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Gryno formulė nustato ryšį tarp dvilypio integralo ir kreivinio integralo antrojo tipo .
∬
D
(
∂
Q
∂
x
−
∂
P
∂
y
)
d
x
d
y
=
∮
L
P
d
x
+
Q
d
y
.
{\displaystyle \iint _{D}({\partial Q \over \partial x}-{\partial P \over \partial y})dxdy=\oint _{L}Pdx+Qdy.}
Su Gryno formule apskaičiuosime kreivinį integralą
∮
L
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
,
{\displaystyle \oint _{L}(x-y)dx+(x+y)dy,}
L - apskritimas
x
2
+
y
2
=
R
2
.
{\displaystyle x^{2}+y^{2}=R^{2}.}
Funkcijos
P
(
x
,
y
)
=
x
−
y
,
{\displaystyle P(x,y)=x-y,}
Q
(
x
,
y
)
=
x
+
y
{\displaystyle Q(x,y)=x+y}
∂
P
∂
y
=
−
1
,
∂
Q
∂
x
=
1
{\displaystyle {\partial P \over \partial y}=-1,\;{\partial Q \over \partial x}=1}
x
2
+
y
2
=
R
2
.
{\displaystyle x^{2}+y^{2}=R^{2}.}
ρ
2
=
R
2
,
{\displaystyle \rho ^{2}=R^{2},}
ρ
=
R
{\displaystyle \rho =R}
∮
L
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
=
∬
D
[
1
−
(
−
1
)
]
d
x
d
y
=
2
∬
D
d
x
d
y
=
2
s
=
2
∫
0
2
π
d
ϕ
∫
0
R
ρ
d
ρ
=
{\displaystyle \oint _{L}(x-y)dx+(x+y)dy=\iint _{D}[1-(-1)]dxdy=2\iint _{D}dxdy=2s=2\int _{0}^{2\pi }d\phi \int _{0}^{R}\rho d\rho =}
=
∫
0
2
π
ρ
2
|
0
R
d
ϕ
=
R
2
∫
0
2
π
d
ϕ
=
R
2
ϕ
|
0
2
π
=
2
π
R
2
.
{\displaystyle =\int _{0}^{2\pi }\rho ^{2}|_{0}^{R}d\phi =R^{2}\int _{0}^{2\pi }d\phi =R^{2}\phi |_{0}^{2\pi }=2\pi R^{2}.}
Taikydami Gryno formulę, apskaičiuokime kreivinį integralą
∫
L
x
y
d
x
+
(
x
2
+
y
2
)
d
y
,
{\displaystyle \int _{L}xydx+(x^{2}+y^{2})dy,}
L - apskritimas
x
2
+
y
2
=
a
x
{\displaystyle x^{2}+y^{2}=ax}
x
2
+
y
2
≤
a
x
{\displaystyle x^{2}+y^{2}\leq ax}
P
(
x
,
y
)
=
x
y
{\displaystyle P(x,y)=xy}
Q
(
x
,
y
)
=
x
2
+
y
2
{\displaystyle Q(x,y)=x^{2}+y^{2}}
∂
P
∂
y
=
x
{\displaystyle {\partial P \over \partial y}=x}
∂
Q
∂
x
=
2
x
{\displaystyle {\partial Q \over \partial x}=2x}
∫
L
x
y
d
x
+
(
x
2
+
y
2
)
d
y
=
∬
D
(
2
x
−
x
)
d
x
d
y
=
∬
D
x
d
x
d
y
.
{\displaystyle \int _{L}xydx+(x^{2}+y^{2})dy=\iint _{D}(2x-x)dxdy=\iint _{D}xdxdy.}
ϕ
{\displaystyle \phi }
−
π
2
{\displaystyle -{\pi \over 2}}
π
2
.
{\displaystyle {\pi \over 2}.}
x
=
ρ
cos
ϕ
,
{\displaystyle x=\rho \cos \phi ,}
ρ
2
=
a
ρ
cos
ϕ
,
{\displaystyle \rho ^{2}=a\rho \cos \phi ,}
ρ
=
a
cos
ϕ
{\displaystyle \rho =a\cos \phi }
∫
D
x
d
x
d
y
=
∬
D
ρ
2
cos
ϕ
d
ρ
d
ϕ
=
∫
−
π
2
π
2
cos
ϕ
d
ϕ
∫
0
a
cos
ϕ
ρ
2
d
ρ
=
∫
−
π
2
π
2
cos
ϕ
ρ
3
3
|
0
a
cos
ϕ
d
ϕ
=
{\displaystyle \int _{D}xdxdy=\iint _{D}\rho ^{2}\cos \phi d\rho d\phi =\int _{-{\pi \over 2}}^{\pi \over 2}\cos \phi d\phi \int _{0}^{a\cos \phi }\rho ^{2}d\rho =\int _{-{\pi \over 2}}^{\pi \over 2}\cos \phi {\rho ^{3} \over 3}|_{0}^{a\cos \phi }d\phi =}
=
a
3
3
∫
−
π
2
π
2
cos
4
ϕ
d
ϕ
=
2
a
3
3
∫
0
π
2
cos
4
ϕ
d
ϕ
=
2
a
3
3
⋅
3
!
!
4
!
!
⋅
π
2
=
π
a
3
8
,
{\displaystyle ={a^{3} \over 3}\int _{-{\pi \over 2}}^{\pi \over 2}\cos ^{4}\phi d\phi ={2a^{3} \over 3}\int _{0}^{\pi \over 2}\cos ^{4}\phi d\phi ={2a^{3} \over 3}\cdot {3!! \over 4!!}\cdot {\pi \over 2}={\pi a^{3} \over 8},}
dvigubu faktorialu .
Plotui apskaičiuoti ploksčios srities naudojamos tokios formulės:
D
=
∮
L
−
y
d
x
=
∮
L
x
d
y
=
1
2
∮
L
x
d
y
−
y
d
x
.
{\displaystyle D=\oint _{L}-ydx=\oint _{L}xdy={1 \over 2}\oint _{L}xdy-ydx.}
∬
D
(
∂
Q
∂
x
−
∂
P
∂
y
)
d
x
d
y
=
∮
L
P
d
x
+
Q
d
y
.
{\displaystyle \iint _{D}({\partial Q \over \partial x}-{\partial P \over \partial y})dxdy=\oint _{L}Pdx+Qdy.}
Pritaikysim Gryno formulę apskaičiavimui srities D (ploksčios figūros ploto). Jei
P
(
x
,
y
)
=
−
y
,
{\displaystyle P(x,y)=-y,}
Q
(
x
,
y
)
=
0.
{\displaystyle Q(x,y)=0.}
∂
P
∂
y
=
−
1
,
∂
Q
∂
x
=
0.
{\displaystyle {\partial P \over \partial y}=-1,\;{\partial Q \over \partial x}=0.}
∬
D
(
0
+
1
)
d
x
d
y
=
∮
L
−
y
d
x
+
0
d
y
.
{\displaystyle \iint _{D}(0+1)dxdy=\oint _{L}-ydx+0dy.}
∬
D
d
x
d
y
{\displaystyle \iint _{D}dxdy}
D , todėl,
D
=
∬
D
d
x
d
y
=
−
∮
L
y
d
x
.
{\displaystyle D=\iint _{D}dxdy=-\oint _{L}ydx.}
Sakykime,
P
(
x
,
y
)
=
0
,
{\displaystyle P(x,y)=0,}
Q
(
x
,
y
)
=
x
,
{\displaystyle Q(x,y)=x,}
D
=
∮
L
x
d
y
.
{\displaystyle D=\oint _{L}xdy.}
Ir, pagaliau, paėmę funkcijas
P
(
x
,
y
)
=
−
1
2
y
,
Q
(
x
,
y
)
=
1
2
x
,
{\displaystyle P(x,y)=-{1 \over 2}y,\;Q(x,y)={1 \over 2}x,}
D
=
∬
D
(
1
2
+
1
2
)
d
x
d
y
=
∬
D
d
x
d
y
=
1
2
∮
L
x
d
y
−
y
d
x
.
{\displaystyle D=\iint _{D}({1 \over 2}+{1 \over 2})dxdy=\iint _{D}dxdy={1 \over 2}\oint _{L}xdy-ydx.}
Pavyzdžiai
Apskaičiuosime plotą apribotą elipse
x
2
a
2
+
y
2
b
2
=
1
,
{\displaystyle {x^{2} \over a^{2}}+{y^{2} \over b^{2}}=1,}
D
=
∮
L
x
d
y
.
{\displaystyle D=\oint _{L}xdy.}
x
=
a
cos
t
,
{\displaystyle x=a\cos t,}
y
=
b
sin
t
,
{\displaystyle y=b\sin t,}
0
≤
t
≤
2
π
,
{\displaystyle 0\leq t\leq 2\pi ,}
d
y
=
b
cos
t
,
{\displaystyle dy=b\cos t,}
D
=
∮
L
x
d
y
=
∫
0
2
π
a
cos
t
b
cos
t
d
t
=
a
b
2
∫
0
2
π
(
1
+
cos
(
2
t
)
)
d
t
=
a
b
2
(
2
π
+
sin
(
2
t
)
2
|
0
2
π
)
=
π
a
b
.
{\displaystyle D=\oint _{L}xdy=\int _{0}^{2\pi }a\cos tb\cos tdt={ab \over 2}\int _{0}^{2\pi }(1+\cos(2t))dt={ab \over 2}(2\pi +{\sin(2t) \over 2}|_{0}^{2\pi })=\pi ab.}
Jėgos darbas padarytas judant kreive plokštumoje apskaičiuojamas pagal formulę
A
=
∫
B
C
P
d
x
+
Q
d
y
.
{\displaystyle A=\int _{BC}Pdx+Qdy.}
A
=
∫
B
C
P
d
x
+
Q
d
y
+
R
d
z
.
{\displaystyle A=\int _{BC}Pdx+Qdy+Rdz.}
Apskaičiuosime darbą jėgos
F
(
x
,
y
)
{\displaystyle F(x,y)}
Pagal sąlyga,
|
F
(
x
,
y
)
|
=
x
2
+
y
2
;
{\displaystyle |F(x,y)|={\sqrt {x^{2}+y^{2}}};}
P
=
−
x
,
{\displaystyle P=-x,}
Q
=
−
y
{\displaystyle Q=-y}
−
{\displaystyle -}
A
=
−
∮
L
x
d
x
+
y
d
y
,
{\displaystyle A=-\oint _{L}xdx+ydy,}
L - elipsė
x
=
a
cos
t
,
{\displaystyle x=a\cos t,}
y
=
b
sin
t
,
{\displaystyle y=b\sin t,}
0
≤
t
≤
2
π
.
{\displaystyle 0\leq t\leq 2\pi .}
A
=
−
∫
0
2
π
a
cos
t
(
−
a
sin
t
)
d
t
+
b
sin
t
b
cos
t
d
t
=
−
∫
0
2
π
(
b
2
−
a
2
)
sin
t
cos
t
d
t
=
{\displaystyle A=-\int _{0}^{2\pi }a\cos t(-a\sin t)dt+b\sin tb\cos tdt=-\int _{0}^{2\pi }(b^{2}-a^{2})\sin t\cos tdt=}
=
a
2
−
b
2
2
∫
0
2
π
sin
(
2
t
)
d
t
=
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
2
π
=
0.
{\displaystyle ={a^{2}-b^{2} \over 2}\int _{0}^{2\pi }\sin(2t)dt={a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{2\pi }=0.}
Jei t keistusi nuo 0 iki
π
2
,
{\displaystyle {\pi \over 2},}
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
π
2
=
a
2
−
b
2
2
.
{\displaystyle {a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{\pi \over 2}={a^{2}-b^{2} \over 2}.}
![{\displaystyle \oint _{L}(x-y)dx+(x+y)dy=\iint _{D}[1-(-1)]dxdy=2\iint _{D}dxdy=2s=2\int _{0}^{2\pi }d\phi \int _{0}^{R}\rho d\rho =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db7fb9ab38401a6634b1f0a6a88616c99461a25a)
