(
6
−
ρ
2
)
=
−
1
2
⋅
(
6
−
ρ
2
)
1
2
+
1
1
2
+
1
+
C
=
{\displaystyle \int \rho {\sqrt {6-\rho ^{2}}}d\rho =\int \rho {\sqrt {6-\rho ^{2}}}{d(6-\rho ^{2}) \over -2\rho }=-{1 \over 2}\int {\sqrt {6-\rho ^{2}}}d(6-\rho ^{2})=-{1 \over 2}\cdot {(6-\rho ^{2})^{{1 \over 2}+1} \over {1 \over 2}+1}+C=}
=
−
1
2
⋅
(
6
−
ρ
2
)
3
2
3
2
+
C
=
−
(
6
−
ρ
2
)
3
3
+
C
,
{\displaystyle =-{1 \over 2}\cdot {(6-\rho ^{2})^{3 \over 2} \over {3 \over 2}}+C=-{{\sqrt {(6-\rho ^{2})^{3}}} \over 3}+C,}
nes
d
(
6
−
ρ
2
)
=
−
2
ρ
d
ρ
,
{\displaystyle d(6-\rho ^{2})=-2\rho d\rho ,}
todėl
d
ρ
=
d
(
6
−
ρ
2
)
−
2
ρ
.
{\displaystyle d\rho ={d(6-\rho ^{2}) \over -2\rho }.}
Apskaičiuosime tūrį kūno V , apriboto paviršiais
x
2
+
y
2
=
z
,
{\displaystyle x^{2}+y^{2}=z,}
z=1, cilindrinėse koordinatėse. Tai yra paraboloidas iš viršaus apribotas plokštuma z=1. Pažymėsime per T erdvės sritį
ρ
ϕ
z
,
{\displaystyle \rho \phi z,}
apribota paviršiais
ρ
2
=
z
,
{\displaystyle \rho ^{2}=z,}
z
=
1
,
{\displaystyle z=1,}
ϕ
=
0
,
{\displaystyle \phi =0,}
ϕ
=
2
π
.
{\displaystyle \phi =2\pi .}
Todėl
V
=
∭
V
d
x
d
y
d
z
=
∭
T
ρ
d
ρ
d
ϕ
d
z
=
∫
0
2
π
d
ϕ
∫
0
1
ρ
d
ρ
∫
ρ
2
1
d
z
=
{\displaystyle V=\iiint _{V}dxdydz=\iiint _{T}\rho d\rho d\phi dz=\int _{0}^{2\pi }d\phi \int _{0}^{1}\rho d\rho \int _{\rho ^{2}}^{1}dz=}
=
∫
0
2
π
d
ϕ
∫
0
1
ρ
(
1
−
ρ
2
)
d
ρ
=
∫
0
2
π
(
ρ
2
2
−
ρ
4
4
)
|
0
1
d
ϕ
=
∫
0
2
π
1
4
d
ϕ
=
π
2
.
{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{1}\rho (1-\rho ^{2})d\rho =\int _{0}^{2\pi }({\rho ^{2} \over 2}-{\rho ^{4} \over 4})|_{0}^{1}d\phi =\int _{0}^{2\pi }{1 \over 4}d\phi ={\pi \over 2}.}
Apskaičiuosime tūrį kūno V , apriboto paviršiais
x
2
+
y
2
=
z
,
{\displaystyle x^{2}+y^{2}=z,}
z=100, cilindrinėse koordinatėse. Tai yra paraboloidas iš viršaus apribotas plokštuma z=100. Pažymėsime per T erdvės sritį
ρ
ϕ
z
,
{\displaystyle \rho \phi z,}
apribota paviršiais
ρ
2
=
z
,
{\displaystyle \rho ^{2}=z,}
z
=
100
,
{\displaystyle z=100,}
ϕ
=
0
,
{\displaystyle \phi =0,}
ϕ
=
2
π
.
{\displaystyle \phi =2\pi .}
Maksimalus spindulys
ρ
=
10
{\displaystyle \rho =10}
. Todėl
V
=
∭
V
d
x
d
y
d
z
=
∭
T
ρ
d
ρ
d
ϕ
d
z
=
∫
0
2
π
d
ϕ
∫
0
10
ρ
d
ρ
∫
ρ
2
100
d
z
=
{\displaystyle V=\iiint _{V}dxdydz=\iiint _{T}\rho d\rho d\phi dz=\int _{0}^{2\pi }d\phi \int _{0}^{10}\rho d\rho \int _{\rho ^{2}}^{100}dz=}
=
∫
0
2
π
d
ϕ
∫
0
10
ρ
(
100
−
ρ
2
)
d
ρ
=
∫
0
2
π
(
100
ρ
2
2
−
ρ
4
4
)
|
0
10
d
ϕ
=
∫
0
2
π
(
50
⋅
10
2
−
10
4
4
)
d
ϕ
=
{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{10}\rho (100-\rho ^{2})d\rho =\int _{0}^{2\pi }({100\rho ^{2} \over 2}-{\rho ^{4} \over 4})|_{0}^{10}d\phi =\int _{0}^{2\pi }(50\cdot 10^{2}-{10^{4} \over 4})d\phi =}
=
∫
0
2
π
(
5000
−
2500
)
d
ϕ
=
2500
∫
0
2
π
d
ϕ
=
5000
π
.
{\displaystyle =\int _{0}^{2\pi }(5000-2500)d\phi =2500\int _{0}^{2\pi }d\phi =5000\pi .}
Pavyzdis . Apskaičiuoti integralą
I
=
∫
V
d
V
,
{\displaystyle I=\int _{V}\;{\mathsf {d}}V,}
paplitusi per tūrį, apribotą plokštumomis xOy ir xOz , cilindru
x
2
+
y
2
=
a
x
{\displaystyle x^{2}+y^{2}=ax}
ir sfera
x
2
+
y
2
+
z
2
=
a
2
.
{\displaystyle x^{2}+y^{2}+z^{2}=a^{2}.}
Kadangi
f
=
1
{\displaystyle f=1}
, integralas skaičiavimu lygus tūriui duoto kūno. Trumpiau tariant, rasime tūrį kūno apriboto išvardintų paviršių.
Sprendimas. Pereidami į cilindrinę koordinačių sistemą, gauname
z
1
=
0
{\displaystyle z_{1}=0}
,
z
2
=
a
2
−
x
2
−
y
2
=
a
2
−
ρ
2
;
{\displaystyle z_{2}={\sqrt {a^{2}-x^{2}-y^{2}}}={\sqrt {a^{2}-\rho ^{2}}};}
ρ
1
=
0
,
{\displaystyle \rho _{1}=0,}
ρ
2
=
a
cos
ϕ
,
{\displaystyle \rho _{2}=a\cos \phi ,}
nes
x
2
+
y
2
=
ρ
2
,
{\displaystyle x^{2}+y^{2}=\rho ^{2},}
o
a
x
=
a
ρ
cos
ϕ
;
{\displaystyle ax=a\rho \cos \phi ;}
ϕ
1
=
0
,
{\displaystyle \phi _{1}=0,}
ϕ
2
=
π
2
.
{\displaystyle \phi _{2}={\frac {\pi }{2}}.}
Randame kūno tūrį:
V
=
I
=
∫
0
π
2
∫
0
a
cos
ϕ
∫
0
a
2
−
ρ
2
ρ
d
z
d
ρ
d
ϕ
=
{\displaystyle V=I=\int _{0}^{\pi \over 2}\int _{0}^{a\cos \phi }\int _{0}^{\sqrt {a^{2}-\rho ^{2}}}\rho {\mathsf {d}}z{\mathsf {d}}\rho {\mathsf {d}}\phi =}
=
∫
0
π
2
[
∫
0
a
cos
ϕ
(
∫
0
a
2
−
ρ
2
ρ
d
z
)
d
ρ
]
d
ϕ
=
∫
0
π
2
[
∫
0
a
cos
ϕ
(
z
|
0
a
2
−
ρ
2
ρ
)
d
ρ
]
d
ϕ
=
∫
0
π
2
[
∫
0
a
cos
ϕ
(
(
a
2
−
ρ
2
−
0
)
ρ
)
d
ρ
]
d
ϕ
=
{\displaystyle =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\left(\int _{0}^{\sqrt {a^{2}-\rho ^{2}}}\rho {\mathsf {d}}z\right){\mathsf {d}}\rho \right]{\mathsf {d}}\phi =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\left(z|_{0}^{\sqrt {a^{2}-\rho ^{2}}}\rho \right){\mathsf {d}}\rho \right]{\mathsf {d}}\phi =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\left(({\sqrt {a^{2}-\rho ^{2}}}-0)\rho \right){\mathsf {d}}\rho \right]{\mathsf {d}}\phi =}
=
∫
0
π
2
[
∫
0
a
cos
ϕ
ρ
a
2
−
ρ
2
d
ρ
]
d
ϕ
=
∫
0
π
2
[
∫
0
a
cos
ϕ
ρ
a
2
−
ρ
2
d
(
a
2
−
ρ
2
)
−
2
ρ
]
d
ϕ
=
−
1
2
∫
0
π
2
[
∫
0
a
cos
ϕ
a
2
−
ρ
2
d
(
a
2
−
ρ
2
)
]
d
ϕ
=
{\displaystyle =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\rho {\sqrt {a^{2}-\rho ^{2}}}\,{\mathsf {d}}\rho \right]{\mathsf {d}}\phi =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\rho {\sqrt {a^{2}-\rho ^{2}}}\,{\frac {{\mathsf {d}}(a^{2}-\rho ^{2})}{-2\rho }}\right]{\mathsf {d}}\phi =-{\frac {1}{2}}\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }{\sqrt {a^{2}-\rho ^{2}}}\,{\mathsf {d}}(a^{2}-\rho ^{2})\right]{\mathsf {d}}\phi =}
=
−
1
2
∫
0
π
2
[
(
a
2
−
ρ
2
)
3
2
3
2
|
0
a
cos
ϕ
]
d
ϕ
=
−
1
2
⋅
2
3
⋅
∫
0
π
2
[
(
a
2
−
(
a
cos
ϕ
)
2
)
3
−
(
a
2
−
0
2
)
3
]
d
ϕ
=
−
1
3
∫
0
π
2
[
(
a
2
(
1
−
cos
2
ϕ
)
)
3
−
a
6
]
d
ϕ
=
{\displaystyle =-{\frac {1}{2}}\int _{0}^{\pi \over 2}\left[{\frac {(a^{2}-\rho ^{2})^{3 \over 2}}{3 \over 2}}|_{0}^{a\cos \phi }\right]{\mathsf {d}}\phi =-{\frac {1}{2}}\cdot {\frac {2}{3}}\cdot \int _{0}^{\pi \over 2}\left[{\sqrt {(a^{2}-(a\cos \phi )^{2})^{3}}}-{\sqrt {(a^{2}-0^{2})^{3}}}\right]{\mathsf {d}}\phi =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[{\sqrt {(a^{2}(1-\cos ^{2}\phi ))^{3}}}-{\sqrt {a^{6}}}\right]{\mathsf {d}}\phi =}
=
−
1
3
∫
0
π
2
[
a
6
(
1
−
cos
2
ϕ
)
3
−
a
3
]
d
ϕ
=
−
1
3
∫
0
π
2
[
a
3
(
sin
2
ϕ
)
3
−
a
3
]
d
ϕ
=
−
1
3
∫
0
π
2
[
a
3
sin
3
ϕ
−
a
3
]
d
ϕ
=
−
a
3
3
∫
0
π
2
[
sin
3
ϕ
−
1
]
d
ϕ
=
{\displaystyle =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[{\sqrt {a^{6}(1-\cos ^{2}\phi )^{3}}}-a^{3}\right]{\mathsf {d}}\phi =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[a^{3}{\sqrt {(\sin ^{2}\phi )^{3}}}-a^{3}\right]{\mathsf {d}}\phi =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[a^{3}\sin ^{3}\phi -a^{3}\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[\sin ^{3}\phi -1\right]{\mathsf {d}}\phi =}
=
−
a
3
3
∫
0
π
2
[
sin
3
ϕ
−
1
]
d
ϕ
=
−
a
3
3
∫
0
π
2
[
1
4
(
3
sin
ϕ
−
sin
(
3
ϕ
)
)
−
1
]
d
ϕ
=
−
a
3
12
∫
0
π
2
[
3
sin
ϕ
−
sin
(
3
ϕ
)
−
4
]
d
ϕ
=
{\displaystyle =-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[\sin ^{3}\phi -1\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[{\frac {1}{4}}(3\sin \phi -\sin(3\phi ))-1\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{12}}\int _{0}^{\pi \over 2}\left[3\sin \phi -\sin(3\phi )-4\right]{\mathsf {d}}\phi =}
=
−
a
3
12
[
−
3
cos
ϕ
+
cos
(
3
ϕ
)
3
−
4
ϕ
]
|
0
π
2
=
−
a
3
12
[
−
3
cos
π
2
+
cos
(
3
⋅
π
2
)
3
−
4
⋅
π
2
−
(
−
3
cos
0
+
cos
(
3
⋅
0
)
3
−
4
⋅
0
)
]
=
{\displaystyle =-{\frac {a^{3}}{12}}\left[-3\cos \phi +{\frac {\cos(3\phi )}{3}}-4\phi \right]|_{0}^{\pi \over 2}=-{\frac {a^{3}}{12}}\left[-3\cos {\frac {\pi }{2}}+{\frac {\cos(3\cdot {\frac {\pi }{2}})}{3}}-4\cdot {\frac {\pi }{2}}-(-3\cos 0+{\frac {\cos(3\cdot 0)}{3}}-4\cdot 0)\right]=}
=
−
a
3
12
[
−
3
⋅
0
+
0
3
−
2
π
−
(
−
3
⋅
1
+
1
3
)
]
=
−
a
3
12
[
−
2
π
+
3
−
1
3
]
=
−
a
3
12
⋅
−
6
π
+
9
−
1
3
=
−
a
3
12
⋅
−
6
π
+
8
3
=
{\displaystyle =-{\frac {a^{3}}{12}}\left[-3\cdot 0+{\frac {0}{3}}-2\pi -(-3\cdot 1+{\frac {1}{3}})\right]=-{\frac {a^{3}}{12}}\left[-2\pi +3-{\frac {1}{3}}\right]=-{\frac {a^{3}}{12}}\cdot {\frac {-6\pi +9-1}{3}}=-{\frac {a^{3}}{12}}\cdot {\frac {-6\pi +8}{3}}=}
=
a
3
12
⋅
6
π
−
8
3
=
a
3
(
6
π
−
8
)
36
=
a
3
(
3
π
−
4
)
18
=
0.301376553
⋅
a
3
.
{\displaystyle ={\frac {a^{3}}{12}}\cdot {\frac {6\pi -8}{3}}={\frac {a^{3}(6\pi -8)}{36}}={\frac {a^{3}(3\pi -4)}{18}}=0.301376553\cdot a^{3}.}
čia
d
(
a
2
−
ρ
2
)
=
−
2
ρ
d
ρ
,
d
ρ
=
d
(
a
2
−
ρ
2
)
−
2
ρ
;
sin
(
3
A
)
=
3
sin
A
−
4
sin
3
A
,
sin
3
A
=
1
4
(
3
sin
A
−
sin
(
3
A
)
)
.
{\displaystyle {\mathsf {d}}(a^{2}-\rho ^{2})=-2\rho \,{\mathsf {d}}\rho ,\;\;{\mathsf {d}}\rho ={\frac {{\mathsf {d}}(a^{2}-\rho ^{2})}{-2\rho }};\quad \sin(3A)=3\sin A-4\sin ^{3}A,\;\;\sin ^{3}A={\frac {1}{4}}(3\sin A-\sin(3A)).}
Kai
a
=
3
{\displaystyle a=3}
, tada
V
=
0.301376553
⋅
a
3
=
0.301376553
⋅
3
3
=
8.137166941.
{\displaystyle V=0.301376553\cdot a^{3}=0.301376553\cdot 3^{3}=8.137166941.}
Kad įsivaizduoti kaip atrodo kūnas, galima pasakyti, kad sferos centras yra (0; 0; 0), o sferos sindulys
R
=
a
{\displaystyle R=a}
. Na, o cilindro pagrindas yra padalintas per pusę ašimi Ox . Cilindro [pagrindo] spindulys
r
=
a
2
{\displaystyle r={\frac {a}{2}}}
, o cilindro skersmuo
d
=
a
{\displaystyle d=a}
. Cilindro pagrindas yra tik ant ašies Ox ir vienas jo pagrindo kraštas liečiasi su koordinačiu pradžios tašku O , o kitas liečiasi su tašku a ant Ox ašies. Sfera, kurios lygtis, priminimui, yra
x
2
+
y
2
+
z
2
=
a
2
{\displaystyle x^{2}+y^{2}+z^{2}=a^{2}}
gaubia iš viršaus, o iš šono apriboja kūną cilindras.
Žinodami cilindro tūrio formulę
V
c
i
l
=
π
r
2
h
,
{\displaystyle V_{cil}=\pi r^{2}h\,,}
palyginsime ar gautas atsakymas neprasilenkia su elementaria logika. Mes surasime pusė cilindro tūrio, nes integravimas vyko pirmame oktante (oktantas yra 1/8 rutulio tūrio). Cilindro spindulys yra r=a/2=3/2=1.5, o cilindro aukštinė h=a=3. Randame palyginąmąjį tūrį:
V
≈
V
c
i
l
2
=
π
r
2
h
2
=
π
(
a
2
)
2
h
2
=
π
(
3
2
)
2
⋅
3
2
=
1
2
⋅
9
4
⋅
3
π
=
27
π
8
=
3.375
π
=
10.60287521.
{\displaystyle V\approx {\frac {V_{cil}}{2}}={\frac {\pi r^{2}h}{2}}={\frac {\pi ({\frac {a}{2}})^{2}h}{2}}={\frac {\pi ({\frac {3}{2}})^{2}\cdot 3}{2}}={\frac {1}{2}}\cdot {\frac {9}{4}}\cdot 3\pi ={\frac {27\pi }{8}}=3.375\pi =10.60287521.}
Pasinaudojant dvigubu faktorialu gauname tą patį atsakymą:
V
=
−
a
3
3
∫
0
π
2
[
sin
3
ϕ
−
1
]
d
ϕ
=
−
a
3
3
(
∫
0
π
2
sin
3
ϕ
d
ϕ
−
∫
0
π
2
d
ϕ
)
=
−
a
3
3
(
(
3
−
1
)
!
!
3
!
!
−
(
π
2
−
0
)
)
=
{\displaystyle V=-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[\sin ^{3}\phi -1\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{3}}(\int _{0}^{\pi \over 2}\sin ^{3}\phi \;{\mathsf {d}}\phi -\int _{0}^{\pi \over 2}{\mathsf {d}}\phi )=-{\frac {a^{3}}{3}}({\frac {(3-1)!!}{3!!}}-({\pi \over 2}-0))=}
=
−
a
3
3
(
2
!
!
3
!
!
−
π
2
)
=
−
a
3
3
(
2
3
−
π
2
)
=
a
3
3
(
π
2
−
2
3
)
=
3
3
3
(
π
2
−
2
3
)
=
9
⋅
0.90412966
=
8.13716694.
{\displaystyle =-{\frac {a^{3}}{3}}({\frac {2!!}{3!!}}-{\pi \over 2})=-{\frac {a^{3}}{3}}({\frac {2}{3}}-{\pi \over 2})={\frac {a^{3}}{3}}({\pi \over 2}-{\frac {2}{3}})={\frac {3^{3}}{3}}({\pi \over 2}-{\frac {2}{3}})=9\cdot 0.90412966=8.13716694.}
Pasinaudodami analitiniu mąstymu, pabandysime parodyti, kad tūris rastas teisingai. Apskritimo spindulys R=3, todėl ketvirtadalis skritulio ploto yra
S
s
k
r
=
π
R
2
4
=
π
⋅
3
2
4
=
9
π
4
=
2.25
π
=
7.068583471.
{\displaystyle S_{skr}={\frac {\pi R^{2}}{4}}={\frac {\pi \cdot 3^{2}}{4}}={\frac {9\pi }{4}}=2.25\pi =7.068583471.}
O kvadrato, kurio kraštinė a=3, plotas yra
S
k
v
=
a
2
=
3
2
=
9.
{\displaystyle S_{kv}=a^{2}=3^{2}=9.}
Dabar randame kvadrato ir 1/4 skritulio santykį:
S
k
v
S
s
k
r
=
9
9
4
⋅
π
=
4
π
=
1.273239545.
{\displaystyle {\frac {S_{kv}}{S_{skr}}}={\frac {9}{{\frac {9}{4}}\cdot \pi }}={\frac {4}{\pi }}=1.273239545.}
Akivaizdu, kad padalinus visą cilindro tūrį iš tūrio, kurį riboja cilindras ir sfera, turėtume gautį santykį didesnį nei kvadrato ir ketvirtadalio skritulio, o santykis yra:
V
c
i
l
2
V
=
10.60287521
8.137166941
=
1.303018027.
{\displaystyle {\frac {\frac {V_{cil}}{2}}{V}}={\frac {10.60287521}{8.137166941}}=1.303018027.}
Taip ir yra, tolstant nuo Ox ašies, z reikšmės mažėja, kas ir užtikrina didesnį santykį.
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
y
{\displaystyle x^{2}+y^{2}=y}
(apskritimas ant plokštumos xOy , kurio centro koordinatės (0; 0.5), o spindulys r=1/2),
z
=
0
{\displaystyle z=0}
(plokštuma ant plokštumos xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
(paraboloidas).
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=\rho \sin \phi ,}
ρ
=
sin
ϕ
.
{\displaystyle \rho =\sin \phi .}
Paraboloido lygtis tampa tokia:
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
Apskaičiuosime kūno tūrį tik viename oktante, todėl
ϕ
{\displaystyle \phi }
kinta nuo 0 iki
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
sin
ϕ
d
ϕ
=
∫
0
π
2
(
sin
4
ϕ
4
−
0
4
4
)
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {\sin ^{4}\phi }{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =}
=
1
4
∫
0
π
2
sin
4
ϕ
d
ϕ
=
1
4
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
1
32
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle ={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi ={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {1}{32}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
1
32
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
1
32
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
1
32
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
1
32
⋅
3
π
2
=
3
π
64
=
0.147262155.
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {1}{32}}\cdot {\frac {3\pi }{2}}={\frac {3\pi }{64}}=0.147262155.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Pasinaudojant dvigubu faktorialu gauname tą patį atsakymą:
V
=
1
4
∫
0
π
2
sin
4
ϕ
d
ϕ
=
1
4
⋅
(
4
−
1
)
!
!
4
!
!
⋅
π
2
=
1
4
⋅
3
!
!
4
!
!
⋅
π
2
=
1
4
⋅
3
⋅
1
4
⋅
2
⋅
π
2
=
1
4
⋅
3
8
⋅
π
2
=
3
π
64
=
0.147262155.
{\displaystyle V={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi ={\frac {1}{4}}\cdot {(4-1)!! \over 4!!}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3!! \over 4!!}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3\cdot 1 \over 4\cdot 2}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3 \over 8}\cdot {\pi \over 2}={\frac {3\pi }{64}}=0.147262155.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
3
π
64
{\displaystyle {\frac {3\pi }{64}}}
padauginti iš 2.
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
x
{\displaystyle x^{2}+y^{2}=x}
(apskritimas ant plokštumos xOy , kurio centro koordinatės (0.5; 0), o spindulys r=1/2),
z
=
0
{\displaystyle z=0}
(plokštuma ant plokštumos xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
(paraboloidas).
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
ρ
cos
ϕ
,
{\displaystyle \rho ^{2}=\rho \cos \phi ,}
ρ
=
cos
ϕ
.
{\displaystyle \rho =\cos \phi .}
Paraboloido lygtis tampa tokia:
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
Apskaičiuosime kūno tūrį tik viename oktante, todėl
ϕ
{\displaystyle \phi }
kinta nuo 0 iki
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
cos
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
cos
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{\cos \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{\cos \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
cos
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
cos
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
cos
ϕ
d
ϕ
=
∫
0
π
2
(
cos
4
ϕ
4
−
0
4
4
)
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{\cos \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{\cos \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{\cos \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {\cos ^{4}\phi }{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =}
=
1
4
∫
0
π
2
cos
4
ϕ
d
ϕ
=
1
4
∫
0
π
2
cos
(
4
ϕ
)
+
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
1
32
(
1
4
⋅
sin
(
4
ϕ
)
+
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle ={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}\cos ^{4}\phi {\mathsf {d}}\phi ={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )+4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {1}{32}}({\frac {1}{4}}\cdot \sin(4\phi )+2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
1
32
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
+
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
+
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})+2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)+2\sin(2\cdot 0)+3\cdot 0)]=}
=
1
32
[
(
1
4
⋅
sin
(
2
π
)
+
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
+
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(2\pi )+2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)+2\sin(0))]=}
=
1
32
[
(
1
4
⋅
0
+
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
+
2
⋅
0
)
]
=
1
32
⋅
3
π
2
=
3
π
64
=
0.147262155.
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot 0+2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0+2\cdot 0)]={\frac {1}{32}}\cdot {\frac {3\pi }{2}}={\frac {3\pi }{64}}=0.147262155.}
kur
cos
4
A
=
cos
(
4
A
)
+
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \cos ^{4}A={\cos(4A)+4\cos(2A)+3 \over 8}.}
Pasinaudojant dvigubu faktorialu gauname tą patį atsakymą:
V
=
1
4
∫
0
π
2
cos
4
ϕ
d
ϕ
=
1
4
⋅
(
4
−
1
)
!
!
4
!
!
⋅
π
2
=
1
4
⋅
3
!
!
4
!
!
⋅
π
2
=
1
4
⋅
3
⋅
1
4
⋅
2
⋅
π
2
=
1
4
⋅
3
8
⋅
π
2
=
3
π
64
=
0.147262155.
{\displaystyle V={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}\cos ^{4}\phi {\mathsf {d}}\phi ={\frac {1}{4}}\cdot {(4-1)!! \over 4!!}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3!! \over 4!!}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3\cdot 1 \over 4\cdot 2}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3 \over 8}\cdot {\pi \over 2}={\frac {3\pi }{64}}=0.147262155.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
3
π
64
{\displaystyle {\frac {3\pi }{64}}}
padauginti iš 2.
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
2
y
{\displaystyle x^{2}+y^{2}=2y}
(apskritimas ant plokštumos xOy , kurio centro koordinatės (0; 1), o spindulys r=1),
z
=
0
{\displaystyle z=0}
(plokštuma ant plokštumos xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
(paraboloidas).
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
2
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=2\rho \sin \phi ,}
ρ
=
2
sin
ϕ
.
{\displaystyle \rho =2\sin \phi .}
Paraboloido lygtis tampa tokia:
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
Apskaičiuosime kūno tūrį tik viename oktante, todėl
ϕ
{\displaystyle \phi }
kinta nuo 0 iki
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
2
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
2
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{2\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{2\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
2
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
2
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
2
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
2
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
16
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{2\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{2\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{2\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(2\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {16\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
4
∫
0
π
2
sin
4
ϕ
d
ϕ
=
4
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
1
2
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle =4\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi =4\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {1}{2}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
1
2
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {1}{2}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
1
2
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {1}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
1
2
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
1
2
⋅
3
π
2
=
3
π
4
=
2.35619449.
{\displaystyle ={\frac {1}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {1}{2}}\cdot {\frac {3\pi }{2}}={\frac {3\pi }{4}}=2.35619449.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
3
π
4
{\displaystyle {\frac {3\pi }{4}}}
padauginti iš 2.
Palyginimui, cilindro tūris viename oktante, kurio spindulys r=1, aukštis
h
=
2
2
=
4
{\displaystyle h=2^{2}=4}
yra lygus
V
c
i
l
1
=
π
⋅
r
2
⋅
h
2
=
π
⋅
1
2
⋅
4
2
=
2
π
=
6.283185307.
{\displaystyle V_{cil1}={\frac {\pi \cdot r^{2}\cdot h}{2}}={\frac {\pi \cdot 1^{2}\cdot 4}{2}}=2\pi =6.283185307.}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
8
y
{\displaystyle x^{2}+y^{2}=8y}
(apskritimas ant plokštumos xOy , kurio centro koordinatės (0; 4), o spindulys r=4),
z
=
0
{\displaystyle z=0}
(plokštuma ant plokštumos xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
(paraboloidas).
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
8
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=8\rho \sin \phi ,}
ρ
=
8
sin
ϕ
.
{\displaystyle \rho =8\sin \phi .}
Paraboloido lygtis tampa tokia:
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
Apskaičiuosime kūno tūrį tik viename oktante, todėl
ϕ
{\displaystyle \phi }
kinta nuo 0 iki
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
8
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
8
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{8\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{8\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
8
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
8
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
8
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
8
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
4096
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{8\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{8\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{8\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(8\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {4096\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
1024
∫
0
π
2
sin
4
ϕ
d
ϕ
=
1024
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
128
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle =1024\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi =1024\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi =128({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
128
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle =128[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
128
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle =128[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
128
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
128
⋅
3
π
2
=
192
π
=
603.1857895.
{\displaystyle =128[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]=128\cdot {\frac {3\pi }{2}}=192\pi =603.1857895.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
192
π
{\displaystyle 192\pi }
padauginti iš 2.
Palyginimui, cilindro tūris viename oktante, kurio spindulys r=4, aukštis
h
=
8
2
=
64
{\displaystyle h=8^{2}=64}
yra lygus
V
c
i
l
1
=
π
⋅
r
2
⋅
h
2
=
π
⋅
4
2
⋅
64
2
=
π
⋅
16
⋅
64
2
=
π
⋅
16
⋅
32
=
512
π
=
1608.495439.
{\displaystyle V_{cil1}={\frac {\pi \cdot r^{2}\cdot h}{2}}={\frac {\pi \cdot 4^{2}\cdot 64}{2}}={\frac {\pi \cdot 16\cdot 64}{2}}=\pi \cdot 16\cdot 32=512\pi =1608.495439.}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
9
y
{\displaystyle x^{2}+y^{2}=9y}
(apskritimas ant plokštumos xOy , kurio centro koordinatės (0; 4.5), o spindulys r=9/2),
z
=
0
{\displaystyle z=0}
(plokštuma ant plokštumos xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
(paraboloidas).
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
9
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=9\rho \sin \phi ,}
ρ
=
9
sin
ϕ
.
{\displaystyle \rho =9\sin \phi .}
Paraboloido lygtis tampa tokia:
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
Apskaičiuosime kūno tūrį tik viename oktante, todėl
ϕ
{\displaystyle \phi }
kinta nuo 0 iki
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
9
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
9
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{9\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{9\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
9
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
9
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
9
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
9
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
6561
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{9\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{9\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{9\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(9\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {6561\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
6561
4
∫
0
π
2
sin
4
ϕ
d
ϕ
=
6561
4
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
6561
32
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle ={\frac {6561}{4}}\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi ={\frac {6561}{4}}\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {6561}{32}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
6561
32
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {6561}{32}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
6561
32
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {6561}{32}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
6561
32
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
6561
32
⋅
3
π
2
=
19683
π
64
=
307.546875
π
=
966.1870031.
{\displaystyle ={\frac {6561}{32}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {6561}{32}}\cdot {\frac {3\pi }{2}}={\frac {19683\pi }{64}}=307.546875\pi =966.1870031.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
19683
π
64
{\displaystyle {\frac {19683\pi }{64}}}
padauginti iš 2.
Palyginimui, cilindro tūris viename oktante, kurio spindulys r=9/2, aukštis
h
=
9
2
=
81
{\displaystyle h=9^{2}=81}
yra lygus
V
c
i
l
1
=
π
⋅
r
2
⋅
h
2
=
π
⋅
(
9
2
)
2
⋅
81
2
=
π
⋅
81
4
⋅
81
2
=
π
⋅
81
⋅
81
4
⋅
2
=
6561
π
8
=
820.125
π
=
2576.498675.
{\displaystyle V_{cil1}={\frac {\pi \cdot r^{2}\cdot h}{2}}={\frac {\pi \cdot \left({\frac {9}{2}}\right)^{2}\cdot 81}{2}}={\frac {\pi \cdot {\frac {81}{4}}\cdot 81}{2}}={\frac {\pi \cdot 81\cdot 81}{4\cdot 2}}={\frac {6561\pi }{8}}=820.125\pi =2576.498675.}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
10
y
{\displaystyle x^{2}+y^{2}=10y}
(apskritimas ant plokštumos xOy , kurio centro koordinatės (0; 5), o spindulys r=5),
z
=
0
{\displaystyle z=0}
(plokštuma ant plokštumos xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
(paraboloidas).
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
10
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=10\rho \sin \phi ,}
ρ
=
10
sin
ϕ
.
{\displaystyle \rho =10\sin \phi .}
Paraboloido lygtis tampa tokia:
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
Apskaičiuosime kūno tūrį tik viename oktante, todėl
ϕ
{\displaystyle \phi }
kinta nuo 0 iki
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
10
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
10
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{10\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{10\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
10
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
10
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
10
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
10
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
10000
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{10\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{10\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{10\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(10\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {10000\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
2500
∫
0
π
2
sin
4
ϕ
d
ϕ
=
2500
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
625
2
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle =2500\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi =2500\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {625}{2}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
625
2
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {625}{2}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
625
2
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {625}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
625
2
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
625
2
⋅
3
π
2
=
1875
π
4
=
468.75
π
=
1472.621556.
{\displaystyle ={\frac {625}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {625}{2}}\cdot {\frac {3\pi }{2}}={\frac {1875\pi }{4}}=468.75\pi =1472.621556.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
1875
π
4
{\displaystyle {\frac {1875\pi }{4}}}
padauginti iš 2.
Palyginimui, cilindro tūris viename oktante, kurio spindulys r=5, aukštis
h
=
10
2
=
100
{\displaystyle h=10^{2}=100}
yra lygus
V
c
i
l
1
=
π
⋅
r
2
⋅
h
2
=
π
⋅
5
2
⋅
100
2
=
π
⋅
25
⋅
100
2
=
π
⋅
25
⋅
50
=
1250
π
=
3926.990817.
{\displaystyle V_{cil1}={\frac {\pi \cdot r^{2}\cdot h}{2}}={\frac {\pi \cdot 5^{2}\cdot 100}{2}}={\frac {\pi \cdot 25\cdot 100}{2}}=\pi \cdot 25\cdot 50=1250\pi =3926.990817.}
Galime pabandyti suprasti ar integravimo budu gautas atsakymas yra teisingas. Kai
x
=
5
{\displaystyle x=5}
ir
y
=
5
{\displaystyle y=5}
, tuomet paraboloido z reikšmė lygi
z
=
x
2
+
y
2
=
5
2
+
5
2
=
25
+
25
=
50.
{\displaystyle z=x^{2}+y^{2}=5^{2}+5^{2}=25+25=50.}
O kai
x
=
0
{\displaystyle x=0}
,
y
=
10
{\displaystyle y=10}
, tuomet paraboloido z reikšmė yra
z
=
x
2
+
y
2
=
0
2
+
10
2
=
0
+
100
=
100.
{\displaystyle z=x^{2}+y^{2}=0^{2}+10^{2}=0+100=100.}
Vadinasi šonuose kažkaip negali būti daugiau, o tiktai didėjant y reikšmei, z reikšmė apskritimo srityje didėja kvadratu. O kai apskritimo srityje y reikšmė mažesnė už 10, tada ir z reikšmė visoje apskritimo (
x
2
+
y
2
=
10
y
{\displaystyle x^{2}+y^{2}=10y}
) srityje yra mažesnės už 100. Taip pat reikia nepamiršti, kad aukščiausiame taške (z=100, y=10, x=0), kur susikerta cilindras su praboloidu, tai nukirtus plokšuma z=100, paraboloido viršų, paraboloido spindulys yra r=10, o centro koordinatės (0; 0), tuo tarpu, apskritimo r=5, o centro koordinatės yra (0; 5). Todėl didesniame apskritime yra mažesnis apskritimas ir todėl to mažesnio apskritimo reikšmės x ir y niekada neduos didesnės z , reikšmės už tą atvejį, kai R=y=10. Cilindru iš praboloido iškerpamas tūris yra tik 2,66667 karto mažesnis už viso cilindro tūrį. Kitaip tariant, jei viso cilindro tūris yra 1, tai tūris, kurį gauname integravimo budu dviejuose oktantuose yra 0.375 visais atvejais. Dar palyginimui, plotas po parabolės
y
=
x
2
{\displaystyle y=x^{2}}
šaka visada lygus 1/3 ploto stačiakampio gretasienio
x
⋅
y
=
x
⋅
y
2
=
x
⋅
x
2
.
{\displaystyle x\cdot y=x\cdot y^{2}=x\cdot x^{2}.}
O tūris po paraboloidu
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
visada lygus 1/2 viso cilindro tūrio.
Dar pastebėjimas, kad z reikšmė yra didesnė, kai
x
=
5
{\displaystyle x=5}
,
y
=
5
{\displaystyle y=5}
, tada
z
=
x
2
+
y
2
=
5
2
+
5
2
=
50
{\displaystyle z=x^{2}+y^{2}=5^{2}+5^{2}=50}
, negu, kai
y
=
5
{\displaystyle y=5}
,
x
=
0
,
{\displaystyle x=0,}
ir tada
z
=
x
2
+
y
2
=
0
2
+
5
2
=
25
{\displaystyle z=x^{2}+y^{2}=0^{2}+5^{2}=25}
. Todėl ant kraštų apskritimo, kurį dalina pusiau Oy ašis, dominuoja didesnės z reikšmės, negu centre, tačiau didžiausia z reikšmė vis tiek, kai y=10, x=0.
Pereidami į polinę koordinačių sistemą rasime tūrį po paraboloidu
z
=
1
+
x
2
+
y
2
,
{\displaystyle z=1+x^{2}+y^{2},}
kurį riboja cilindrinis paviršius
x
2
+
y
2
=
1.
{\displaystyle x^{2}+y^{2}=1.}
V
=
∫
0
π
2
d
ϕ
∫
0
1
ρ
d
ρ
∫
0
1
+
ρ
2
d
z
=
∫
0
π
2
d
ϕ
∫
0
1
ρ
(
1
+
ρ
2
)
d
ρ
=
∫
0
π
2
d
ϕ
∫
0
1
(
ρ
+
ρ
3
)
d
ρ
=
∫
0
π
2
d
ϕ
(
ρ
2
2
+
ρ
4
4
)
|
0
1
=
{\displaystyle V=\int _{0}^{\pi \over 2}d\phi \int _{0}^{1}\rho d\rho \int _{0}^{1+\rho ^{2}}dz=\int _{0}^{\pi \over 2}d\phi \int _{0}^{1}\rho (1+\rho ^{2})d\rho =\int _{0}^{\pi \over 2}d\phi \int _{0}^{1}(\rho +\rho ^{3})d\rho =\int _{0}^{\pi \over 2}d\phi ({\frac {\rho ^{2}}{2}}+{\frac {\rho ^{4}}{4}})|_{0}^{1}=}
=
∫
0
π
2
(
1
2
2
+
1
4
4
)
d
ϕ
=
(
1
2
+
1
4
)
ϕ
|
0
π
2
=
2
+
1
4
ϕ
|
0
π
2
=
3
4
⋅
π
2
=
3
π
8
=
1.178097245.
{\displaystyle =\int _{0}^{\pi \over 2}({\frac {1^{2}}{2}}+{\frac {1^{4}}{4}})d\phi =({\frac {1}{2}}+{\frac {1}{4}})\phi |_{0}^{\pi \over 2}={\frac {2+1}{4}}\phi |_{0}^{\pi \over 2}={\frac {3}{4}}\cdot {\pi \over 2}={\frac {3\pi }{8}}=1.178097245.}
Šis tūris keturiuose oktantuose yra lygus
3
π
2
=
4.71238898.
{\displaystyle {\frac {3\pi }{2}}=4.71238898.}
Trilypio integralo taikymas mechanikoje [ keisti ]
Kūno masės centro koordinatės[ keisti ]
Kai tam tikros masės tankis lygus
γ
(
x
,
y
,
z
)
,
{\displaystyle \gamma (x,y,z),}
tai to kūno masės centro koordinatės apskaičiuojamos pagal formules
x
c
=
∭
V
x
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
∭
V
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
,
y
c
=
∭
V
y
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
∭
V
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
,
z
c
=
∭
V
z
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
∭
V
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
.
{\displaystyle x_{c}={\iiint _{V}x\gamma (x,y,z)dxdydz \over \iiint _{V}\gamma (x,y,z)dxdydz},\;y_{c}={\iiint _{V}y\gamma (x,y,z)dxdydz \over \iiint _{V}\gamma (x,y,z)dxdydz},\;z_{c}={\iiint _{V}z\gamma (x,y,z)dxdydz \over \iiint _{V}\gamma (x,y,z)dxdydz}.}
Pavyzdžiai
Kūną riboja paviršiai
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
ir
z
=
4.
{\displaystyle z=4.}
Apskaičiuokime to kūno masės centro koordinates, kai
γ
=
c
o
n
s
t
.
{\displaystyle \gamma =const.}
Kadangi kūnas simteriškas plokštumų xOy ir yOz atžvilgiu, tai
x
c
=
y
c
=
0.
{\displaystyle x_{c}=y_{c}=0.}
Rasime
z
c
{\displaystyle z_{c}}
koordinatę. Pagal sąlygą,
γ
=
c
o
n
s
t
,
{\displaystyle \gamma =const,}
todėl iš formulių išplaukia, kad
z
c
=
∭
V
z
d
z
d
y
d
z
∭
V
d
x
d
y
d
z
.
{\displaystyle z_{c}={\iiint _{V}zdzdydz \over \iiint _{V}dxdydz}.}
Integralus apskaičiuosime pakeisdami juos kartotiniais cilindrinėje koordinačių sistemoje.
z
c
=
∫
0
2
π
d
ϕ
∫
0
2
ρ
d
ρ
∫
ρ
2
4
z
d
z
∫
0
2
π
d
ϕ
∫
0
2
ρ
d
ρ
∫
ρ
2
4
d
z
=
1
2
∫
0
2
π
d
ϕ
∫
0
2
(
16
−
ρ
4
)
ρ
d
ρ
∫
0
2
π
d
ϕ
∫
0
2
(
4
−
ρ
2
)
ρ
d
ρ
=
{\displaystyle z_{c}={\int _{0}^{2\pi }d\phi \int _{0}^{2}\rho d\rho \int _{\rho ^{2}}^{4}z\;dz \over \int _{0}^{2\pi }d\phi \int _{0}^{2}\rho d\rho \int _{\rho ^{2}}^{4}dz}={{1 \over 2}\int _{0}^{2\pi }d\phi \int _{0}^{2}(16-\rho ^{4})\rho d\rho \over \int _{0}^{2\pi }d\phi \int _{0}^{2}(4-\rho ^{2})\rho d\rho }=}
=
1
2
∫
0
2
π
(
16
ρ
2
2
−
ρ
6
6
)
|
0
2
d
ϕ
∫
0
2
π
(
4
ρ
2
2
−
ρ
4
4
)
|
0
2
d
ϕ
=
1
2
∫
0
2
π
(
32
−
32
3
)
d
ϕ
∫
0
2
π
(
8
−
4
)
d
ϕ
=
1
2
⋅
64
3
⋅
2
π
4
⋅
2
π
=
8
3
.
{\displaystyle ={{1 \over 2}\int _{0}^{2\pi }(16{\rho ^{2} \over 2}-{\rho ^{6} \over 6})|_{0}^{2}d\phi \over \int _{0}^{2\pi }(4{\rho ^{2} \over 2}-{\rho ^{4} \over 4})|_{0}^{2}d\phi }={{1 \over 2}\int _{0}^{2\pi }(32-{32 \over 3})d\phi \over \int _{0}^{2\pi }(8-4)d\phi }={{1 \over 2}\cdot {64 \over 3}\cdot 2\pi \over 4\cdot 2\pi }={8 \over 3}.}
Kūno inercijos momentai[ keisti ]
Taško M (x; y; z), kurio masė m , inercijos momentai koordinačių plokštumų xOy , xOz ir yOz atžvilgiu išreiškiami formulėmis
I
x
O
y
=
z
2
m
,
{\displaystyle I_{xOy}=z^{2}m,}
I
x
O
z
=
y
2
m
,
{\displaystyle I_{xOz}=y^{2}m,}
I
y
O
z
=
x
2
m
,
{\displaystyle I_{yOz}=x^{2}m,}
ašių Ox , Oy , Oz atžvilgiu - formulėmis
I
x
x
=
(
y
2
+
z
2
)
m
,
{\displaystyle I_{xx}=(y^{2}+z^{2})m,}
I
y
y
=
(
x
2
+
z
2
)
m
,
{\displaystyle I_{yy}=(x^{2}+z^{2})m,}
I
z
z
=
(
x
2
+
y
2
)
m
,
{\displaystyle I_{zz}=(x^{2}+y^{2})m,}
koordinačių pradžios atžvilgiu - formule
I
0
=
(
x
2
+
y
2
+
z
2
)
m
.
{\displaystyle I_{0}=(x^{2}+y^{2}+z^{2})m.}
Kūno inercijos momentai išreiškiami atitinkamais trilypiais integralais. Pavyzdžiui, tam tikros masės kūno, kurio tankis
γ
(
x
,
y
,
z
)
,
{\displaystyle \gamma (x,y,z),}
inercijos momentas plokštumos xOy atžvilgiu apskaičiuojamas pagal formulę
I
x
O
y
=
∭
V
z
2
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
,
{\displaystyle I_{xOy}=\iiint _{V}z^{2}\gamma (x,y,z)dxdydz,}
ašies Oz atžvilgiu - pagal formulę
I
z
z
=
∭
V
(
x
2
+
y
2
)
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
{\displaystyle I_{zz}=\iiint _{V}(x^{2}+y^{2})\gamma (x,y,z)dxdydz}
ir t. t.
Pavyzdžiai
Apskaičiuokime kūno, kurį riboja paraboloidas
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
ir plokštuma
z
=
4
{\displaystyle z=4}
(žr. auksčiau pateiktą pavyzdį apie paraboloido masės centro skaičiavimą), inercijos momentą ašies, einančios per jo masės centrą statmenai to paraboloido sukimosi ašiai, atžvilgiu (
γ
=
1
{\displaystyle \gamma =1}
).
Koordinačių ašis parinkime taip, kad jų pradžios taškas sutaptų su paraboloido masės centru, o ašis Ox būtų statmena paraboloido sukimosi ašiai. Tuomet turėsime rasti
I
x
x
{\displaystyle I_{xx}}
(arba
I
y
y
{\displaystyle I_{yy}}
).
Paraboloido lygtis tokioje koordinačių sistemoje yra
z
+
8
3
=
x
2
+
y
2
,
{\displaystyle z+{8 \over 3}=x^{2}+y^{2},}
o jo projekcija plokštumoje xOy - sritis, apribota apskritimo
x
2
+
y
2
=
4.
{\displaystyle x^{2}+y^{2}=4.}
Taikome formulę
I
x
x
=
∭
V
(
y
2
+
z
2
)
d
x
d
y
d
z
.
{\displaystyle I_{xx}=\iiint _{V}(y^{2}+z^{2})dxdydz.}
Tuomet
J
x
x
=
∫
0
2
π
d
ϕ
∫
0
2
ρ
d
ρ
∫
ρ
2
−
8
/
3
4
/
3
(
ρ
2
sin
2
ϕ
+
z
2
)
d
z
=
∫
0
2
π
d
ϕ
∫
0
2
ρ
(
z
ρ
2
sin
2
ϕ
+
z
3
3
)
|
ρ
2
−
8
3
4
3
d
ρ
=
{\displaystyle J_{xx}=\int _{0}^{2\pi }d\phi \int _{0}^{2}\rho d\rho \int _{\rho ^{2}-8/3}^{4/3}(\rho ^{2}\sin ^{2}\phi +z^{2})dz=\int _{0}^{2\pi }d\phi \int _{0}^{2}\rho (z\rho ^{2}\sin ^{2}\phi +{z^{3} \over 3})|_{\rho ^{2}-{8 \over 3}}^{4 \over 3}d\rho =}
=
∫
0
2
π
d
ϕ
∫
0
2
[
(
4
ρ
3
−
ρ
5
)
sin
2
ϕ
+
64
9
ρ
−
ρ
7
3
+
8
ρ
5
3
−
64
9
ρ
3
]
d
ρ
=
{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{2}[(4\rho ^{3}-\rho ^{5})\sin ^{2}\phi +{64 \over 9}\rho -{\rho ^{7} \over 3}+{8\rho ^{5} \over 3}-{64 \over 9}\rho ^{3}]d\rho =}
=
∫
0
2
π
[
(
ρ
4
−
ρ
6
6
)
sin
2
ϕ
+
32
9
ρ
2
−
ρ
8
24
+
4
ρ
6
9
−
16
9
ρ
4
]
|
0
2
d
ϕ
=
{\displaystyle =\int _{0}^{2\pi }[(\rho ^{4}-{\rho ^{6} \over 6})\sin ^{2}\phi +{32 \over 9}\rho ^{2}-{\rho ^{8} \over 24}+{4\rho ^{6} \over 9}-{16 \over 9}\rho ^{4}]|_{0}^{2}d\phi =}
=
∫
0
2
π
(
16
3
sin
2
ϕ
+
128
9
−
64
6
+
256
9
−
256
9
)
d
ϕ
=
∫
0
2
π
(
16
3
sin
2
ϕ
+
32
9
)
d
ϕ
=
{\displaystyle =\int _{0}^{2\pi }({16 \over 3}\sin ^{2}\phi +{128 \over 9}-{64 \over 6}+{256 \over 9}-{256 \over 9})d\phi =\int _{0}^{2\pi }({16 \over 3}\sin ^{2}\phi +{32 \over 9})d\phi =}
=
16
3
∫
0
2
π
1
−
cos
(
2
ϕ
)
2
d
ϕ
+
64
π
9
=
16
π
3
−
8
3
∫
0
2
π
cos
(
2
ϕ
)
d
(
2
ϕ
)
2
+
64
π
9
=
112
π
9
−
4
3
sin
(
2
ϕ
)
|
0
2
π
=
112
π
9
.
{\displaystyle ={16 \over 3}\int _{0}^{2\pi }{1-\cos(2\phi ) \over 2}d\phi +{64\pi \over 9}={16\pi \over 3}-{8 \over 3}\int _{0}^{2\pi }\cos(2\phi ){d(2\phi ) \over 2}+{64\pi \over 9}={112\pi \over 9}-{4 \over 3}\sin(2\phi )|_{0}^{2\pi }={112\pi \over 9}.}
Apskaičiuosime kūno sritį V , kuri apribota paviršiais
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
ir
z
=
1
{\displaystyle z=1}
inercijos momentą Oz ašies atžvilgiu
∭
V
(
x
2
+
y
2
)
d
z
d
y
d
z
.
{\displaystyle \iiint _{V}(x^{2}+y^{2})dzdydz.}
Taip kaip V į plokštumą xOy projektuojasi į skritulį
x
2
+
y
2
≤
1
,
{\displaystyle x^{2}+y^{2}\leq 1,}
tai koordinatė
ϕ
{\displaystyle \phi }
kinta ribose 0 ir
2
π
{\displaystyle 2\pi }
, koordinatė
ρ
{\displaystyle \rho }
- nuo
ρ
=
0
{\displaystyle \rho =0}
iki
ρ
=
1
{\displaystyle \rho =1}
. Nuolatinei reikšmei
ρ
{\displaystyle \rho }
(
0
≤
ρ
≤
1
)
{\displaystyle (0\leq \rho \leq 1)}
erdvėje Oxyz atitinka cilindras
x
2
+
y
2
=
ρ
2
.
{\displaystyle x^{2}+y^{2}=\rho ^{2}.}
Apžiurinėdami susikirtimą šito cilindro su sritimi V , gauname kitimą koordinčių z nuo reikšmės taškams gulinčių ant paraboloido
z
=
x
2
+
y
2
,
{\displaystyle z=x^{2}+y^{2},}
iki reikšmių taškams, gulinčių ant plokštumos
z
=
1
{\displaystyle z=1}
, t. y. nuo
z
=
ρ
2
{\displaystyle z=\rho ^{2}}
iki
z
=
1.
{\displaystyle z=1.}
Pritaikę formulę turime
I
z
z
=
∭
V
(
x
2
+
y
2
)
d
x
d
y
d
z
=
∫
0
2
π
d
ϕ
∫
0
1
d
ρ
∫
ρ
2
1
ρ
2
⋅
ρ
d
z
=
{\displaystyle I_{zz}=\iiint _{V}(x^{2}+y^{2})dxdydz=\int _{0}^{2\pi }d\phi \int _{0}^{1}d\rho \int _{\rho ^{2}}^{1}\rho ^{2}\cdot \rho dz=}
=
∫
0
2
π
d
ϕ
∫
0
1
ρ
3
z
|
ρ
2
1
d
ρ
=
∫
0
2
π
(
ρ
4
4
−
ρ
6
6
)
|
0
1
d
ϕ
=
1
12
∫
0
2
π
d
ϕ
=
π
6
.
{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{1}\rho ^{3}z|_{\rho ^{2}}^{1}d\rho =\int _{0}^{2\pi }({\rho ^{4} \over 4}-{\rho ^{6} \over 6})|_{0}^{1}d\phi ={1 \over 12}\int _{0}^{2\pi }d\phi ={\pi \over 6}.}
Trilypis integralas sferinėse koordinatėse[ keisti ]
x
=
ρ
sin
θ
cos
ϕ
,
y
=
ρ
sin
θ
sin
ϕ
,
z
=
ρ
cos
θ
(
0
≤
ρ
<
∞
,
0
≤
ϕ
≤
2
π
,
0
≤
θ
≤
π
)
.
{\displaystyle x=\rho \sin \theta \cos \phi ,\;y=\rho \sin \theta \sin \phi ,\;z=\rho \cos \theta \;(0\leq \rho <\infty ,\;0\leq \phi \leq 2\pi ,\;0\leq \theta \leq \pi ).}
∭
V
f
(
x
,
y
,
z
)
d
x
d
y
d
z
=
∭
T
f
[
ρ
sin
θ
cos
ϕ
,
ρ
sin
θ
sin
ϕ
,
ρ
cos
θ
]
ρ
2
sin
θ
d
ρ
d
ϕ
d
θ
.
{\displaystyle \iiint _{V}f(x,y,z)dxdydz=\iiint _{T}f[\rho \sin \theta \cos \phi ,\;\rho \sin \theta \sin \phi ,\;\rho \cos \theta ]\rho ^{2}\sin \theta d\rho d\phi d\theta .}
x
2
+
y
2
+
z
2
=
ρ
2
sin
2
θ
cos
2
ϕ
+
ρ
2
sin
2
θ
sin
2
ϕ
+
ρ
2
cos
2
θ
=
ρ
2
sin
2
θ
+
ρ
2
cos
2
θ
=
ρ
2
.
{\displaystyle x^{2}+y^{2}+z^{2}=\rho ^{2}\sin ^{2}\theta \cos ^{2}\phi +\rho ^{2}\sin ^{2}\theta \sin ^{2}\phi +\rho ^{2}\cos ^{2}\theta =\rho ^{2}\sin ^{2}\theta +\rho ^{2}\cos ^{2}\theta =\rho ^{2}.}
Pavyzdžiai
Apskaičiuosime rutulio
x
2
+
y
2
+
z
2
≤
R
2
{\displaystyle x^{2}+y^{2}+z^{2}\leq R^{2}}
tūrį V :
V
=
∭
V
d
x
d
y
d
z
=
∭
T
ρ
2
sin
θ
d
ρ
d
θ
d
ϕ
=
∫
0
R
d
ρ
∫
0
π
d
θ
∫
0
2
π
ρ
2
sin
θ
d
ϕ
=
{\displaystyle V=\iiint _{V}dxdydz=\iiint _{T}\rho ^{2}\sin \theta d\rho d\theta d\phi =\int _{0}^{R}d\rho \int _{0}^{\pi }d\theta \int _{0}^{2\pi }\rho ^{2}\sin \theta d\phi =}
=
2
π
∫
0
R
ρ
2
d
ρ
∫
0
π
sin
θ
d
θ
=
−
2
π
∫
0
R
ρ
2
cos
θ
|
0
π
d
ρ
=
4
π
∫
0
R
ρ
2
d
ρ
=
4
π
ρ
3
3
|
0
R
=
4
π
R
3
3
.
{\displaystyle =2\pi \int _{0}^{R}\rho ^{2}d\rho \int _{0}^{\pi }\sin \theta d\theta =-2\pi \int _{0}^{R}\rho ^{2}\cos \theta |_{0}^{\pi }d\rho =4\pi \int _{0}^{R}\rho ^{2}d\rho =4\pi {\rho ^{3} \over 3}|_{0}^{R}={4\pi R^{3} \over 3}.}
Apskaičiuosime rutulio
x
2
+
y
2
+
z
2
≤
R
2
{\displaystyle x^{2}+y^{2}+z^{2}\leq R^{2}}
inercijos momentą koordinačių pradžios atžvilgiu. Kadangi
x
2
+
y
2
+
z
2
=
ρ
2
,
{\displaystyle x^{2}+y^{2}+z^{2}=\rho ^{2},}
gauname
I
0
=
∭
V
(
x
2
+
y
2
+
z
2
)
d
x
d
y
d
z
=
∭
T
ρ
2
ρ
2
sin
θ
d
ρ
d
θ
d
ϕ
=
∫
0
R
ρ
4
d
ρ
∫
0
π
sin
θ
d
θ
∫
0
2
π
d
ϕ
=
{\displaystyle I_{0}=\iiint _{V}(x^{2}+y^{2}+z^{2})dxdydz=\iiint _{T}\rho ^{2}\rho ^{2}\sin \theta d\rho d\theta d\phi =\int _{0}^{R}\rho ^{4}d\rho \int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =}
=
2
π
∫
0
R
ρ
4
d
ρ
∫
0
π
sin
θ
d
θ
=
2
π
∫
0
R
ρ
4
(
−
cos
θ
)
|
0
π
d
ρ
=
2
π
∫
0
R
ρ
4
(
1
+
1
)
d
ρ
=
4
π
ρ
5
5
|
0
R
=
4
π
R
5
5
.
{\displaystyle =2\pi \int _{0}^{R}\rho ^{4}d\rho \int _{0}^{\pi }\sin \theta d\theta =2\pi \int _{0}^{R}\rho ^{4}(-\cos \theta )|_{0}^{\pi }d\rho =2\pi \int _{0}^{R}\rho ^{4}(1+1)d\rho =4\pi {\rho ^{5} \over 5}|_{0}^{R}={4\pi R^{5} \over 5}.}
Nustatysime masės centro koordinates viršutinės pusės vienalyčio rutulio V spindulio R esančio centre koordinačių pradžios.
Duotas pusrutulis apribotas paviršiais
z
=
R
2
−
x
2
−
y
2
{\displaystyle z={\sqrt {R^{2}-x^{2}-y^{2}}}}
ir
z
=
0.
{\displaystyle z=0.}
Dėl pusrutulio simetrijos
x
c
=
y
c
=
0.
{\displaystyle x_{c}=y_{c}=0.}