Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
18
y
{\displaystyle x^{2}+y^{2}=18y}
xOy , kurio centro koordinatės (0; 9), o spindulys r=9),
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
18
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=18\rho \sin \phi ,}
ρ
=
18
sin
ϕ
.
{\displaystyle \rho =18\sin \phi .}
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
ϕ
{\displaystyle \phi }
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
18
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
18
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}^{18\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{18\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
18
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
18
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
18
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
18
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
104976
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{18\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{18\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{18\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(18\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {104976\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
26244
∫
0
π
2
sin
4
ϕ
d
ϕ
=
26244
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
6561
2
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle =26244\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi =26244\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {6561}{2}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
6561
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 {6561}{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)]=}
=
6561
2
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {6561}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
6561
2
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
6561
2
⋅
3
π
2
=
19683
π
4
=
15458.99205.
{\displaystyle ={\frac {6561}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {6561}{2}}\cdot {\frac {3\pi }{2}}={\frac {19683\pi }{4}}=15458.99205.}
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
4
{\displaystyle {\frac {19683\pi }{4}}}
Palyginimui cilindro viename oktante tūris yra
V
c
i
l
1
=
π
r
2
h
2
=
π
⋅
9
2
⋅
18
2
2
=
π
⋅
81
⋅
324
2
=
13122
π
=
41223.9788.
{\displaystyle V_{cil1}={\frac {\pi r^{2}h}{2}}={\frac {\pi \cdot 9^{2}\cdot 18^{2}}{2}}={\frac {\pi \cdot 81\cdot 324}{2}}=13122\pi =41223.9788.}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
14
y
{\displaystyle x^{2}+y^{2}=14y}
xOy , kurio centro koordinatės (0; 7), o spindulys r=7),
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
14
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=14\rho \sin \phi ,}
ρ
=
14
sin
ϕ
.
{\displaystyle \rho =14\sin \phi .}
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
ϕ
{\displaystyle \phi }
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
14
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
14
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}^{14\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{14\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
14
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
14
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
14
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
14
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
38416
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{14\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{14\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{14\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(14\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {38416\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
9604
∫
0
π
2
sin
4
ϕ
d
ϕ
=
9604
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
2401
2
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle =9604\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi =9604\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {2401}{2}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
2401
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 {2401}{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)]=}
=
2401
2
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {2401}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
2401
2
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
2401
2
⋅
3
π
2
=
7203
π
4
=
5657.222971.
{\displaystyle ={\frac {2401}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {2401}{2}}\cdot {\frac {3\pi }{2}}={\frac {7203\pi }{4}}=5657.222971.}
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
7203
π
4
{\displaystyle {\frac {7203\pi }{4}}}
Palyginimui cilindro viename oktante tūris yra
V
c
i
l
1
=
π
r
2
h
2
=
π
⋅
7
2
⋅
14
2
2
=
π
⋅
49
⋅
196
2
=
4802
π
=
15085.92792.
{\displaystyle V_{cil1}={\frac {\pi r^{2}h}{2}}={\frac {\pi \cdot 7^{2}\cdot 14^{2}}{2}}={\frac {\pi \cdot 49\cdot 196}{2}}=4802\pi =15085.92792.}
Rasime kūno tūrį V , kurį išpjauna praboloidas
z
=
1
−
x
2
−
y
2
.
{\displaystyle z=1-x^{2}-y^{2}.}
z lygi 1, kai reikšmės x ir y lygios 0. Iš šonų kūna riboja begalinė prizmė (stačiakampis gretasienis) su kraštinėmis x=1, y=1.
V
=
∫
0
1
d
x
∫
0
1
d
y
∫
0
1
−
x
2
−
y
2
d
z
=
∫
0
1
d
x
∫
0
1
d
y
z
|
0
1
−
x
2
−
y
2
=
∫
0
1
d
x
∫
0
1
(
(
1
−
x
2
−
y
2
)
−
0
)
d
y
=
{\displaystyle V=\int _{0}^{1}dx\int _{0}^{1}dy\int _{0}^{1-x^{2}-y^{2}}dz=\int _{0}^{1}dx\int _{0}^{1}dy\;z|_{0}^{1-x^{2}-y^{2}}=\int _{0}^{1}dx\int _{0}^{1}((1-x^{2}-y^{2})-0)dy=}
=
∫
0
1
d
x
∫
0
1
(
1
−
x
2
−
y
2
)
d
y
=
∫
0
1
d
x
(
y
−
x
2
y
−
y
3
3
)
|
0
1
=
∫
0
1
(
1
−
x
2
⋅
1
−
1
3
3
)
d
x
=
(
x
−
x
3
3
−
x
3
)
|
0
1
=
1
−
1
3
3
−
1
3
=
1
3
=
0.3
(
3
)
.
{\displaystyle =\int _{0}^{1}dx\int _{0}^{1}(1-x^{2}-y^{2})dy=\int _{0}^{1}dx(y-x^{2}y-{\frac {y^{3}}{3}})|_{0}^{1}=\int _{0}^{1}(1-x^{2}\cdot 1-{\frac {1^{3}}{3}})dx=(x-{\frac {x^{3}}{3}}-{\frac {x}{3}})|_{0}^{1}=1-{\frac {1^{3}}{3}}-{\frac {1}{3}}={\frac {1}{3}}=0.3(3).}
Kad gauti tūrį keturiuose oktantuose, reikia padauginti iš keturių, tuomet tūris bus lygus
V
4
=
4
⋅
1
3
=
4
3
.
{\displaystyle V_{4}=4\cdot {\frac {1}{3}}={\frac {4}{3}}.}
Tūris, kurį gausime atėmę šį tūrį is cilindro tūrio, kurio aukštis h=1 ir spindulys r=1, yra tūris po paraboloidu
V
p
p
=
V
c
i
l
−
V
4
=
π
r
2
h
−
∫
0
1
d
x
∫
0
1
d
y
∫
0
1
−
x
2
−
y
2
d
z
=
π
⋅
1
2
⋅
1
−
4
3
=
π
−
4
3
=
3.141592654
−
1.333333333
=
1.80825932.
{\displaystyle V_{pp}=V_{cil}-V_{4}=\pi r^{2}h-\int _{0}^{1}dx\int _{0}^{1}dy\int _{0}^{1-x^{2}-y^{2}}dz=\pi \cdot 1^{2}\cdot 1-{\frac {4}{3}}=\pi -{\frac {4}{3}}=3.141592654-1.333333333=1.80825932.}
Rodos šis skaičiavimo būdas neteisingas, nors ir naudojamas matematikos knygoje(-se).
Rasime kūno tūrį V , esantį po paraboloidu
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
x arba y lygios
2
{\displaystyle {\sqrt {2}}}
V
=
∫
0
2
d
x
∫
0
2
d
y
∫
0
x
2
+
y
2
d
z
=
∫
0
2
d
x
∫
0
2
d
y
z
|
0
x
2
+
y
2
=
∫
0
2
d
x
∫
0
2
(
(
x
2
+
y
2
)
−
0
)
d
y
=
{\displaystyle V=\int _{0}^{\sqrt {2}}dx\int _{0}^{\sqrt {2}}dy\int _{0}^{x^{2}+y^{2}}dz=\int _{0}^{\sqrt {2}}dx\int _{0}^{\sqrt {2}}dy\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{\sqrt {2}}dx\int _{0}^{\sqrt {2}}((x^{2}+y^{2})-0)dy=}
=
∫
0
2
d
x
∫
0
2
(
x
2
+
y
2
)
d
y
=
∫
0
2
d
x
(
x
2
y
+
y
3
3
)
|
0
2
=
∫
0
2
(
x
2
⋅
2
+
(
2
)
3
3
)
d
x
=
{\displaystyle =\int _{0}^{\sqrt {2}}dx\int _{0}^{\sqrt {2}}(x^{2}+y^{2})dy=\int _{0}^{\sqrt {2}}dx(x^{2}y+{\frac {y^{3}}{3}})|_{0}^{\sqrt {2}}=\int _{0}^{\sqrt {2}}(x^{2}\cdot {\sqrt {2}}+{\frac {({\sqrt {2}})^{3}}{3}})dx=}
=
(
2
x
3
3
+
2
3
⋅
x
3
)
|
0
2
=
2
(
2
)
3
3
+
2
3
⋅
2
3
=
(
2
)
4
3
+
2
4
⋅
1
3
=
{\displaystyle =({\sqrt {2}}{\frac {x^{3}}{3}}+{\sqrt {2^{3}}}\cdot {\frac {x}{3}})|_{0}^{\sqrt {2}}={\sqrt {2}}{\frac {({\sqrt {2}})^{3}}{3}}+{\sqrt {2^{3}}}\cdot {\frac {\sqrt {2}}{3}}={\frac {({\sqrt {2}})^{4}}{3}}+{\sqrt {2^{4}}}\cdot {\frac {1}{3}}=}
=
4
3
+
4
3
=
8
3
=
2.6666667.
{\displaystyle ={\frac {4}{3}}+{\frac {4}{3}}={\frac {8}{3}}=2.6666667.}
Kad gauti tūrį keturiuose oktantuose, reikia padauginti iš keturių, tuomet tūris bus lygus
V
4
=
4
⋅
8
3
=
32
3
=
10.66666667.
{\displaystyle V_{4}=4\cdot {\frac {8}{3}}={\frac {32}{3}}=10.66666667.}
Pavyzdis . Rasti kūno tūrį V , išpjaunamą iš begalinės prizmės su kraštais
x
=
±
2
,
y
=
±
2
{\displaystyle x=\pm 2,\;y=\pm 2}
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)}
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.}
x ir y yra 0, tai
z
1
=
4
{\displaystyle z_{1}=4}
z
2
=
−
2
,
{\displaystyle z_{2}=-2,}
x
=
2
{\displaystyle x=2}
y
=
0
{\displaystyle y=0}
z
1
=
4
−
x
2
−
y
2
=
4
−
2
2
−
0
2
=
0
,
{\displaystyle z_{1}=4-x^{2}-y^{2}=4-2^{2}-0^{2}=0,}
z
2
=
2
2
+
0
2
4
−
2
=
−
1.
{\displaystyle z_{2}={\frac {2^{2}+0^{2}}{4}}-2=-1.}
V
=
∫
0
2
d
x
∫
0
2
d
y
∫
x
2
+
y
2
4
−
2
4
−
x
2
−
y
2
d
z
=
∫
0
2
d
x
∫
0
2
d
y
z
|
x
2
+
y
2
4
−
2
4
−
x
2
−
y
2
=
∫
0
2
d
x
∫
0
2
[
(
4
−
x
2
−
y
2
)
−
(
x
2
+
y
2
4
−
2
)
]
d
y
=
{\displaystyle V=\int _{0}^{2}{\mathsf {d}}x\int _{0}^{2}{\mathsf {d}}y\int _{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}{\mathsf {d}}z=\int _{0}^{2}{\mathsf {d}}x\int _{0}^{2}{\mathsf {d}}y\;z|_{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}=\int _{0}^{2}{\mathsf {d}}x\int _{0}^{2}[(4-x^{2}-y^{2})-({\frac {x^{2}+y^{2}}{4}}-2)]{\mathsf {d}}y=}
=
∫
0
2
d
x
∫
0
2
(
6
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x
2
−
y
2
−
x
2
+
y
2
4
)
d
y
=
∫
0
2
d
x
(
6
y
−
x
2
y
−
y
3
3
−
x
2
y
4
−
y
3
4
⋅
3
)
|
0
2
=
∫
0
2
(
6
⋅
2
−
x
2
⋅
1
−
2
3
3
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x
2
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1
4
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2
3
12
)
d
x
=
{\displaystyle =\int _{0}^{2}{\mathsf {d}}x\int _{0}^{2}(6-x^{2}-y^{2}-{\frac {x^{2}+y^{2}}{4}}){\mathsf {d}}y=\int _{0}^{2}{\mathsf {d}}x(6y-x^{2}y-{\frac {y^{3}}{3}}-{\frac {x^{2}y}{4}}-{\frac {y^{3}}{4\cdot 3}})|_{0}^{2}=\int _{0}^{2}(6\cdot 2-x^{2}\cdot 1-{\frac {2^{3}}{3}}-{\frac {x^{2}\cdot 1}{4}}-{\frac {2^{3}}{12}}){\mathsf {d}}x=}
=
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0
2
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12
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8
3
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4
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8
12
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x
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0
2
(
12
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12
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8
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4
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8
12
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2
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x
2
4
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d
x
=
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0
2
(
144
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32
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8
12
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2
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x
2
4
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d
x
=
{\displaystyle =\int _{0}^{2}(12-x^{2}-{\frac {8}{3}}-{\frac {x^{2}}{4}}-{\frac {8}{12}}){\mathsf {d}}x=\int _{0}^{2}({\frac {12\cdot 12-8\cdot 4-8}{12}}-x^{2}-{\frac {x^{2}}{4}}){\mathsf {d}}x=\int _{0}^{2}({\frac {144-32-8}{12}}-x^{2}-{\frac {x^{2}}{4}}){\mathsf {d}}x=}
=
∫
0
2
(
104
12
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x
2
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x
2
4
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d
x
=
(
104
x
12
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3
3
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x
3
4
⋅
3
)
|
0
2
=
104
⋅
2
12
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2
3
3
−
2
3
12
=
208
−
4
⋅
8
−
8
12
=
168
12
=
14.
{\displaystyle =\int _{0}^{2}({\frac {104}{12}}-x^{2}-{\frac {x^{2}}{4}}){\mathsf {d}}x=({\frac {104x}{12}}-{\frac {x^{3}}{3}}-{\frac {x^{3}}{4\cdot 3}})|_{0}^{2}={\frac {104\cdot 2}{12}}-{\frac {2^{3}}{3}}-{\frac {2^{3}}{12}}={\frac {208-4\cdot 8-8}{12}}={\frac {168}{12}}=14.}
Kad gauti tūrį visuose 8-iuose oktantuose, reikia
14
{\displaystyle 14}
Palyginimui, stačiakampio gretasienio tūris, kurio kraštinės a=2, b=2, c=6 yra lygus
V
b
i
g
=
2
⋅
2
⋅
6
=
24.
{\displaystyle V_{big}=2\cdot 2\cdot 6=24.}
![{\displaystyle ={\frac {6561}{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)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/277143e02fa07e743e2585c07d3e64e8e536f79c)
![{\displaystyle ={\frac {6561}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b36805b23a70edbb3ff115944d974e5ea9939bb1)
![{\displaystyle ={\frac {6561}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {6561}{2}}\cdot {\frac {3\pi }{2}}={\frac {19683\pi }{4}}=15458.99205.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/528d62bc39ed46d09811adf6c589a02cc6208e85)
![{\displaystyle ={\frac {2401}{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)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49ad1e7991ed61026ea45ccd940b026b752a70a5)
![{\displaystyle ={\frac {2401}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16f1439fbdc22f6102126f51b7aea6f4f821e450)
![{\displaystyle ={\frac {2401}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {2401}{2}}\cdot {\frac {3\pi }{2}}={\frac {7203\pi }{4}}=5657.222971.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2526de05142cc08f9c58e7a3d520b65f5708fb99)
![{\displaystyle V=\int _{0}^{2}{\mathsf {d}}x\int _{0}^{2}{\mathsf {d}}y\int _{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}{\mathsf {d}}z=\int _{0}^{2}{\mathsf {d}}x\int _{0}^{2}{\mathsf {d}}y\;z|_{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}=\int _{0}^{2}{\mathsf {d}}x\int _{0}^{2}[(4-x^{2}-y^{2})-({\frac {x^{2}+y^{2}}{4}}-2)]{\mathsf {d}}y=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee9dc2c27267d5d0dd7c44d6d9ecb4ec31f23f10)
