[ keisti ] Tarkime, kad srityje D paviršių nusako lygtis z=z(x, y), funkcijos išvestinės
z
x
′
{\displaystyle z_{x}'}
z
y
′
{\displaystyle z_{y}'}
D . Paviršiaus dalies, kurios projekcija plokštumoje xOy yra sritis D , plotas apskaičiuojamas pagal formulę
S
=
∬
D
1
+
(
z
x
′
)
2
+
(
z
y
′
)
2
d
x
d
y
=
∬
D
1
+
(
∂
z
∂
x
)
2
+
(
∂
z
∂
y
)
2
d
x
d
y
.
{\displaystyle S=\iint _{D}{\sqrt {1+(z_{x}')^{2}+(z_{y}')^{2}}}dxdy=\iint _{D}{\sqrt {1+({\partial z \over \partial x})^{2}+({\partial z \over \partial y})^{2}}}dxdy.}
1. Apskaičiuosime
∬
S
d
S
{\displaystyle \iint _{S}dS}
z
=
x
2
+
y
2
,
1
≤
z
≤
2
{\displaystyle z={\sqrt {x^{2}+y^{2}}},\;1\leq z\leq 2}
S projektuojasi į plokštumą XOY srityje D , kuri yra žiedas
1
≤
x
2
+
y
2
≤
4.
{\displaystyle 1\leq x^{2}+y^{2}\leq 4.}
z
=
x
2
+
y
2
,
z
x
′
=
x
x
2
+
y
2
,
z
y
′
=
y
x
2
+
y
2
{\displaystyle z={\sqrt {x^{2}+y^{2}}},\;z_{x}'={x \over {\sqrt {x^{2}+y^{2}}}},\;z_{y}'={y \over {\sqrt {x^{2}+y^{2}}}}}
∬
S
d
S
=
∬
D
1
+
x
2
x
2
+
y
2
+
y
2
x
2
+
y
2
d
x
d
y
=
∬
D
2
d
x
d
y
=
2
∫
0
2
π
d
ϕ
∫
1
2
ρ
d
ρ
=
2
∫
0
2
π
ρ
2
2
|
1
2
d
ϕ
=
{\displaystyle \iint _{S}dS=\iint _{D}{\sqrt {1+{x^{2} \over x^{2}+y^{2}}+{y^{2} \over x^{2}+y^{2}}}}dxdy=\iint _{D}{\sqrt {2}}dxdy={\sqrt {2}}\int _{0}^{2\pi }d\phi \int _{1}^{2}\rho d\rho ={\sqrt {2}}\int _{0}^{2\pi }{\rho ^{2} \over 2}|_{1}^{2}d\phi =}
=
2
∫
0
2
π
(
2
2
2
−
1
2
2
)
d
ϕ
=
2
∫
0
2
π
3
2
d
ϕ
=
2
⋅
3
2
ϕ
|
0
2
π
=
3
2
2
(
2
π
−
0
)
=
3
π
2
=
13.32864881.
{\displaystyle ={\sqrt {2}}\int _{0}^{2\pi }({2^{2} \over 2}-{1^{2} \over 2})d\phi ={\sqrt {2}}\int _{0}^{2\pi }{3 \over 2}d\phi ={\sqrt {2}}\cdot {3 \over 2}\phi |_{0}^{2\pi }={3{\sqrt {2}} \over 2}(2\pi -0)=3\pi {\sqrt {2}}=13.32864881.}
Patikriname. Kūgio paviršiaus ploto formulė be pagrindo yra
S
=
π
R
l
,
{\displaystyle S=\pi Rl,}
R yra pagrindo spindulys, o l yra apotema ir didžiojo kūgio yra lygi
l
d
=
R
2
+
H
2
=
2
2
+
2
2
=
8
=
2
2
.
{\displaystyle l_{d}={\sqrt {R^{2}+H^{2}}}={\sqrt {2^{2}+2^{2}}}={\sqrt {8}}=2{\sqrt {2}}.}
l
m
=
r
2
+
h
2
=
1
2
+
1
2
=
2
.
{\displaystyle l_{m}={\sqrt {r^{2}+h^{2}}}={\sqrt {1^{2}+1^{2}}}={\sqrt {2}}.}
S
3
=
S
d
−
S
m
=
π
R
l
d
−
π
r
l
m
=
π
⋅
2
⋅
2
2
−
π
⋅
1
⋅
2
=
4
π
2
−
π
2
=
3
π
2
=
13.32864881.
{\displaystyle S_{3}=S_{d}-S_{m}=\pi Rl_{d}-\pi rl_{m}=\pi \cdot 2\cdot 2{\sqrt {2}}-\pi \cdot 1\cdot {\sqrt {2}}=4\pi {\sqrt {2}}-\pi {\sqrt {2}}=3\pi {\sqrt {2}}=13.32864881.}
2. Rasime plotą dalies kanoninio paviršiaus
z
=
x
2
+
y
2
,
{\displaystyle z={\sqrt {x^{2}+y^{2}}},}
x
=
0
,
{\displaystyle x=0,}
y
=
0
,
{\displaystyle y=0,}
x
+
y
=
1
,
{\displaystyle x+y=1,}
x
+
y
=
2
{\displaystyle x+y=2}
f
(
x
,
y
)
=
x
2
+
y
2
{\displaystyle f(x,y)={\sqrt {x^{2}+y^{2}}}}
D , esanti projekcija šios dalies į plokšumą XOY , tenkina tolydumo sąlyga, apskaičiuojame paviršių pagal formulę. Be to
f
x
′
(
x
,
y
)
=
x
x
2
+
y
2
,
f
y
′
(
x
,
y
)
=
y
x
2
+
y
2
,
{\displaystyle f_{x}'(x,y)={x \over {\sqrt {x^{2}+y^{2}}}},\;f_{y}'(x,y)={y \over {\sqrt {x^{2}+y^{2}}}},}
1
+
f
x
′
2
(
x
,
y
)
+
f
y
′
2
(
x
,
y
)
=
2
,
{\displaystyle {\sqrt {1+f_{x}'^{2}(x,y)+f_{y}'^{2}(x,y)}}={\sqrt {2}},}
S
=
∬
D
2
d
D
=
2
∬
D
d
D
=
2
D
=
3
2
2
.
{\displaystyle S=\iint _{D}{\sqrt {2}}dD={\sqrt {2}}\iint _{D}dD={\sqrt {2}}D={3 \over 2}{\sqrt {2}}.}
D randame kaip trikampių skirtumą:
S
Δ
1
=
2
2
2
=
∫
0
2
(
2
−
x
)
d
x
=
2
;
S
Δ
2
=
1
2
2
=
∫
0
1
(
1
−
x
)
d
x
=
1
2
;
D
=
S
Δ
1
−
S
Δ
2
=
2
−
1
2
=
3
2
.
{\displaystyle S_{\Delta 1}={2^{2} \over 2}=\int _{0}^{2}(2-x)dx=2;\;S_{\Delta 2}={1^{2} \over 2}=\int _{0}^{1}(1-x)dx={1 \over 2};\;D=S_{\Delta 1}-S_{\Delta 2}=2-{\frac {1}{2}}={3 \over 2}.}
Vaizdas:1315pav.jpg 13.15. Paraboliniai cilindrai
y
=
x
,
y
=
2
x
{\displaystyle y={\sqrt {x}},\;y=2{\sqrt {x}}}
z
=
0
{\displaystyle z=0}
x
+
z
=
4
{\displaystyle x+z=4}
Plokštumos lygtį parašykime taip:
z
=
4
−
x
.
{\displaystyle z=4-x.}
z
x
′
=
−
1
,
z
y
′
=
0
,
1
+
z
x
′
2
+
z
y
′
2
=
2
.
{\displaystyle z_{x}'=-1,\;z_{y}'=0,\;{\sqrt {1+z_{x}'^{2}+z_{y}'^{2}}}={\sqrt {2}}.}
S
=
∬
D
2
d
x
d
y
=
2
∫
0
4
d
x
∫
x
2
x
d
y
=
2
∫
0
4
(
2
x
−
x
)
d
x
=
2
⋅
2
3
x
3
2
|
0
4
=
{\displaystyle S=\iint _{D}{\sqrt {2}}dxdy={\sqrt {2}}\int _{0}^{4}dx\int _{\sqrt {x}}^{2{\sqrt {x}}}dy={\sqrt {2}}\int _{0}^{4}(2{\sqrt {x}}-{\sqrt {x}})dx={\sqrt {2}}\cdot {2 \over 3}x^{3 \over 2}|_{0}^{4}=}
=
2
2
3
⋅
8
=
16
2
3
=
7.542472333.
{\displaystyle ={2{\sqrt {2}} \over 3}\cdot 8={16{\sqrt {2}} \over 3}=7.542472333.}
Raskime plotą tos ritinio
y
2
+
z
2
=
a
2
{\displaystyle y^{2}+z^{2}=a^{2}}
x
2
+
y
2
=
a
2
.
{\displaystyle x^{2}+y^{2}=a^{2}.}
Iš paviršiaus lygties
y
2
+
z
2
=
a
2
{\displaystyle y^{2}+z^{2}=a^{2}}
z
=
a
2
−
y
2
.
{\displaystyle z={\sqrt {a^{2}-y^{2}}}.}
yOx .
Vadinasi:
z
y
′
=
−
y
a
2
−
y
2
,
{\displaystyle z_{y}'=-{y \over {\sqrt {a^{2}-y^{2}}}},}
z
x
′
=
0
;
{\displaystyle z_{x}'=0;}
1
+
z
y
′
2
+
z
x
′
2
=
1
+
(
−
y
a
2
−
y
2
)
2
+
0
2
=
1
+
y
2
a
2
−
y
2
=
(
a
2
−
y
2
)
+
y
2
a
2
−
y
2
=
a
a
2
−
y
2
.
{\displaystyle {\sqrt {1+z_{y}'^{2}+z_{x}'^{2}}}={\sqrt {1+\left(-{y \over {\sqrt {a^{2}-y^{2}}}}\right)^{2}+0^{2}}}={\sqrt {1+{\frac {y^{2}}{a^{2}-y^{2}}}}}={\sqrt {\frac {(a^{2}-y^{2})+y^{2}}{a^{2}-y^{2}}}}={a \over {\sqrt {a^{2}-y^{2}}}}.}
Tuomet
S
=
∬
D
a
a
2
−
y
2
d
x
d
y
;
{\displaystyle S=\iint _{D}{a \over {\sqrt {a^{2}-y^{2}}}}{\mathsf {d}}x{\mathsf {d}}y;}
D yra ketvirtis skritulio, apriboto apskritimo
x
2
+
y
2
=
a
2
.
{\displaystyle x^{2}+y^{2}=a^{2}.}
S
=
a
∫
0
a
d
y
a
2
−
y
2
∫
0
a
2
−
y
2
d
x
=
a
∫
0
a
d
y
a
2
−
y
2
x
|
0
a
2
−
y
2
=
a
∫
0
a
d
y
a
2
−
y
2
(
a
2
−
y
2
−
0
)
=
a
∫
0
a
d
y
=
a
y
|
0
a
=
a
(
a
−
0
)
=
a
2
.
{\displaystyle S=a\int _{0}^{a}{dy \over {\sqrt {a^{2}-y^{2}}}}\int _{0}^{\sqrt {a^{2}-y^{2}}}dx=a\int _{0}^{a}{dy \over {\sqrt {a^{2}-y^{2}}}}x|_{0}^{\sqrt {a^{2}-y^{2}}}=a\int _{0}^{a}{dy \over {\sqrt {a^{2}-y^{2}}}}({\sqrt {a^{2}-y^{2}}}-0)=a\int _{0}^{a}dy=ay|_{0}^{a}=a(a-0)=a^{2}.}
Palyginimui, kvadrato plotas yra
S
k
=
a
2
.
{\displaystyle S_{k}=a^{2}.}
xOy ašį, o tos projekcijos plotas yra
π
r
2
4
=
π
a
2
4
.
{\displaystyle {\frac {\pi r^{2}}{4}}={\frac {\pi a^{2}}{4}}.}
Vaizdas:1316pav.jpg 13.16. Raskime plotą tos ritinio
x
2
+
y
2
=
a
2
{\displaystyle x^{2}+y^{2}=a^{2}}
y
2
+
z
2
=
a
2
.
{\displaystyle y^{2}+z^{2}=a^{2}.}
Iš paviršiaus lygties
x
2
+
y
2
=
a
2
{\displaystyle x^{2}+y^{2}=a^{2}}
x
=
a
2
−
y
2
.
{\displaystyle x={\sqrt {a^{2}-y^{2}}}.}
yOz .
Vadinasi:
x
y
′
=
−
y
a
2
−
y
2
,
{\displaystyle x_{y}'=-{y \over {\sqrt {a^{2}-y^{2}}}},}
x
z
′
=
0
,
{\displaystyle x_{z}'=0,}
1
+
x
y
′
2
+
x
z
′
2
=
a
a
2
−
y
2
.
{\displaystyle {\sqrt {1+x_{y}'^{2}+x_{z}'^{2}}}={a \over {\sqrt {a^{2}-y^{2}}}}.}
S
=
8
∬
D
a
a
2
−
y
2
d
y
d
z
;
{\displaystyle S=8\iint _{D}{a \over {\sqrt {a^{2}-y^{2}}}}dydz;}
D yra ketvirtis skritulio, apriboto apskritimo
y
2
+
z
2
=
a
2
.
{\displaystyle y^{2}+z^{2}=a^{2}.}
S
=
8
a
∫
0
a
d
y
a
2
−
y
2
∫
0
a
2
−
y
2
d
z
=
8
a
∫
0
a
d
y
=
8
a
2
.
{\displaystyle S=8a\int _{0}^{a}{dy \over {\sqrt {a^{2}-y^{2}}}}\int _{0}^{\sqrt {a^{2}-y^{2}}}dz=8a\int _{0}^{a}dy=8a^{2}.}
Palyginimui, kvadrato plotas yra
S
k
=
a
2
,
{\displaystyle S_{k}=a^{2},}
S
K
=
8
⋅
a
2
.
{\displaystyle S_{K}=8\cdot a^{2}.}
Paaiškinimui, ritinio spindulys r=a, o aukštinė h=a, jei skaičiuoti tik teigiamas reikšmes (viename oktante) arba h=2a su ritinio puse(mis) esančia(-iomis) kituose oktantuose, kai z , x reikšmės neigiamos. Pirmo ritinio pagrindas yra plokštumoje xOy , o centras yra (0; 0; 0), o antro ritinio pagrindas yra plokštumoje zOy , o centras (0; 0; 0) (jei neigiamos z ir x reikšmės "nepratesiamos"). 3. Apskaičiuosime plotą tos dalies plokštumos
6
x
+
3
y
+
2
z
=
12
,
{\displaystyle 6x+3y+2z=12,}
Taip kaip funkcija
z
=
6
−
3
x
−
(
3
/
2
)
y
{\displaystyle z=6-3x-(3/2)y}
D , esanti projekcija šios dalies paviršiaus į plokštumą Oxy , tenkina suformuluotas auksčiau salygas, tai ieškomą plotą galima apskaičiuoti pagal formule. Turime
f
x
′
(
x
,
y
)
=
−
3
,
{\displaystyle f_{x}'(x,y)=-3,}
f
y
′
(
x
,
y
)
=
−
3
/
2
;
{\displaystyle f_{y}'(x,y)=-3/2;}
1
+
f
x
′
2
(
x
,
y
)
+
f
y
′
2
(
x
,
y
)
=
1
+
9
+
9
/
4
=
7
/
2.
{\displaystyle {\sqrt {1+f_{x}'^{2}(x,y)+f_{y}'^{2}(x,y)}}={\sqrt {1+9+9/4}}=7/2.}
D yra trikampis, apribotas ašimis Ox , Oy ir tiese 6x+3y=12, gaunamos iš lygties duotos plokštumos kai z=0. Išdėstę integravimo ribas dvilypiam integrale, gauname
S
=
∫
0
2
d
x
∫
0
4
−
2
x
7
2
d
y
=
7
2
∫
0
2
y
|
0
4
−
2
x
d
x
=
7
2
∫
0
2
(
4
−
2
x
)
d
x
=
{\displaystyle S=\int _{0}^{2}dx\int _{0}^{4-2x}{7 \over 2}dy={7 \over 2}\int _{0}^{2}y|_{0}^{4-2x}dx={7 \over 2}\int _{0}^{2}(4-2x)dx=}
=
7
2
(
4
x
−
x
2
)
|
0
2
=
7
2
⋅
4
=
14.
{\displaystyle ={7 \over 2}(4x-x^{2})|_{0}^{2}={7 \over 2}\cdot 4=14.}
Šį plotą galima surasti ir klasikiniu budu. Ieškomas plotas yra trikampis ABC , kurio taškai yra A(2; 0; 0), B(0; 4; 0) ir C(0; 0; 6). Pavadiname atkarpas AB=a, BC=b, CA=c; OA=d=2, OB=e=4, OC=f=6. Koordinačių pradžios taškas yra O(0; 0; 0). Randame trikampio ABC kraštinių ilgius:
a
=
d
2
+
e
2
=
2
2
+
4
2
=
4
+
16
=
20
=
2
5
=
4.472135955
;
{\displaystyle a={\sqrt {d^{2}+e^{2}}}={\sqrt {2^{2}+4^{2}}}={\sqrt {4+16}}={\sqrt {20}}=2{\sqrt {5}}=4.472135955;}
b
=
e
2
+
f
2
=
4
2
+
6
2
=
16
+
36
=
52
=
2
13
=
7.211102551
;
{\displaystyle b={\sqrt {e^{2}+f^{2}}}={\sqrt {4^{2}+6^{2}}}={\sqrt {16+36}}={\sqrt {52}}=2{\sqrt {13}}=7.211102551;}
c
=
d
2
+
f
2
=
2
2
+
6
2
=
4
+
36
=
40
=
2
10
=
6.32455532.
{\displaystyle c={\sqrt {d^{2}+f^{2}}}={\sqrt {2^{2}+6^{2}}}={\sqrt {4+36}}={\sqrt {40}}=2{\sqrt {10}}=6.32455532.}
Toliau randame trikampio ABC pusperimetrį
p
=
a
+
b
+
c
2
=
2
5
+
2
13
+
2
10
2
=
5
+
13
+
10
=
9.003896913
{\displaystyle p={\frac {a+b+c}{2}}={\frac {2{\sqrt {5}}+2{\sqrt {13}}+2{\sqrt {10}}}{2}}={\sqrt {5}}+{\sqrt {13}}+{\sqrt {10}}=9.003896913}
S
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
=
9.003896913
(
9.003896913
−
20
)
(
9.003896913
−
52
)
(
9.003896913
−
40
)
=
{\displaystyle S={\sqrt {p(p-a)(p-b)(p-c)}}={\sqrt {9.003896913(9.003896913-{\sqrt {20}})(9.003896913-{\sqrt {52}})(9.003896913-{\sqrt {40}})}}=}
=
9.003896913
⋅
4.531760958
⋅
1.792794362
⋅
2.679341593
=
196
=
14.
{\displaystyle ={\sqrt {9.003896913\cdot 4.531760958\cdot 1.792794362\cdot 2.679341593}}={\sqrt {196}}=14.}
Vaizdas:Pavirsinisintris427.jpg 427. Apskaičiuoti plotą plokštumos
2
x
+
3
y
+
z
=
6
{\displaystyle 2x+3y+z=6}
Randame:
z
=
6
−
2
x
−
3
y
;
{\displaystyle z=6-2x-3y;}
∂
z
∂
x
=
∂
(
6
−
2
x
−
3
y
)
∂
x
=
−
2
;
{\displaystyle {\frac {\partial z}{\partial x}}={\frac {\partial (6-2x-3y)}{\partial x}}=-2;}
∂
z
∂
y
=
∂
(
6
−
2
x
−
3
y
)
∂
y
=
−
3
;
{\displaystyle {\frac {\partial z}{\partial y}}={\frac {\partial (6-2x-3y)}{\partial y}}=-3;}
1
+
(
z
x
′
(
x
,
y
)
)
2
+
(
z
y
′
(
x
,
y
)
)
2
=
1
+
(
−
2
)
2
+
(
−
3
)
2
=
1
+
4
+
9
=
14
.
{\displaystyle {\sqrt {1+(z_{x}'(x,y))^{2}+(z_{y}'(x,y))^{2}}}={\sqrt {1+(-2)^{2}+(-3)^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}.}
Sritis
σ
{\displaystyle \sigma }
Ox , Oy ir tiese 2x+3y=6, y=(6-2x)/3 gaunama iš lygties duotos plokštumos kai z=0. Išdėstę integravimo ribas dvilypiam integrale, gauname
S
=
∫
0
3
d
x
∫
0
6
−
2
x
3
14
d
y
=
14
∫
0
3
d
x
y
|
0
6
−
2
x
3
=
14
∫
0
3
6
−
2
x
3
d
x
=
14
3
(
6
x
−
x
2
)
|
0
3
=
{\displaystyle S=\int _{0}^{3}dx\int _{0}^{6-2x \over 3}{\sqrt {14}}dy={\sqrt {14}}\int _{0}^{3}dx\;y|_{0}^{6-2x \over 3}={\sqrt {14}}\int _{0}^{3}{6-2x \over 3}dx={\frac {\sqrt {14}}{3}}(6x-x^{2})|_{0}^{3}=}
=
14
3
(
6
⋅
3
−
3
2
)
=
14
3
(
18
−
9
)
=
14
3
⋅
9
=
3
14
=
11.22497216.
{\displaystyle ={\frac {\sqrt {14}}{3}}(6\cdot 3-3^{2})={\frac {\sqrt {14}}{3}}(18-9)={\frac {\sqrt {14}}{3}}\cdot 9=3{\sqrt {14}}=11.22497216.}
Šį plotą galima surasti ir klasikiniu budu. Ieškomas plotas yra trikampis ABC , kurio taškai yra A(3; 0; 0), B(0; 2; 0) ir C(0; 0; 6). Pavadiname atkarpas AB=a, BC=b, CA=c; OA=d=3, OB=e=2, OC=f=6. Koordinačių pradžios taškas yra O(0; 0; 0). Randame trikampio ABC kraštinių ilgius:
a
=
d
2
+
e
2
=
3
2
+
2
2
=
9
+
4
=
13
=
3
,
605551275
;
{\displaystyle a={\sqrt {d^{2}+e^{2}}}={\sqrt {3^{2}+2^{2}}}={\sqrt {9+4}}={\sqrt {13}}=3,605551275;}
b
=
e
2
+
f
2
=
2
2
+
6
2
=
4
+
36
=
40
=
2
10
=
6
,
32455532
;
{\displaystyle b={\sqrt {e^{2}+f^{2}}}={\sqrt {2^{2}+6^{2}}}={\sqrt {4+36}}={\sqrt {40}}=2{\sqrt {10}}=6,32455532;}
c
=
d
2
+
f
2
=
3
2
+
6
2
=
9
+
36
=
45
=
3
5
=
6
,
708203933.
{\displaystyle c={\sqrt {d^{2}+f^{2}}}={\sqrt {3^{2}+6^{2}}}={\sqrt {9+36}}={\sqrt {45}}=3{\sqrt {5}}=6,708203933.}
Toliau randame trikampio ABC pusperimetrį
p
=
a
+
b
+
c
2
=
13
+
40
+
45
2
=
8
,
319155264
{\displaystyle p={\frac {a+b+c}{2}}={\frac {{\sqrt {13}}+{\sqrt {40}}+{\sqrt {45}}}{2}}=8,319155264}
S
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
=
8
,
319155264
(
8
,
319155264
−
13
)
(
8
,
319155264
−
40
)
(
8
,
319155264
−
45
)
=
{\displaystyle S={\sqrt {p(p-a)(p-b)(p-c)}}={\sqrt {8,319155264(8,319155264-{\sqrt {13}})(8,319155264-{\sqrt {40}})(8,319155264-{\sqrt {45}})}}=}
=
8.319155264
⋅
4.713603989
⋅
1.994599944
⋅
1.610951332
=
126
=
11.22497216.
{\displaystyle ={\sqrt {8.319155264\cdot 4.713603989\cdot 1.994599944\cdot 1.610951332}}={\sqrt {126}}=11.22497216.}
