Pereiti prie turinio
Paviršinis integralas antrojo tipo
∬
S
P
d
y
d
z
+
Q
d
z
d
x
+
R
d
x
d
y
=
∬
σ
(
−
p
P
−
q
Q
+
R
)
d
x
d
y
,
{\displaystyle \iint _{S}Pdydz+Qdzdx+Rdxdy=\iint _{\sigma }(-pP-qQ+R)dxdy,}
kur
p
=
∂
z
∂
x
,
q
=
∂
z
∂
y
.
{\displaystyle p={\frac {\partial z}{\partial x}},\quad q={\frac {\partial z}{\partial y}}.}
Vaizdas:Ris424ir425.jpg 425. Apskaičiuoti integralą
∬
S
x
d
y
d
z
+
y
d
z
d
x
+
z
d
x
d
y
{\displaystyle \iint _{S}xdydz+ydzdx+zdxdy}
x
+
2
z
=
2
,
{\displaystyle x+2z=2,}
y
=
4
{\displaystyle y=4}
Pagal apibrėžimą
∬
S
x
d
y
d
z
+
y
d
z
d
x
+
z
d
x
d
y
=
∬
S
x
d
y
d
z
+
∬
S
y
d
z
d
x
+
∬
S
z
d
x
d
y
.
{\displaystyle \iint _{S}xdydz+ydzdx+zdxdy=\iint _{S}xdydz+\iint _{S}ydzdx+\iint _{S}zdxdy.}
Todėl
∬
S
x
d
y
d
z
=
∬
σ
1
(
2
−
2
z
)
d
y
d
z
=
2
∫
0
4
d
y
∫
0
1
(
1
−
z
)
d
z
=
2
∫
0
4
d
y
(
z
−
z
2
2
)
|
0
1
=
2
∫
0
4
(
1
−
1
2
2
)
d
y
=
{\displaystyle \iint _{S}xdydz=\iint _{\sigma _{1}}(2-2z)dydz=2\int _{0}^{4}dy\int _{0}^{1}(1-z)dz=2\int _{0}^{4}dy(z-{\frac {z^{2}}{2}})|_{0}^{1}=2\int _{0}^{4}(1-{\frac {1^{2}}{2}})dy=}
=
2
⋅
(
y
−
y
2
)
|
0
4
=
2
(
4
−
4
2
)
=
2
(
4
−
2
)
=
4.
{\displaystyle =2\cdot (y-{\frac {y}{2}})|_{0}^{4}=2(4-{\frac {4}{2}})=2(4-2)=4.}
Pagal formulę
∬
S
P
(
x
;
y
;
x
)
d
y
d
z
=
∬
σ
P
(
f
(
y
;
z
)
;
y
;
z
)
d
y
d
z
,
{\displaystyle \iint _{S}P(x;y;x)dydz=\iint _{\sigma }P(f(y;z);y;z)dydz,}
∬
S
y
d
z
d
x
=
0
,
{\displaystyle \iint _{S}ydzdx=0,}
nes plokštuma S lygiagreti ašiai Oy ,
∬
S
z
d
x
d
y
=
∬
σ
2
(
1
−
x
2
)
d
x
d
y
=
∫
0
4
d
y
∫
0
2
(
1
−
x
2
)
d
x
=
∫
0
4
d
y
(
x
−
x
2
4
)
|
0
2
=
∫
0
4
(
2
−
2
2
4
)
d
y
=
{\displaystyle \iint _{S}zdxdy=\iint _{\sigma _{2}}\left(1-{\frac {x}{2}}\right)dxdy=\int _{0}^{4}dy\int _{0}^{2}\left(1-{\frac {x}{2}}\right)dx=\int _{0}^{4}dy\left(x-{\frac {x^{2}}{4}}\right)|_{0}^{2}=\int _{0}^{4}\left(2-{\frac {2^{2}}{4}}\right)dy=}
=
∫
0
4
(
2
−
1
)
d
y
=
y
|
0
4
=
4
{\displaystyle =\int _{0}^{4}\left(2-1\right)dy=y|_{0}^{4}=4}
pagal formulę
∬
S
R
(
x
;
y
;
x
)
d
x
d
y
=
∬
σ
R
(
x
;
y
;
f
(
x
;
y
)
)
d
x
d
y
.
{\displaystyle \iint _{S}R(x;y;x)dxdy=\iint _{\sigma }R(x;y;f(x;y))dxdy.}
Todėl,
∬
S
x
d
y
d
z
+
y
d
z
d
x
+
z
d
x
d
y
=
4
+
0
+
4
=
8.
{\displaystyle \iint _{S}xdydz+ydzdx+zdxdy=4+0+4=8.}
Vaizdas:Pavirsinisintris427.jpg 427. Apskaičiuoti integralą
∬
S
−
x
d
y
d
z
+
z
d
z
d
x
+
5
d
x
d
y
{\displaystyle \iint _{S}-xdydz+zdzdx+5dxdy}
2
x
+
3
y
+
z
=
6
{\displaystyle 2x+3y+z=6}
Paviršiui S
z
=
6
−
2
x
−
3
y
,
{\displaystyle z=6-2x-3y,}
p
=
−
2
,
{\displaystyle p=-2,}
q
=
−
3
{\displaystyle q=-3}
tai pagal formulę
∬
S
P
d
y
d
z
+
Q
d
z
d
x
+
R
d
x
d
y
=
∬
σ
(
−
p
P
−
q
Q
+
R
)
d
x
d
y
{\displaystyle \iint _{S}Pdydz+Qdzdx+Rdxdy=\iint _{\sigma }(-pP-qQ+R)dxdy}
∬
S
−
x
d
y
d
z
+
z
d
z
d
x
+
5
d
x
d
y
=
∬
σ
[
2
x
−
3
(
6
−
2
x
−
3
y
)
+
5
]
d
x
d
y
=
∬
σ
(
2
x
+
6
x
+
9
y
−
18
+
5
)
d
x
d
y
=
∬
σ
(
8
x
+
9
y
−
13
)
d
x
d
y
,
{\displaystyle \iint _{S}-xdydz+zdzdx+5dxdy=\iint _{\sigma }[2x-3(6-2x-3y)+5]dxdy=\iint _{\sigma }(2x+6x+9y-18+5)dxdy=\iint _{\sigma }(8x+9y-13)dxdy,}
kur
σ
{\displaystyle \sigma }
S į plokštumą xOy .
Apskaičiuodami dvilypį integralą, randame:
∬
S
−
x
d
y
d
z
+
z
d
z
d
x
+
5
d
x
d
y
=
∫
0
3
d
x
∫
0
6
−
2
x
3
(
8
x
+
9
y
−
13
)
d
y
=
∫
0
3
d
x
(
8
x
y
+
9
y
2
2
−
13
y
)
|
0
6
−
2
x
3
=
{\displaystyle \iint _{S}-xdydz+zdzdx+5dxdy=\int _{0}^{3}dx\int _{0}^{6-2x \over 3}(8x+9y-13)dy=\int _{0}^{3}dx(8xy+{\frac {9y^{2}}{2}}-13y)|_{0}^{6-2x \over 3}=}
=
∫
0
3
(
8
x
⋅
6
−
2
x
3
+
9
(
6
−
2
x
3
)
2
2
−
13
⋅
6
−
2
x
3
)
d
x
=
∫
0
3
(
48
x
−
16
x
2
3
+
(
6
−
2
x
)
2
2
−
78
−
26
x
3
)
d
x
=
{\displaystyle =\int _{0}^{3}(8x\cdot {6-2x \over 3}+{\frac {9({6-2x \over 3})^{2}}{2}}-13\cdot {6-2x \over 3})dx=\int _{0}^{3}({48x-16x^{2} \over 3}+{\frac {(6-2x)^{2}}{2}}-{78-26x \over 3})dx=}
=
∫
0
3
(
16
x
−
16
x
2
3
+
36
−
24
x
+
4
x
2
2
−
26
+
26
x
3
)
d
x
=
∫
0
3
(
16
x
−
16
x
2
3
+
18
−
12
x
+
2
x
2
−
26
+
26
x
3
)
d
x
=
{\displaystyle =\int _{0}^{3}(16x-{16x^{2} \over 3}+{\frac {36-24x+4x^{2}}{2}}-26+{26x \over 3})dx=\int _{0}^{3}(16x-{16x^{2} \over 3}+18-12x+2x^{2}-26+{26x \over 3})dx=}
=
∫
0
3
(
4
x
−
16
x
2
3
+
2
x
2
−
8
+
26
x
3
)
d
x
=
(
4
x
2
2
−
16
x
3
9
+
2
x
3
3
−
8
x
+
26
x
2
6
)
|
0
3
=
{\displaystyle =\int _{0}^{3}(4x-{16x^{2} \over 3}+2x^{2}-8+{26x \over 3})dx=({\frac {4x^{2}}{2}}-{16x^{3} \over 9}+{\frac {2x^{3}}{3}}-8x+{26x^{2} \over 6})|_{0}^{3}=}
=
(
2
x
2
−
16
x
3
9
+
2
x
3
3
−
8
x
+
13
x
2
3
)
|
0
3
=
(
2
⋅
3
2
−
16
⋅
3
3
9
+
2
⋅
3
3
3
−
8
⋅
3
+
13
⋅
3
2
3
)
−
0
=
2
⋅
9
−
16
⋅
27
9
+
2
⋅
27
3
−
24
+
13
⋅
9
3
=
{\displaystyle =(2x^{2}-{16x^{3} \over 9}+{\frac {2x^{3}}{3}}-8x+{13x^{2} \over 3})|_{0}^{3}=(2\cdot 3^{2}-{16\cdot 3^{3} \over 9}+{\frac {2\cdot 3^{3}}{3}}-8\cdot 3+{13\cdot 3^{2} \over 3})-0=2\cdot 9-{16\cdot 27 \over 9}+{\frac {2\cdot 27}{3}}-24+{13\cdot 9 \over 3}=}
=
18
−
16
⋅
3
+
2
⋅
9
−
24
+
13
⋅
3
=
18
−
48
+
18
−
24
+
39
=
36
+
39
−
72
=
39
−
36
=
3.
{\displaystyle =18-16\cdot 3+2\cdot 9-24+13\cdot 3=18-48+18-24+39=36+39-72=39-36=3.}
Ši 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.}
Vadinasi šiame pavyzdyje ieškomas buvo ne trikampio plotas.
![{\displaystyle \iint _{S}-xdydz+zdzdx+5dxdy=\iint _{\sigma }[2x-3(6-2x-3y)+5]dxdy=\iint _{\sigma }(2x+6x+9y-18+5)dxdy=\iint _{\sigma }(8x+9y-13)dxdy,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9055a0b1a0e202c5e2b979c6de594b4345532a5a)
