Šis straipsni yra apie pirmojo ir antrojo tipo kreivinius integralus .
Pirmojo tipo kreivinis integralas
Pirmojo tipo kreivinis integralas naudojamas dvimačio ar trimačio lanko masės apskaičiavimui. Galima apskaičiuoti masę kai ji pastovi ar kai kinta pagal tam tikrą funkciją. Jeigu masė pastovi tai jos skaičiavimas sutampa su lanko ilgio skaičiavimu.
d
s
=
1
+
y
′
2
d
x
,
{\displaystyle ds={\sqrt {1+y'^{2}}}dx,}
kai kreivė L apibrėžta lygtimi y=y(x), o
a
≤
x
≤
b
.
{\displaystyle a\leq x\leq b.}
∫
L
f
(
x
,
y
)
d
s
=
∫
a
b
f
(
x
,
y
(
x
)
)
1
+
(
y
′
(
x
)
)
2
d
x
.
{\displaystyle \int _{L}f(x,y)ds=\int _{a}^{b}f(x,y(x)){\sqrt {1+(y'(x))^{2}}}dx.}
Kai kreivė L apibrėžta parametrinėmis lygtimis
x
=
x
(
t
)
,
{\displaystyle x=x(t),}
y
=
y
(
t
)
,
{\displaystyle y=y(t),}
t
∈
[
t
0
;
T
]
,
{\displaystyle t\in [t_{0};T],}
tai
d
s
=
x
t
′
2
+
y
t
′
2
d
t
,
{\displaystyle ds={\sqrt {x_{t}'^{2}+y_{t}'^{2}}}dt,}
todėl
∫
L
f
(
x
,
y
)
d
s
=
∫
t
0
T
f
(
x
(
t
)
,
y
(
t
)
)
x
t
′
2
+
y
t
′
2
d
t
.
{\displaystyle \int _{L}f(x,y)ds=\int _{t_{0}}^{T}f(x(t),y(t)){\sqrt {x_{t}'^{2}+y_{t}'^{2}}}dt.}
Kai prametrinėmis lygtimis
x
=
x
(
t
)
,
{\displaystyle x=x(t),}
y
=
y
(
t
)
,
{\displaystyle y=y(t),}
z
=
z
(
t
)
,
{\displaystyle z=z(t),}
t
∈
[
t
0
;
T
]
{\displaystyle t\in [t_{0};T]}
apibrėžta erdvinė kreivė L , tai
∫
L
f
(
x
,
y
,
z
)
d
s
=
∫
t
0
T
f
(
x
(
t
)
,
y
(
t
)
,
z
(
t
)
)
x
t
′
2
+
y
t
′
2
+
z
t
′
2
d
t
.
{\displaystyle \int _{L}f(x,y,z)ds=\int _{t_{0}}^{T}f(x(t),y(t),z(t)){\sqrt {x_{t}'^{2}+y_{t}'^{2}+z_{t}'^{2}}}dt.}
Kai kreivė L polinėje koordinačių sistemoje apibrėžta lygtimi
ρ
=
ρ
(
ϕ
)
,
{\displaystyle \rho =\rho (\phi ),}
ϕ
∈
[
α
;
β
]
{\displaystyle \phi \in [\alpha ;\beta ]}
tai
d
s
=
ρ
2
+
ρ
′
2
d
ϕ
{\displaystyle ds={\sqrt {\rho ^{2}+\rho '^{2}}}d\phi }
ir
∫
L
f
(
x
,
y
)
d
s
=
∫
α
β
f
(
ρ
cos
ϕ
,
ρ
sin
ϕ
)
ρ
2
+
ρ
′
2
d
ϕ
.
{\displaystyle \int _{L}f(x,y)ds=\int _{\alpha }^{\beta }f(\rho \cos \phi ,\rho \sin \phi ){\sqrt {\rho ^{2}+\rho '^{2}}}d\phi .}
Pavyzdžiai
Apskaičiuokime integralą
∫
L
(
x
+
y
)
d
s
,
{\displaystyle \int _{L}(x+{\sqrt {y}})ds,}
kai L - prabolės
y
=
1
2
x
2
{\displaystyle y={1 \over 2}x^{2}}
lankas nuo taško (0; 0) iki taško (1; 1/2).
Remdamiesi sąlyga
y
=
1
2
x
2
,
{\displaystyle y={1 \over 2}x^{2},}
randame y'=x,
d
s
=
1
+
x
2
.
{\displaystyle ds={\sqrt {1+x^{2}}}.}
Pritaikę pirmą formulę, gauname
∫
L
(
x
+
y
)
d
s
=
∫
0
1
(
x
+
1
2
x
)
1
+
x
2
d
x
=
(
1
+
1
2
)
∫
0
1
x
1
+
x
2
d
x
=
{\displaystyle \int _{L}(x+{\sqrt {y}})ds=\int _{0}^{1}(x+{1 \over {\sqrt {2}}}x){\sqrt {1+x^{2}}}dx=(1+{1 \over {\sqrt {2}}})\int _{0}^{1}x{\sqrt {1+x^{2}}}dx=}
=
2
+
1
2
∫
0
1
1
+
x
2
d
(
1
+
x
2
)
2
=
2
+
1
3
2
(
1
+
x
2
)
3
|
0
1
=
2
+
1
3
2
(
2
2
−
1
)
=
1
6
(
2
+
3
2
)
,
{\displaystyle ={{\sqrt {2}}+1 \over {\sqrt {2}}}\int _{0}^{1}{\sqrt {1+x^{2}}}{d(1+x^{2}) \over 2}={{\sqrt {2}}+1 \over 3{\sqrt {2}}}{\sqrt {(1+x^{2})^{3}}}|_{0}^{1}={{\sqrt {2}}+1 \over 3{\sqrt {2}}}(2{\sqrt {2}}-1)={1 \over 6}(2+3{\sqrt {2}}),}
kur
d
(
1
+
x
2
)
=
2
x
d
x
;
{\displaystyle d(1+x^{2})=2xdx;}
d
x
=
d
(
1
+
x
2
)
2
x
.
{\displaystyle dx={d(1+x^{2}) \over 2x}.}
Apskaičiuokime kreivės
y
=
x
3
2
,
0
≤
x
≤
4
{\displaystyle y=x^{3 \over 2},\;0\leq x\leq 4}
lanko ilgį.
Randame
y
′
=
3
2
x
1
2
,
1
+
y
′
2
=
1
+
9
4
x
.
{\displaystyle y'={3 \over 2}x^{1 \over 2},\;{\sqrt {1+y'^{2}}}={\sqrt {1+{9 \over 4}x}}.}
Tuomet
L
=
∫
0
4
1
+
9
4
x
d
x
=
4
9
∫
0
4
(
1
+
9
4
x
)
1
2
d
(
1
+
9
4
x
)
=
4
9
⋅
2
3
(
1
+
9
4
x
)
3
2
|
0
4
=
8
27
(
10
10
−
1
)
≈
9
,
0734.
{\displaystyle L=\int _{0}^{4}{\sqrt {1+{9 \over 4}x}}dx={4 \over 9}\int _{0}^{4}(1+{9 \over 4}x)^{1 \over 2}d(1+{9 \over 4}x)={4 \over 9}\cdot {2 \over 3}(1+{9 \over 4}x)^{3 \over 2}|_{0}^{4}={8 \over 27}(10{\sqrt {10}}-1)\approx 9,0734.}
Apskaičiuosime lanko ilgį pusiaukūbinės parabolės
y
=
x
3
/
2
,
{\displaystyle y=x^{3/2},}
jei
0
≤
x
≤
5.
{\displaystyle 0\leq x\leq 5.}
Iš lygties
y
=
x
3
/
2
{\displaystyle y=x^{3/2}}
randame:
y
′
=
3
2
x
1
2
.
{\displaystyle y'={3 \over 2}x^{1 \over 2}.}
Iš pirmos formulės gausime
L
=
∫
a
b
1
+
f
′
2
(
x
)
d
x
=
∫
0
5
1
+
y
′
2
d
x
=
∫
0
5
1
+
9
x
4
d
x
=
4
9
∫
0
5
1
+
9
x
4
d
(
1
+
9
x
4
)
=
{\displaystyle L=\int _{a}^{b}{\sqrt {1+f'^{2}(x)}}dx=\int _{0}^{5}{\sqrt {1+y'^{2}}}dx=\int _{0}^{5}{\sqrt {1+{9x \over 4}}}dx={4 \over 9}\int _{0}^{5}{\sqrt {1+{9x \over 4}}}d(1+{9x \over 4})=}
=
4
9
(
1
+
9
x
4
)
3
2
3
2
|
0
5
=
8
27
(
1
+
9
x
4
)
3
2
|
0
5
=
8
27
[
(
4
4
+
45
4
)
3
2
−
(
1
+
0
)
3
2
]
=
8
27
[
(
7
2
)
3
−
1
]
=
335
27
≈
12
,
4074
;
{\displaystyle ={{4 \over 9}(1+{9x \over 4})^{3 \over 2} \over {3 \over 2}}|_{0}^{5}={8 \over 27}(1+{9x \over 4})^{3 \over 2}|_{0}^{5}={8 \over 27}[({4 \over 4}+{45 \over 4})^{3 \over 2}-(1+0)^{3 \over 2}]={8 \over 27}[({7 \over 2})^{3}-1]={335 \over 27}\approx 12,4074;}
kur
d
(
1
+
9
x
4
)
=
9
4
d
x
{\displaystyle d(1+{9x \over 4})={9 \over 4}dx}
;
d
x
=
4
9
d
(
1
+
9
x
4
)
{\displaystyle dx={4 \over 9}d(1+{9x \over 4})}
.
Apskaičiuosime lanko ilgį pusiaukūbinės parabolės
y
=
x
3
2
,
{\displaystyle y=x^{3 \over 2},}
jei
1
≤
x
≤
5.
{\displaystyle 1\leq x\leq 5.}
Iš lygties
y
=
x
3
/
2
{\displaystyle y=x^{3/2}}
randame:
y
′
=
(
x
3
2
)
′
=
3
2
x
1
2
.
{\displaystyle y'=(x^{3 \over 2})'={3 \over 2}x^{1 \over 2}.}
Gausime
L
=
∫
1
5
1
+
y
′
2
d
x
=
∫
1
5
1
+
(
3
x
2
)
2
d
x
=
∫
1
5
1
+
9
x
4
d
x
=
4
9
∫
1
5
1
+
9
x
4
d
(
1
+
9
x
4
)
=
{\displaystyle L=\int _{1}^{5}{\sqrt {1+y'^{2}}}dx=\int _{1}^{5}{\sqrt {1+({3{\sqrt {x}} \over 2})^{2}}}dx=\int _{1}^{5}{\sqrt {1+{9x \over 4}}}dx={4 \over 9}\int _{1}^{5}{\sqrt {1+{9x \over 4}}}d(1+{9x \over 4})=}
=
4
9
⋅
(
1
+
9
x
4
)
3
2
3
2
|
1
5
=
8
27
(
1
+
9
x
4
)
3
2
|
1
5
=
8
27
[
(
4
4
+
45
4
)
3
2
−
(
1
+
9
4
)
3
2
]
=
8
27
[
(
7
2
)
3
−
(
1
+
2
,
25
)
3
]
=
{\displaystyle ={4 \over 9}\cdot {(1+{9x \over 4})^{3 \over 2} \over {3 \over 2}}|_{1}^{5}={8 \over 27}(1+{9x \over 4})^{3 \over 2}|_{1}^{5}={8 \over 27}[({4 \over 4}+{45 \over 4})^{3 \over 2}-(1+{9 \over 4})^{3 \over 2}]={8 \over 27}[({7 \over 2})^{3}-{\sqrt {(1+2,25)^{3}}}]=}
=
8
27
⋅
343
8
−
8
27
⋅
34
,
328125
≈
12
,
7037037
−
1
,
73600617
≈
10
,
96769753
;
{\displaystyle ={8 \over 27}\cdot {343 \over 8}-{8 \over 27}\cdot {\sqrt {34,328125}}\approx 12,7037037-1,73600617\approx 10,96769753;}
kur
d
(
1
+
9
x
4
)
=
9
4
d
x
{\displaystyle d(1+{9x \over 4})={9 \over 4}dx}
;
d
x
=
4
9
d
(
1
+
9
x
4
)
{\displaystyle dx={4 \over 9}d(1+{9x \over 4})}
.
Palyginimui, linijos ilgis nuo taško (1; 1) iki taško (5;
5
3
/
2
{\displaystyle 5^{3/2}}
) yra
c
=
(
5
−
1
)
2
+
(
5
3
/
2
−
1
)
2
=
16
+
(
125
−
1
)
2
=
16
+
10
,
18033989
2
=
119
,
6393202
=
{\displaystyle c={\sqrt {(5-1)^{2}+(5^{3/2}-1)^{2}}}={\sqrt {16+({\sqrt {125}}-1)^{2}}}={\sqrt {16+10,18033989^{2}}}={\sqrt {119,6393202}}=}
=
10
,
093797606.
{\displaystyle =10,093797606.}
Apskaičiuosime parabolės
y
=
x
2
{\displaystyle y=x^{2}}
lanko ilgį, kai
0
≤
x
≤
5.
{\displaystyle 0\leq x\leq 5.}
Randame
y
′
=
(
x
2
)
′
=
2
x
{\displaystyle y'=(x^{2})'=2x}
. Gauname
L
=
∫
a
b
1
+
f
′
2
(
x
)
d
x
=
∫
0
5
1
+
y
′
2
d
x
=
∫
0
5
1
+
(
2
x
)
2
d
x
=
∫
0
5
1
+
4
x
2
d
x
=
{\displaystyle L=\int _{a}^{b}{\sqrt {1+f'^{2}(x)}}dx=\int _{0}^{5}{\sqrt {1+y'^{2}}}dx=\int _{0}^{5}{\sqrt {1+(2x)^{2}}}dx=\int _{0}^{5}{\sqrt {1+4x^{2}}}dx=}
=
1
4
(
2
x
4
x
2
+
1
+
sinh
−
1
(
2
x
)
)
|
0
5
=
1
4
(
2
x
4
x
2
+
1
+
(
e
2
x
−
e
−
2
x
2
)
−
1
)
|
0
5
=
{\displaystyle ={1 \over 4}(2x{\sqrt {4x^{2}+1}}+\sinh ^{-1}(2x))|_{0}^{5}={1 \over 4}(2x{\sqrt {4x^{2}+1}}+({e^{2x}-e^{-2x} \over 2})^{-1})|_{0}^{5}=}
=
1
4
(
2
x
4
x
2
+
1
+
2
e
2
x
−
e
−
2
x
)
|
0
5
=
1
4
(
10
100
+
1
+
2
e
10
−
e
−
10
)
−
1
4
(
0
0
+
1
+
2
e
0
−
e
0
)
=
{\displaystyle ={1 \over 4}(2x{\sqrt {4x^{2}+1}}+{2 \over e^{2x}-e^{-2x}})|_{0}^{5}={1 \over 4}(10{\sqrt {100+1}}+{2 \over e^{10}-e^{-10}})-{1 \over 4}(0{\sqrt {0+1}}+{2 \over e^{0}-e^{0}})=}
=
1
4
(
10
100
+
1
+
2
e
10
−
e
−
10
)
=
1
4
(
100
,
4987562
+
2
22026
,
46579
−
0
,
000045399
≈
{\displaystyle ={1 \over 4}(10{\sqrt {100+1}}+{2 \over e^{10}-e^{-10}})={1 \over 4}(100,4987562+{2 \over 22026,46579-0,000045399}\approx }
≈
1
4
(
100
,
4987562
+
2
22026
,
46574
)
≈
25
,
12471175.
{\displaystyle \approx {1 \over 4}(100,4987562+{2 \over 22026,46574})\approx 25,12471175.}
Apskaičiuosime parabolės
y
=
x
2
{\displaystyle y=x^{2}}
lanko ilgį, kai
1
≤
x
≤
5.
{\displaystyle 1\leq x\leq 5.}
Randame
y
′
=
(
x
2
)
′
=
2
x
{\displaystyle y'=(x^{2})'=2x}
. Gauname
L
=
∫
1
5
1
+
(
2
x
)
2
d
x
=
∫
1
5
1
+
4
x
2
d
x
=
1
4
(
2
x
4
x
2
+
1
+
sinh
−
1
(
2
x
)
)
|
1
5
=
{\displaystyle L=\int _{1}^{5}{\sqrt {1+(2x)^{2}}}dx=\int _{1}^{5}{\sqrt {1+4x^{2}}}dx={1 \over 4}(2x{\sqrt {4x^{2}+1}}+\sinh ^{-1}(2x))|_{1}^{5}=}
=
1
4
(
2
x
4
x
2
+
1
+
(
e
2
x
−
e
−
2
x
2
)
−
1
)
|
1
5
=
1
4
(
2
x
4
x
2
+
1
+
2
e
2
x
−
e
−
2
x
)
|
1
5
=
{\displaystyle ={1 \over 4}(2x{\sqrt {4x^{2}+1}}+({e^{2x}-e^{-2x} \over 2})^{-1})|_{1}^{5}={1 \over 4}(2x{\sqrt {4x^{2}+1}}+{2 \over e^{2x}-e^{-2x}})|_{1}^{5}=}
=
1
4
(
10
100
+
1
+
2
e
10
−
e
−
10
)
−
1
4
(
2
4
+
1
+
2
e
2
−
e
−
2
)
≈
{\displaystyle ={1 \over 4}(10{\sqrt {100+1}}+{2 \over e^{10}-e^{-10}})-{1 \over 4}(2{\sqrt {4+1}}+{2 \over e^{2}-e^{-2}})\approx }
≈
25
,
12471175
−
1
4
(
4
,
472135955
+
2
7
,
389056099
−
0
,
135335283
)
≈
{\displaystyle \approx 25,12471175-{1 \over 4}(4,472135955+{2 \over 7,389056099-0,135335283})\approx }
≈
25
,
12471175
−
1
4
(
4
,
472135955
+
2
7
,
253720816
)
≈
{\displaystyle \approx 25,12471175-{1 \over 4}(4,472135955+{2 \over 7,253720816})\approx }
≈
25
,
12471175
−
1
4
(
4
,
472135955
+
0
,
275720564
)
≈
{\displaystyle \approx 25,12471175-{1 \over 4}(4,472135955+0,275720564)\approx }
≈
25
,
12471175
−
1
,
18696413
≈
23
,
93774762.
{\displaystyle \approx 25,12471175-1,18696413\approx 23,93774762.}
Apskaičiuosime parabolės
y
=
x
2
{\displaystyle y=x^{2}}
lanko ilgį, kai
0
≤
x
≤
4.
{\displaystyle 0\leq x\leq 4.}
Randame
y
′
=
(
x
2
)
′
=
2
x
{\displaystyle y'=(x^{2})'=2x}
. Pasinaudodami integralų lentele
∫
x
2
+
a
2
d
x
=
x
2
a
2
+
x
2
+
a
2
2
ln
|
x
+
x
2
+
a
2
|
{\displaystyle \int {\sqrt {x^{2}+a^{2}}}\;{\mathsf {d}}x={\frac {x}{2}}{\sqrt {a^{2}+x^{2}}}+{\frac {a^{2}}{2}}\ln \left|x+{\sqrt {x^{2}+a^{2}}}\right|}
, gauname
L
=
∫
0
4
1
+
y
′
2
d
x
=
∫
0
4
1
+
(
2
x
)
2
d
x
=
∫
0
4
1
+
4
x
2
d
x
=
4
∫
0
4
1
4
+
x
2
d
x
=
{\displaystyle L=\int _{0}^{4}{\sqrt {1+y'^{2}}}dx=\int _{0}^{4}{\sqrt {1+(2x)^{2}}}dx=\int _{0}^{4}{\sqrt {1+4x^{2}}}dx={\sqrt {4}}\int _{0}^{4}{\sqrt {{1 \over 4}+x^{2}}}dx=}
=
2
(
x
2
0
,
25
+
x
2
+
0
,
25
2
ln
|
x
+
x
2
+
0
,
25
|
)
|
0
4
=
{\displaystyle =2({x \over 2}{\sqrt {0,25+x^{2}}}+{\frac {0,25}{2}}\ln \left|x+{\sqrt {x^{2}+0,25}}\right|)|_{0}^{4}=}
=
2
(
4
2
0
,
25
+
4
2
+
0
,
25
2
ln
|
4
+
4
2
+
0
,
25
|
)
−
2
(
0
2
0
,
25
+
0
2
+
0
,
25
2
ln
|
0
+
0
2
+
0
,
25
|
)
=
{\displaystyle =2({4 \over 2}{\sqrt {0,25+4^{2}}}+{\frac {0,25}{2}}\ln \left|4+{\sqrt {4^{2}+0,25}}\right|)-2({0 \over 2}{\sqrt {0,25+0^{2}}}+{\frac {0,25}{2}}\ln \left|0+{\sqrt {0^{2}+0,25}}\right|)=}
=
2
(
2
16
,
25
+
0
,
125
ln
|
4
+
16
,
25
|
)
−
2
(
0
+
0
,
125
ln
|
0
+
0
,
25
|
)
=
{\displaystyle =2(2{\sqrt {16,25}}+0,125\ln \left|4+{\sqrt {16,25}}\right|)-2(0+0,125\ln \left|0+{\sqrt {0,25}}\right|)=}
≈
2
(
2
⋅
4
,
031128874
+
0
,
125
ln
|
8
,
031128874
|
)
−
2
(
0
,
125
⋅
ln
|
0
,
5
|
)
≈
{\displaystyle \approx 2(2\cdot 4,031128874+0,125\ln |8,031128874|)-2(0,125\cdot \ln |0,5|)\approx }
≈
2
(
8
,
062257748
+
0
,
125
⋅
2
,
0833251
)
−
2
(
0
,
125
⋅
(
−
0
,
69314718
)
)
=
{\displaystyle \approx 2(8,062257748+0,125\cdot 2,0833251)-2(0,125\cdot (-0,69314718))=}
=
2
(
8
,
062257748
+
0
,
260415637
)
−
2
(
−
0
,
086643397
)
=
16
,
64534677
+
0
,
173286795
=
16
,
81863357.
{\displaystyle =2(8,062257748+0,260415637)-2(-0,086643397)=16,64534677+0,173286795=16,81863357.}
Palyginimui, linijos ilgis nuo taško (0; 0) iki taško (4; 16) yra
c
=
a
2
+
b
2
=
4
2
+
16
2
=
272
=
16
,
4924225.
{\displaystyle c={\sqrt {a^{2}+b^{2}}}={\sqrt {4^{2}+16^{2}}}={\sqrt {272}}=16,4924225.}
Apskaičiuosime parabolės
y
=
x
2
{\displaystyle y=x^{2}}
lanko ilgį, kai
1
≤
x
≤
4.
{\displaystyle 1\leq x\leq 4.}
Randame
y
′
=
(
x
2
)
′
=
2
x
{\displaystyle y'=(x^{2})'=2x}
. Pasinaudodami integralų lentele gauname
L
=
∫
1
4
1
+
y
′
2
d
x
=
∫
1
4
1
+
(
2
x
)
2
d
x
=
∫
1
4
1
+
4
x
2
d
x
=
4
∫
1
4
1
4
+
x
2
d
x
=
{\displaystyle L=\int _{1}^{4}{\sqrt {1+y'^{2}}}dx=\int _{1}^{4}{\sqrt {1+(2x)^{2}}}dx=\int _{1}^{4}{\sqrt {1+4x^{2}}}dx={\sqrt {4}}\int _{1}^{4}{\sqrt {{1 \over 4}+x^{2}}}dx=}
=
2
(
x
2
0
,
25
+
x
2
+
0
,
25
2
ln
|
x
+
x
2
+
0
,
25
|
)
|
1
4
=
{\displaystyle =2({x \over 2}{\sqrt {0,25+x^{2}}}+{\frac {0,25}{2}}\ln \left|x+{\sqrt {x^{2}+0,25}}\right|)|_{1}^{4}=}
=
2
(
4
2
0
,
25
+
4
2
+
0
,
25
2
ln
|
4
+
4
2
+
0
,
25
|
)
−
2
(
1
2
0
,
25
+
1
2
+
0
,
25
2
ln
|
1
+
1
2
+
0
,
25
|
)
=
{\displaystyle =2({4 \over 2}{\sqrt {0,25+4^{2}}}+{\frac {0,25}{2}}\ln \left|4+{\sqrt {4^{2}+0,25}}\right|)-2({1 \over 2}{\sqrt {0,25+1^{2}}}+{\frac {0,25}{2}}\ln \left|1+{\sqrt {1^{2}+0,25}}\right|)=}
=
2
(
2
16
,
25
+
0
,
125
ln
|
4
+
16
,
25
|
)
−
2
(
1
2
1
,
25
+
0
,
125
ln
|
1
+
1
,
25
|
)
=
{\displaystyle =2(2{\sqrt {16,25}}+0,125\ln \left|4+{\sqrt {16,25}}\right|)-2({1 \over 2}{\sqrt {1,25}}+0,125\ln \left|1+{\sqrt {1,25}}\right|)=}
≈
2
(
2
⋅
4
,
031128874
+
0
,
125
ln
|
8
,
031128874
|
)
−
(
1
,
118033989
+
0
,
25
⋅
ln
|
1
+
1
,
118033989
|
)
≈
{\displaystyle \approx 2(2\cdot 4,031128874+0,125\ln \left|8,031128874\right|)-(1,118033989+0,25\cdot \ln \left|1+1,118033989\right|)\approx }
≈
2
(
8
,
062257748
+
0
,
125
⋅
2
,
0833251
)
−
(
1
,
118033989
+
0
,
25
⋅
0
,
750488294
)
=
{\displaystyle \approx 2(8,062257748+0,125\cdot 2,0833251)-(1,118033989+0,25\cdot 0,750488294)=}
=
2
(
8
,
062257748
+
0
,
260415637
)
−
(
1
,
118033989
+
0
,
187622073
)
=
16
,
64534677
−
1
,
305656063
=
15
,
33969071.
{\displaystyle =2(8,062257748+0,260415637)-(1,118033989+0,187622073)=16,64534677-1,305656063=15,33969071.}
Palyginimui, linijos ilgis nuo taško (1; 1) iki taško (4; 16) yra
c
=
3
2
+
15
2
=
234
=
15
,
29705854.
{\displaystyle c={\sqrt {3^{2}+15^{2}}}={\sqrt {234}}=15,29705854.}
Apskaičiuosime parabolės
y
=
(
x
+
1
)
3
{\displaystyle y={\sqrt {(x+1)^{3}}}}
lanko ilgį, kai
4
≤
x
≤
12.
{\displaystyle 4\leq x\leq 12.}
Randame
y
′
=
3
2
x
+
1
,
d
s
=
1
+
(
d
y
d
x
)
2
d
x
=
1
+
9
4
(
x
+
1
)
d
x
=
13
4
+
9
4
x
d
x
.
{\displaystyle y'={3 \over 2}{\sqrt {x+1}},\;ds={\sqrt {1+({dy \over dx})^{2}}}dx={\sqrt {1+{9 \over 4}(x+1)}}dx={\sqrt {{13 \over 4}+{9 \over 4}x}}dx.}
Tada iš integralų lentelės
L
=
∫
L
d
s
=
∫
4
12
13
4
+
9
4
x
d
x
=
∫
4
12
1
4
⋅
13
+
9
x
d
x
=
1
9
⋅
1
2
∫
4
12
13
+
9
x
d
(
13
+
9
x
)
=
{\displaystyle L=\int _{L}ds=\int _{4}^{12}{\sqrt {{13 \over 4}+{9 \over 4}x}}\;dx=\int _{4}^{12}{\sqrt {1 \over 4}}\cdot {\sqrt {13+9x}}dx={1 \over 9}\cdot {1 \over 2}\int _{4}^{12}{\sqrt {13+9x}}d(13+9x)=}
=
1
18
⋅
(
13
+
9
x
)
1
2
+
1
3
2
|
4
12
=
{\displaystyle ={1 \over 18}\cdot {(13+9x)^{{1 \over 2}+1} \over {3 \over 2}}|_{4}^{12}=}
=
2
18
⋅
3
(
13
+
9
x
)
3
|
4
12
=
1
27
[
(
13
+
9
⋅
12
)
3
−
(
13
+
9
⋅
4
)
3
]
=
{\displaystyle ={2 \over 18\cdot 3}{\sqrt {(13+9x)^{3}}}|_{4}^{12}={1 \over 27}[{\sqrt {(13+9\cdot 12)^{3}}}-{\sqrt {(13+9\cdot 4)^{3}}}]=}
=
1
27
[
1771561
−
117649
]
=
1
27
⋅
(
1331
−
343
)
=
988
27
=≈
36
,
59259259.
{\displaystyle ={1 \over 27}[{\sqrt {1771561}}-{\sqrt {117649}}]={1 \over 27}\cdot (1331-343)={988 \over 27}=\approx 36,59259259.}
Palyginimui, atkrapos ilgis nuo taško (4;
(
4
+
1
)
3
/
2
{\displaystyle (4+1)^{3/2}}
) iki taško (12;
(
12
+
1
)
3
/
2
{\displaystyle (12+1)^{3/2}}
) yra
c
=
(
12
−
4
)
2
+
[
(
12
+
1
)
3
−
(
4
+
1
)
3
]
2
=
{\displaystyle c={\sqrt {(12-4)^{2}+[{\sqrt {(12+1)^{3}}}-{\sqrt {(4+1)^{3}}}]^{2}}}=}
=
8
2
+
(
2197
−
125
)
2
=
64
+
1273
,
906493
=
36
,
57740413.
{\displaystyle ={\sqrt {8^{2}+({\sqrt {2197}}-{\sqrt {125}})^{2}}}={\sqrt {64+1273,906493}}=36,57740413.}
Apskaičiuosime kreivės
y
=
x
{\displaystyle y={\sqrt {x}}}
lanko ilgį, kai
1
≤
x
≤
16.
{\displaystyle 1\leq x\leq 16.}
Randame
y
′
=
(
x
1
2
)
′
=
1
2
⋅
1
x
.
{\displaystyle y'=(x^{1 \over 2})'={1 \over 2}\cdot {1 \over {\sqrt {x}}}.}
Gauname
L
=
∫
1
16
1
+
(
y
′
)
2
d
x
=
∫
1
16
1
+
(
1
2
x
)
2
d
x
=
∫
1
16
1
+
1
4
x
d
x
=
1
2
∫
1
16
4
+
1
x
d
x
=
{\displaystyle L=\int _{1}^{16}{\sqrt {1+(y')^{2}}}dx=\int _{1}^{16}{\sqrt {1+({1 \over 2{\sqrt {x}}})^{2}}}dx=\int _{1}^{16}{\sqrt {1+{1 \over 4x}}}dx={1 \over 2}\int _{1}^{16}{\sqrt {4+{1 \over x}}}dx=}
=
1
2
[
x
1
x
+
4
+
1
4
ln
(
4
x
(
1
x
+
4
+
2
)
+
1
)
]
|
1
16
=
{\displaystyle ={1 \over 2}[x{\sqrt {{1 \over x}+4}}+{1 \over 4}\ln(4x({\sqrt {{1 \over x}+4}}+2)+1)]|_{1}^{16}=}
=
1
2
[
16
1
16
+
4
+
1
4
ln
(
4
⋅
16
(
1
16
+
4
+
2
)
+
1
)
]
−
1
2
[
1
1
1
+
4
+
1
4
ln
(
4
(
1
1
+
4
+
2
)
+
1
)
]
=
{\displaystyle ={1 \over 2}[16{\sqrt {{1 \over 16}+4}}+{1 \over 4}\ln(4\cdot 16({\sqrt {{1 \over 16}+4}}+2)+1)]-{1 \over 2}[1{\sqrt {{1 \over 1}+4}}+{1 \over 4}\ln(4({\sqrt {{1 \over 1}+4}}+2)+1)]=}
=
1
2
[
16
⋅
2
,
015564437
+
1
4
ln
(
64
(
2
,
015564437
+
2
)
+
1
)
]
−
1
2
[
5
+
1
4
ln
(
4
(
5
+
2
)
+
1
)
]
=
{\displaystyle ={1 \over 2}[16\cdot 2,015564437+{1 \over 4}\ln(64(2,015564437+2)+1)]-{1 \over 2}[{\sqrt {5}}+{1 \over 4}\ln(4({\sqrt {5}}+2)+1)]=}
=
1
2
[
32
,
24903099
+
1
4
ln
(
256
,
996124
+
1
)
]
−
1
2
[
2
,
236067978
+
1
4
ln
(
17
,
94427191
)
]
=
{\displaystyle ={1 \over 2}[32,24903099+{1 \over 4}\ln(256,996124+1)]-{1 \over 2}[2,236067978+{1 \over 4}\ln(17,94427191)]=}
=
1
2
[
32
,
24903099
+
1
4
⋅
5
,
552944561
]
−
1
2
[
2
,
236067978
+
2
,
88727095
4
]
=
{\displaystyle ={1 \over 2}[32,24903099+{1 \over 4}\cdot 5,552944561]-{1 \over 2}[2,236067978+{2,88727095 \over 4}]=}
=
16
,
81863357
−
1
,
478942858
=
15
,
33969071.
{\displaystyle =16,81863357-1,478942858=15,33969071.}
Palyginimui, tiesės ilgis nuo taško (1; 1) iki taško (16; 4) yra
c
=
(
16
−
1
)
2
+
(
4
−
1
)
2
=
225
+
9
=
15
,
29705854.
{\displaystyle c={\sqrt {(16-1)^{2}+(4-1)^{2}}}={\sqrt {225+9}}=15,29705854.}
Apskaičiuosime kreivinį integralą
∫
A
B
y
d
l
,
{\displaystyle \int _{AB}y\;dl,}
kur AB - parabolės
y
2
=
2
x
{\displaystyle y^{2}=2x}
lankas nuo taško (0; 0) iki taško (2; 2).
Turime
y
=
2
x
,
y
′
=
1
2
x
,
d
l
=
1
+
y
′
2
d
x
=
1
+
1
2
x
d
x
.
{\displaystyle y={\sqrt {2x}},\;y'={1 \over {\sqrt {2x}}},\;dl={\sqrt {1+y'^{2}}}dx={\sqrt {1+{1 \over 2x}}}dx.}
Pagal pirmą formulę gauname
∫
A
B
y
d
l
=
∫
0
2
2
x
1
+
1
2
x
d
x
=
∫
0
2
2
x
+
1
d
x
=
∫
0
2
2
x
+
1
d
(
2
x
+
1
)
2
=
(
2
x
+
1
)
3
2
3
|
0
2
=
{\displaystyle \int _{AB}ydl=\int _{0}^{2}{\sqrt {2x}}{\sqrt {1+{1 \over 2x}}}dx=\int _{0}^{2}{\sqrt {2x+1}}dx=\int _{0}^{2}{\sqrt {2x+1}}{d(2x+1) \over 2}={(2x+1)^{3 \over 2} \over 3}|_{0}^{2}=}
=
5
5
−
1
3
.
{\displaystyle ={5{\sqrt {5}}-1 \over 3}.}
Apskaičiuosime kreivės lanko ilgį , kai
y
=
1
−
ln
cos
x
;
{\displaystyle y=1-\ln \cos x;}
0
≤
x
≤
π
3
.
{\displaystyle 0\leq x\leq {\pi \over 3}.}
Iš integralų lnetelės :
l
=
∫
0
π
3
1
+
sin
2
x
cos
2
x
d
x
=
∫
0
π
3
d
x
cos
x
=
ln
|
tan
(
x
2
+
π
4
)
|
|
0
π
3
=
ln
|
tan
5
π
12
|
−
ln
|
tan
π
4
|
=
{\displaystyle l=\int _{0}^{\pi \over 3}{\sqrt {1+{\sin ^{2}x \over \cos ^{2}x}}}dx=\int _{0}^{\pi \over 3}{dx \over \cos x}=\ln \left|\tan \left({\frac {x}{2}}+{\frac {\pi }{4}}\right)\right||_{0}^{\pi \over 3}=\ln |\tan {5\pi \over 12}|-\ln |\tan {\pi \over 4}|=}
=
1
,
316957897.
{\displaystyle =1,316957897.}
Apskaičiuosime integralą
∫
A
B
x
2
d
s
,
{\displaystyle \int _{AB}x^{2}ds,}
kur AB - dalis logoritminės kreivės
y
=
ln
x
{\displaystyle y=\ln x}
nuo
x
=
1
{\displaystyle x=1}
iki
x
=
2.
{\displaystyle x=2.}
Pagal pirmą formulę
∫
A
B
x
2
d
s
=
∫
1
2
x
2
1
+
1
x
2
d
x
=
∫
1
2
x
x
2
+
1
d
x
=
∫
1
2
x
x
2
+
1
d
(
x
2
+
1
)
2
x
=
1
3
(
1
+
x
2
)
3
2
|
x
=
1
x
=
2
=
{\displaystyle \int _{AB}x^{2}ds=\int _{1}^{2}x^{2}{\sqrt {1+{1 \over x^{2}}}}dx=\int _{1}^{2}x{\sqrt {x^{2}+1}}dx=\int _{1}^{2}x{\sqrt {x^{2}+1}}{d(x^{2}+1) \over 2x}={1 \over 3}(1+x^{2})^{3 \over 2}|_{x=1}^{x=2}=}
=
1
3
(
5
5
−
2
2
)
,
{\displaystyle ={1 \over 3}(5{\sqrt {5}}-2{\sqrt {2}}),}
kur
d
(
x
2
+
1
)
=
2
x
d
x
.
{\displaystyle d(x^{2}+1)=2xdx.}
Apskaičiuokime kreivės
y
2
=
4
9
x
2
,
{\displaystyle y^{2}={4 \over 9}x^{2},}
(
x
∈
[
3
;
8
]
)
{\displaystyle (x\in [3;8])}
lanko L masę, kai tankis kreivės taške yra tiesiog proporcingas to taško ordinatei (y ) ir atvirkščiai proporcingas kvadratinei šakniai iš to taško abscisės (
1
/
x
1
/
2
{\displaystyle 1/x^{1/2}}
), be to, taške
(
4
;
16
3
)
{\displaystyle (4;{16 \over 3})}
jo (tankio) reikšmė lygi 8 g/cm.
Kreivės lanko masę, kai to lanko tankis lygus
γ
(
x
,
y
)
{\displaystyle \gamma (x,y)}
, apskaičiuosime pagal formulę
m
=
∫
L
γ
(
x
,
y
)
d
s
.
{\displaystyle m=\int _{L}\gamma (x,y)ds.}
Pagal uždavinio sąlyga, tankis lygus
γ
(
x
,
y
)
=
k
y
x
;
{\displaystyle \gamma (x,y)={ky \over {\sqrt {x}}};}
čia k - proporcingumo koeficientas. Kadangi
γ
=
8
{\displaystyle \gamma =8}
, kai
x
=
4
,
{\displaystyle x=4,}
y
=
16
3
,
{\displaystyle y={16 \over 3},}
tai iš lygybės
8
=
k
16
3
4
=
8
k
3
{\displaystyle 8=k{{16 \over 3} \over {\sqrt {4}}}={8k \over 3}}
gauname: k=3. Tuomet, pagal formule,
m
=
∫
L
γ
(
x
,
y
)
d
s
=
3
∫
L
y
x
d
s
.
{\displaystyle m=\int _{L}\gamma (x,y)ds=3\int _{L}{y \over {\sqrt {x}}}ds.}
Norėdami apskaičiuoti šį integralą, taikysime pirmą formulę. Iš sąlygos
y
=
2
3
x
x
{\displaystyle y={2 \over 3}x{\sqrt {x}}}
turime
y
′
=
x
{\displaystyle y'={\sqrt {x}}}
ir
d
s
=
1
+
(
y
′
)
2
d
x
=
1
+
x
d
x
.
{\displaystyle ds={\sqrt {1+(y')^{2}}}dx={\sqrt {1+x}}dx.}
Tuomet
m
=
3
∫
3
8
2
3
x
x
x
1
+
x
d
x
=
2
∫
3
8
x
1
+
x
d
x
=
4
∫
2
3
(
t
2
−
1
)
t
2
d
t
=
4
(
t
5
5
−
t
3
3
)
|
2
3
=
{\displaystyle m=3\int _{3}^{8}{{2 \over 3}x{\sqrt {x}} \over {\sqrt {x}}}{\sqrt {1+x}}dx=2\int _{3}^{8}x{\sqrt {1+x}}dx=4\int _{2}^{3}(t^{2}-1)t^{2}dt=4({t^{5} \over 5}-{t^{3} \over 3})|_{2}^{3}=}
=
4
[
243
5
−
9
−
(
32
5
−
8
3
)
]
=
4
(
211
5
−
19
3
)
=
2152
15
(
g
)
=
143.4
(
6
)
(
g
)
,
{\displaystyle =4[{243 \over 5}-9-({32 \over 5}-{8 \over 3})]=4({211 \over 5}-{19 \over 3})={2152 \over 15}\;(g)=143.4(6)\;(g),}
kur
1
+
x
=
t
,
{\displaystyle {\sqrt {1+x}}=t,}
1
+
x
=
t
2
,
{\displaystyle 1+x=t^{2},}
x
=
t
2
−
1
,
{\displaystyle x=t^{2}-1,}
d
x
=
2
t
d
t
,
{\displaystyle dx=2tdt,}
t
1
=
2
,
{\displaystyle t_{1}=2,}
t
2
=
3.
{\displaystyle t_{2}=3.}
Arba galėjome apskaičiuoti integruodami dalimis :
m
=
2
∫
3
8
x
1
+
x
d
x
=
2
x
⋅
2
3
(
1
+
x
)
3
2
|
3
8
−
2
∫
3
8
2
3
(
1
+
x
)
3
2
d
x
=
{\displaystyle m=2\int _{3}^{8}x{\sqrt {1+x}}dx=2x\cdot {2 \over 3}(1+x)^{3 \over 2}|_{3}^{8}-2\int _{3}^{8}{2 \over 3}(1+x)^{3 \over 2}dx=}
=
4
3
x
(
1
+
x
)
1
+
x
|
3
8
−
4
3
⋅
2
5
(
1
+
x
)
5
2
|
3
8
=
(
32
3
⋅
9
⋅
3
−
4
⋅
4
⋅
2
)
−
(
8
15
⋅
3
5
−
8
15
⋅
2
5
)
=
{\displaystyle ={4 \over 3}x(1+x){\sqrt {1+x}}|_{3}^{8}-{4 \over 3}\cdot {2 \over 5}(1+x)^{5 \over 2}|_{3}^{8}=({32 \over 3}\cdot 9\cdot 3-4\cdot 4\cdot 2)-({8 \over 15}\cdot 3^{5}-{8 \over 15}\cdot 2^{5})=}
=
(
288
−
32
)
−
(
1994
15
−
256
15
)
=
256
−
1688
15
=
2152
15
(
g
)
=
143.4
(
6
)
(
g
)
,
{\displaystyle =(288-32)-({1994 \over 15}-{256 \over 15})=256-{1688 \over 15}={2152 \over 15}\;(g)=143.4(6)\;(g),}
kur
u
=
x
,
{\displaystyle u=x,}
d
v
=
1
+
x
,
{\displaystyle dv={\sqrt {1+x}},}
d
u
=
d
x
,
{\displaystyle du=dx,}
v
=
∫
(
1
+
x
)
0.5
d
x
=
2
3
(
1
+
x
)
3
2
.
{\displaystyle v=\int (1+x)^{0.5}dx={2 \over 3}(1+x)^{3 \over 2}.}
Apskaičiuokime integralą
∫
L
x
d
s
,
{\displaystyle \int _{L}xds,}
kai L - pirmoji cikloidės
x
=
a
(
t
−
sin
t
)
,
{\displaystyle x=a(t-\sin t),}
y
=
a
(
1
−
cos
t
)
{\displaystyle y=a(1-\cos t)}
arka.
Taikome antrą formulę. Randame:
x
t
′
=
a
(
1
−
cos
t
)
,
{\displaystyle x_{t}'=a(1-\cos t),}
y
t
′
=
a
sin
t
,
{\displaystyle y_{t}'=a\sin t,}
d
s
=
x
t
′
2
+
y
t
′
2
d
t
=
a
2
(
1
−
cos
t
)
2
+
a
2
sin
2
t
d
t
=
a
1
−
2
cos
t
+
cos
2
+
sin
2
t
d
t
=
{\displaystyle ds={\sqrt {x_{t}'^{2}+y_{t}'^{2}}}dt={\sqrt {a^{2}(1-\cos t)^{2}+a^{2}\sin ^{2}t}}dt=a{\sqrt {1-2\cos t+\cos ^{2}+\sin ^{2}t}}dt=}
=
a
2
−
2
cos
t
d
t
=
a
2
−
2
(
1
−
2
sin
2
t
2
)
d
t
=
2
a
sin
t
2
d
t
.
{\displaystyle =a{\sqrt {2-2\cos t}}dt=a{\sqrt {2-2(1-2\sin ^{2}{t \over 2})}}dt=2a\sin {t \over 2}dt.}
Tuomet
∫
L
d
s
=
2
a
2
∫
0
2
π
(
t
−
sin
t
)
sin
t
2
d
t
=
2
a
2
(
∫
0
2
π
t
sin
t
2
d
t
−
∫
0
2
π
sin
t
sin
t
2
d
t
)
.
{\displaystyle \int _{L}ds=2a^{2}\int _{0}^{2\pi }(t-\sin t)\sin {t \over 2}dt=2a^{2}(\int _{0}^{2\pi }t\sin {t \over 2}dt-\int _{0}^{2\pi }\sin t\sin {t \over 2}dt).}
Pirmąjį integralą integruojame dalimis , pažymėdami
u
=
t
,
{\displaystyle u=t,}
d
v
=
sin
t
2
d
t
,
{\displaystyle dv=\sin {t \over 2}dt,}
d
u
=
d
t
,
{\displaystyle du=dt,}
v
=
∫
sin
t
2
d
t
=
−
2
cos
t
2
,
{\displaystyle v=\int \sin {t \over 2}dt=-2\cos {t \over 2},}
gauname
∫
0
2
π
t
sin
t
2
d
t
=
−
2
(
t
cos
t
2
)
|
0
2
π
+
2
∫
0
2
π
cos
t
2
d
t
=
−
2
(
2
π
(
−
1
)
)
+
4
sin
t
2
|
0
2
π
=
4
π
.
{\displaystyle \int _{0}^{2\pi }t\sin {t \over 2}dt=-2(t\cos {t \over 2})|_{0}^{2\pi }+2\int _{0}^{2\pi }\cos {t \over 2}dt=-2(2\pi (-1))+4\sin {t \over 2}|_{0}^{2\pi }=4\pi .}
Antrąjį integralą apskaičiuojame taikydami formulę
sin
t
sin
t
2
=
(
2
sin
t
2
cos
t
2
)
sin
t
2
=
2
sin
2
t
2
cos
t
2
,
{\displaystyle \sin t\sin {t \over 2}=(2\sin {t \over 2}\cos {t \over 2})\sin {t \over 2}=2\sin ^{2}{t \over 2}\cos {t \over 2},}
d
(
sin
t
2
)
=
cos
t
2
d
(
t
2
)
,
d
(
t
2
)
=
1
2
d
t
,
d
t
=
2
d
(
t
2
)
,
{\displaystyle d(\sin {t \over 2})=\cos {t \over 2}d({t \over 2}),\;d({t \over 2})={1 \over 2}dt,\;dt=2d({t \over 2}),}
reiškia
∫
0
2
π
sin
t
sin
t
2
d
t
=
∫
0
2
π
2
sin
2
t
2
cos
t
2
2
d
(
t
2
)
=
4
∫
0
2
π
sin
2
t
2
d
(
sin
t
2
)
=
4
3
sin
3
t
2
|
0
2
π
=
0.
{\displaystyle \int _{0}^{2\pi }\sin t\sin {t \over 2}dt=\int _{0}^{2\pi }2\sin ^{2}{t \over 2}\cos {t \over 2}\;2d({t \over 2})=4\int _{0}^{2\pi }\sin ^{2}{t \over 2}\;d(\sin {t \over 2})={4 \over 3}\sin ^{3}{t \over 2}|_{0}^{2\pi }=0.}
Todėl bendras integralas lygus:
∫
L
x
d
s
=
2
a
2
(
4
π
−
0
)
=
8
π
a
2
.
{\displaystyle \int _{L}xds=2a^{2}(4\pi -0)=8\pi a^{2}.}
cikloidė
Apskaičiuokime cikloidės
x
=
a
(
t
−
sin
t
)
,
{\displaystyle x=a(t-\sin t),}
y
=
a
(
1
−
cos
t
)
{\displaystyle y=a(1-\cos t)}
(
a
>
0
)
{\displaystyle (a>0)}
pirmosios arkos ilgį.
Pirmoji cikloidės arka gaunama, kai parametras t kinta nuo 0 iki
2
π
.
{\displaystyle 2\pi .}
Randame:
x
t
′
=
a
(
1
−
cos
t
)
,
{\displaystyle x_{t}'=a(1-\cos t),}
y
t
′
=
a
sin
t
,
{\displaystyle y_{t}'=a\sin t,}
x
t
′
2
+
y
t
′
2
=
a
2
(
1
−
2
cos
t
+
cos
2
t
)
+
a
2
sin
2
t
=
2
a
2
(
1
−
cos
t
)
=
{\displaystyle {\sqrt {x_{t}'^{2}+y_{t}'^{2}}}={\sqrt {a^{2}(1-2\cos t+\cos ^{2}t)+a^{2}\sin ^{2}t}}={\sqrt {2a^{2}(1-\cos t)}}=}
=
4
a
2
sin
2
t
t
2
=
2
a
|
sin
t
2
|
=
2
a
sin
t
2
,
{\displaystyle ={\sqrt {4a^{2}\sin ^{2}t{t \over 2}}}=2a|\sin {t \over 2}|=2a\sin {t \over 2},}
nes
sin
t
2
≥
0
,
{\displaystyle \sin {t \over 2}\geq 0,}
kai
t
∈
[
0
;
2
π
]
.
{\displaystyle t\in [0;2\pi ].}
Tuomet
L
=
2
a
∫
0
2
π
sin
t
2
d
t
=
4
a
∫
0
2
π
sin
t
2
d
(
t
/
2
)
=
−
4
a
cos
t
2
|
0
2
π
=
8
a
.
{\displaystyle L=2a\int _{0}^{2\pi }\sin {t \over 2}dt=4a\int _{0}^{2\pi }\sin {t \over 2}d(t/2)=-4a\cos {t \over 2}|_{0}^{2\pi }=8a.}
Apskaičiuosime kreivinį integralą
∫
A
B
y
2
d
l
,
{\displaystyle \int _{AB}y^{2}dl,}
kur AB - dalis apskritimo
x
=
a
cos
t
,
{\displaystyle x=a\cos t,}
y
=
a
sin
t
,
{\displaystyle y=a\sin t,}
0
≤
t
≤
π
2
.
{\displaystyle 0\leq t\leq {\pi \over 2}.}
Kadangi
y
2
=
a
2
sin
2
t
,
{\displaystyle y^{2}=a^{2}\sin ^{2}t,}
d
l
=
a
2
sin
2
t
+
a
2
cos
2
t
d
t
=
a
d
t
,
{\displaystyle dl={\sqrt {a^{2}\sin ^{2}t+a^{2}\cos ^{2}t}}dt=a\;dt,}
tai pagal antrą formulę gauname
∫
A
B
y
2
d
l
=
∫
0
π
/
2
a
2
sin
2
t
⋅
a
d
t
=
a
3
2
∫
0
π
2
(
1
−
cos
(
2
t
)
)
d
t
=
a
3
2
(
t
−
sin
(
2
t
)
2
)
|
0
π
2
=
a
3
π
4
.
{\displaystyle \int _{AB}y^{2}dl=\int _{0}^{\pi /2}a^{2}\sin ^{2}t\cdot a\;dt={a^{3} \over 2}\int _{0}^{\pi \over 2}(1-\cos(2t))dt={a^{3} \over 2}(t-{\sin(2t) \over 2})|_{0}^{\pi \over 2}={a^{3}\pi \over 4}.}
Rasime lanko AB ilgį susuktos linijos
x
=
cos
t
,
{\displaystyle x=\cos t,}
y
=
sin
t
,
{\displaystyle y=\sin t,}
z
=
2
t
,
{\displaystyle z=2t,}
0
≤
t
≤
π
.
{\displaystyle 0\leq t\leq \pi .}
Pagal trečią formulę:
s
=
∫
0
π
sin
2
t
+
cos
2
t
+
4
d
t
=
5
π
.
{\displaystyle s=\int _{0}^{\pi }{\sqrt {\sin ^{2}t+\cos ^{2}t+4}}dt={\sqrt {5}}\pi .}
Reikia apskaičiuoti integralą
∫
A
B
(
x
2
+
y
2
+
z
2
)
d
s
{\displaystyle \int _{AB}(x^{2}+y^{2}+z^{2})ds}
pagal vien1 viją susuktos linijos:
x
=
cos
t
,
{\displaystyle x=\cos t,}
y
=
sin
t
,
{\displaystyle y=\sin t,}
z
=
t
,
{\displaystyle z=t,}
0
≤
t
≤
2
π
.
{\displaystyle 0\leq t\leq 2\pi .}
Pagal trečią formulę gauname:
∫
A
B
(
x
2
+
y
2
+
z
2
)
d
s
=
∫
0
2
π
(
cos
2
t
+
sin
2
t
+
t
2
)
(
−
sin
2
t
)
2
+
(
cos
t
)
2
+
1
d
t
=
2
∫
0
2
π
(
1
+
t
2
)
d
t
=
{\displaystyle \int _{AB}(x^{2}+y^{2}+z^{2})ds=\int _{0}^{2\pi }(\cos ^{2}t+\sin ^{2}t+t^{2}){\sqrt {(-\sin ^{2}t)^{2}+(\cos t)^{2}+1}}dt={\sqrt {2}}\int _{0}^{2\pi }(1+t^{2})dt=}
=
2
(
t
+
t
3
3
|
0
2
π
=
2
2
π
(
1
+
4
π
2
3
)
.
{\displaystyle ={\sqrt {2}}(t+{t^{3} \over 3}|_{0}^{2\pi }=2{\sqrt {2}}\pi (1+{4\pi ^{2} \over 3}).}
Rasime lanko ilgį kardiodės
ρ
=
a
(
1
−
cos
ϕ
)
,
{\displaystyle \rho =a(1-\cos \phi ),}
a
>
0.
{\displaystyle a>0.}
Pagal ketvirtą formulę turime:
s
=
∫
α
β
ρ
2
+
ρ
′
2
d
ϕ
=
2
a
∫
0
π
(
1
−
cos
ϕ
)
2
+
sin
2
ϕ
d
ϕ
=
{\displaystyle s=\int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+\rho '^{2}}}d\phi =2a\int _{0}^{\pi }{\sqrt {(1-\cos \phi )^{2}+\sin ^{2}\phi }}d\phi =}
=
2
a
∫
0
π
1
−
2
cos
ϕ
+
cos
2
ϕ
+
sin
2
ϕ
d
ϕ
=
2
a
∫
0
π
2
(
1
−
cos
ϕ
)
d
ϕ
=
2
a
∫
0
π
4
sin
2
ϕ
2
d
ϕ
=
{\displaystyle =2a\int _{0}^{\pi }{\sqrt {1-2\cos \phi +\cos ^{2}\phi +\sin ^{2}\phi }}d\phi =2a\int _{0}^{\pi }{\sqrt {2(1-\cos \phi )}}d\phi =2a\int _{0}^{\pi }{\sqrt {4\sin ^{2}{\phi \over 2}}}d\phi =}
=
4
a
∫
0
π
sin
ϕ
2
=
−
8
a
cos
ϕ
2
|
0
π
=
−
8
a
(
0
−
1
)
=
8
a
.
{\displaystyle =4a\int _{0}^{\pi }\sin {\phi \over 2}=-8a\cos {\phi \over 2}|_{0}^{\pi }=-8a(0-1)=8a.}
Rasime kreivės lanko ilgį, kai
ρ
=
sin
3
ϕ
3
,
0
≤
ϕ
≤
π
2
.
{\displaystyle \rho =\sin ^{3}{\phi \over 3},\;0\leq \phi \leq {\pi \over 2}.}
Pagal ketvirtą formulę:
l
=
∫
0
π
3
sin
6
ϕ
3
+
9
sin
4
ϕ
3
⋅
cos
2
ϕ
3
⋅
1
9
d
ϕ
=
∫
0
π
2
sin
2
ϕ
3
sin
2
ϕ
3
+
cos
2
ϕ
3
d
ϕ
=
{\displaystyle l=\int _{0}^{\pi \over 3}{\sqrt {\sin ^{6}{\phi \over 3}+9\sin ^{4}{\phi \over 3}\cdot \cos ^{2}{\phi \over 3}\cdot {1 \over 9}}}d\phi =\int _{0}^{\pi \over 2}\sin ^{2}{\phi \over 3}{\sqrt {\sin ^{2}{\phi \over 3}+\cos ^{2}{\phi \over 3}}}d\phi =}
=
1
2
∫
0
π
2
(
1
+
cos
2
ϕ
3
)
d
ϕ
=
1
2
⋅
π
2
+
1
2
⋅
3
2
sin
2
ϕ
3
|
0
π
2
=
π
4
+
3
4
⋅
3
2
=
π
4
+
3
3
8
.
{\displaystyle ={1 \over 2}\int _{0}^{\pi \over 2}(1+\cos {2\phi \over 3})d\phi ={1 \over 2}\cdot {\pi \over 2}+{1 \over 2}\cdot {3 \over 2}\sin {2\phi \over 3}|_{0}^{\pi \over 2}={\pi \over 4}+{3 \over 4}\cdot {{\sqrt {3}} \over 2}={\pi \over 4}+{3{\sqrt {3}} \over 8}.}
Apskaičiuokime
∫
L
(
x
+
y
)
d
s
,
{\displaystyle \int _{L}(x+y)ds,}
kai L - apskritimas
x
2
+
y
2
=
a
y
,
{\displaystyle x^{2}+y^{2}=ay,}
(
a
>
0
)
.
{\displaystyle (a>0).}
Integralą apskaičiuokime, Dekatro koordinates pakeitę polinėmis. Kreivės L lygtis šioje koordinačių sistemoje yra
ρ
=
a
sin
ϕ
,
{\displaystyle \rho =a\sin \phi ,}
ϕ
∈
[
0
;
π
]
.
{\displaystyle \phi \in [0;\pi ].}
Randame
ρ
′
=
a
cos
ϕ
,
{\displaystyle \rho '=a\cos \phi ,}
d
s
=
ρ
2
+
ρ
ϕ
′
2
d
ϕ
=
a
sin
2
ϕ
+
a
cos
2
ϕ
d
ϕ
=
a
d
ϕ
.
{\displaystyle ds={\sqrt {\rho ^{2}+\rho _{\phi }'^{2}}}d\phi ={\sqrt {a\sin ^{2}\phi +a\cos ^{2}\phi }}d\phi =a\;d\phi .}
Tuomet
∫
L
(
x
+
y
)
d
s
=
a
∫
0
π
(
ρ
sin
ϕ
+
ρ
cos
ϕ
)
d
ϕ
=
a
2
∫
0
π
(
sin
2
ϕ
+
sin
ϕ
cos
ϕ
)
d
ϕ
=
{\displaystyle \int _{L}(x+y)ds=a\int _{0}^{\pi }(\rho \sin \phi +\rho \cos \phi )d\phi =a^{2}\int _{0}^{\pi }(\sin ^{2}\phi +\sin \phi \cos \phi )d\phi =}
=
a
2
∫
0
π
(
1
−
cos
(
2
ϕ
)
2
+
sin
(
2
ϕ
)
2
)
d
ϕ
=
π
2
−
1
4
sin
(
2
ϕ
)
|
0
π
−
1
4
cos
(
2
ϕ
)
|
0
π
=
π
2
−
0
−
0
=
π
2
.
{\displaystyle =a^{2}\int _{0}^{\pi }({1-\cos(2\phi ) \over 2}+{\sin(2\phi ) \over 2})d\phi ={\pi \over 2}-{1 \over 4}\sin(2\phi )|_{0}^{\pi }-{1 \over 4}\cos(2\phi )|_{0}^{\pi }={\pi \over 2}-0-0={\pi \over 2}.}
Archimedo spiralė.
Apskaičiuosime ilgį pirmos vijos Archimedo spiralės:
ρ
=
a
ϕ
.
{\displaystyle \rho =a\phi .}
Pirma vija spiralės pasidaro, keičiantis poliariniui kampui
ϕ
{\displaystyle \phi }
nuo 0 iki
2
π
.
{\displaystyle 2\pi .}
Todėl pagal ketvirtą formulę ieškomas ilgis lanko yra
L
=
∫
α
β
ρ
(
ϕ
)
+
ρ
′
2
(
ϕ
)
d
ϕ
=
∫
0
2
π
a
2
ϕ
2
+
a
2
d
ϕ
=
a
∫
0
2
π
ϕ
2
+
1
d
ϕ
=
{\displaystyle L=\int _{\alpha }^{\beta }{\sqrt {\rho (\phi )+\rho '^{2}(\phi )}}d\phi =\int _{0}^{2\pi }{\sqrt {a^{2}\phi ^{2}+a^{2}}}d\phi =a\int _{0}^{2\pi }{\sqrt {\phi ^{2}+1}}d\phi =}
=
a
[
π
4
π
2
+
1
+
1
2
ln
(
2
π
+
4
π
2
+
1
)
]
=
{\displaystyle =a[\pi {\sqrt {4\pi ^{2}+1}}+{1 \over 2}\ln(2\pi +{\sqrt {4\pi ^{2}+1}})]=}
=
a
[
19
,
9876454
+
1
,
268648751
]
=
a
21
,
25629415.
{\displaystyle =a[19,9876454+1,268648751]=a21,25629415.}
Sukimo paviršiaus plotas
Plotas sukant kokia nors funkcija (pavyzdžiui, parabolę ) aplink Ox ašį apskaičiuojamas pagal formule:
S
=
2
π
∫
a
b
f
(
x
)
1
+
f
′
2
(
x
)
d
x
.
{\displaystyle S=2\pi \int _{a}^{b}f(x){\sqrt {1+f'^{2}(x)}}dx.}
Jeigu paviršius gaunamas sukimu aplink ašį Ox kreive AB , nusakomos parametrinėmis lygtimis
x
=
x
(
t
)
,
{\displaystyle x=x(t),}
y
=
y
(
t
)
,
{\displaystyle y=y(t),}
α
≤
t
≤
β
,
{\displaystyle \alpha \leq t\leq \beta ,}
ir
y
(
t
)
≥
0
,
{\displaystyle y(t)\geq 0,}
x(t) keičiasi nuo a iki b , keičiantis t nuo
α
{\displaystyle \alpha }
iki
β
{\displaystyle \beta }
,
x
(
α
)
=
a
,
{\displaystyle x(\alpha )=a,}
x
(
β
)
=
b
,
{\displaystyle x(\beta )=b,}
tai, pirmoje lygtyje pakeite
x
=
x
(
t
)
,
{\displaystyle x=x(t),}
gauname
S
=
2
π
∫
α
β
y
(
t
)
1
+
(
d
y
d
x
)
2
d
x
=
2
π
∫
α
β
y
(
t
)
x
′
2
(
t
)
+
y
′
2
(
t
)
d
t
.
{\displaystyle S=2\pi \int _{\alpha }^{\beta }y(t){\sqrt {1+({dy \over dx})^{2}}}dx=2\pi \int _{\alpha }^{\beta }y(t){\sqrt {x'^{2}(t)+y'^{2}(t)}}dt.}
Pagaliau, jeigu kreivė užduota lygtimi poliarinėse koordinatėse:
ρ
=
ρ
(
ϕ
)
,
{\displaystyle \rho =\rho (\phi ),}
α
≤
ϕ
≤
β
,
{\displaystyle \alpha \leq \phi \leq \beta ,}
kur
ρ
(
ϕ
)
{\displaystyle \rho (\phi )}
turi netrūkią išvestine ant
[
α
,
β
]
,
{\displaystyle [\alpha ,\beta ],}
tai šis atvejis susiveda į parametrinį uždavima kreivės
x
=
ρ
(
ϕ
)
cos
ϕ
,
{\displaystyle x=\rho (\phi )\cos \phi ,}
y
=
ρ
(
ϕ
)
sin
ϕ
,
{\displaystyle y=\rho (\phi )\sin \phi ,}
α
≤
ϕ
≤
β
,
{\displaystyle \alpha \leq \phi \leq \beta ,}
ir antra formulė priima pavidalą
S
=
2
π
∫
α
β
ρ
(
ϕ
)
sin
ϕ
ρ
2
(
ϕ
)
+
ρ
′
2
(
ϕ
)
d
ϕ
.
{\displaystyle S=2\pi \int _{\alpha }^{\beta }\rho (\phi )\sin \phi {\sqrt {\rho ^{2}(\phi )+\rho '^{2}(\phi )}}d\phi .}
Pavyzdžiai
Apskaičiuosime plotą S paviršiaus rutulinio pusiaujo, atsiradusio dėl sukimo pusiauapskritimio,
f
(
x
)
=
R
2
−
x
2
,
−
R
<
a
≤
x
≤
b
<
R
,
{\displaystyle f(x)={\sqrt {R^{2}-x^{2}}},\;-R<a\leq x\leq b<R,}
aplink ašį Ox . Pagal pirmą formulę gauname
S
=
2
π
∫
a
b
R
2
−
x
2
1
+
x
2
R
2
−
x
2
d
x
=
2
π
∫
a
b
R
2
−
x
2
R
R
2
−
x
2
d
x
=
2
π
∫
a
b
R
d
x
=
{\displaystyle S=2\pi \int _{a}^{b}{\sqrt {R^{2}-x^{2}}}{\sqrt {1+{x^{2} \over R^{2}-x^{2}}}}dx=2\pi \int _{a}^{b}{\sqrt {R^{2}-x^{2}}}{\sqrt {R \over R^{2}-x^{2}}}dx=2\pi \int _{a}^{b}Rdx=}
=
2
π
R
(
b
−
a
)
=
2
π
R
h
.
{\displaystyle =2\pi R(b-a)=2\pi Rh.}
Apskaičiuosime plotą S , paviršiaus, gauto sukimu vienos arkos cikloidės
x
=
a
(
t
−
sin
t
)
,
{\displaystyle x=a(t-\sin t),}
y
=
(
1
−
cos
t
)
,
{\displaystyle y=(1-\cos t),}
0
≤
t
≤
2
π
,
{\displaystyle 0\leq t\leq 2\pi ,}
aplink ašį Ox . Pagal antrą formulę turime
S
=
2
π
∫
0
2
π
a
(
1
−
cos
t
)
(
a
(
1
−
cos
t
)
)
2
+
(
a
sin
t
)
2
d
t
=
{\displaystyle S=2\pi \int _{0}^{2\pi }a(1-\cos t){\sqrt {(a(1-\cos t))^{2}+(a\sin t)^{2}}}dt=}
=
2
π
∫
0
2
π
a
(
1
−
cos
t
)
a
2
(
1
−
2
cos
t
+
cos
2
t
)
+
a
2
sin
2
t
d
t
=
2
π
∫
0
2
π
a
(
1
−
cos
t
)
a
2
(
2
−
2
cos
t
)
d
t
=
{\displaystyle =2\pi \int _{0}^{2\pi }a(1-\cos t){\sqrt {a^{2}(1-2\cos t+\cos ^{2}t)+a^{2}\sin ^{2}t}}dt=2\pi \int _{0}^{2\pi }a(1-\cos t){\sqrt {a^{2}(2-2\cos t)}}dt=}
=
2
2
π
a
2
∫
0
2
π
(
1
−
cos
t
)
3
2
d
t
=
2
2
π
a
2
∫
0
2
π
(
2
sin
2
t
2
)
3
2
d
t
=
8
π
a
2
∫
0
2
π
sin
3
t
2
d
t
=
{\displaystyle =2{\sqrt {2}}\pi a^{2}\int _{0}^{2\pi }(1-\cos t)^{3 \over 2}dt=2{\sqrt {2}}\pi a^{2}\int _{0}^{2\pi }(2\sin ^{2}{t \over 2})^{3 \over 2}dt=8\pi a^{2}\int _{0}^{2\pi }\sin ^{3}{t \over 2}dt=}
=
32
π
a
2
∫
0
π
2
sin
3
t
2
d
(
t
2
)
=
32
π
a
2
⋅
2
!
!
3
!
!
=
64
3
π
a
2
,
{\displaystyle =32\pi a^{2}\int _{0}^{\pi \over 2}\sin ^{3}{t \over 2}d({t \over 2})=32\pi a^{2}\cdot {2!! \over 3!!}={64 \over 3}\pi a^{2},}
kur pasinaudojome dvigubu faktorialu .
Arba galima buvo šį integralą sudorot paprastai:
S
=
8
π
a
2
∫
0
2
π
sin
3
t
2
d
t
=
8
π
a
2
∫
0
2
π
3
sin
t
2
−
sin
3
t
2
4
d
t
=
{\displaystyle S=8\pi a^{2}\int _{0}^{2\pi }\sin ^{3}{t \over 2}dt=8\pi a^{2}\int _{0}^{2\pi }{3\sin {t \over 2}-\sin {3t \over 2} \over 4}dt=}
=
4
π
a
2
∫
0
2
π
3
sin
t
2
d
(
t
2
)
−
4
3
π
a
2
∫
0
2
π
sin
3
t
2
d
(
3
t
2
)
=
−
12
π
a
2
cos
t
2
|
0
2
π
+
4
3
π
a
2
cos
3
t
2
|
0
2
π
=
{\displaystyle =4\pi a^{2}\int _{0}^{2\pi }3\sin {t \over 2}d({t \over 2})-{4 \over 3}\pi a^{2}\int _{0}^{2\pi }\sin {3t \over 2}d({3t \over 2})=-12\pi a^{2}\cos {t \over 2}|_{0}^{2\pi }+{4 \over 3}\pi a^{2}\cos {3t \over 2}|_{0}^{2\pi }=}
=
−
12
π
a
2
⋅
(
−
2
)
+
4
3
π
a
2
⋅
(
−
2
)
=
64
3
π
a
2
.
{\displaystyle =-12\pi a^{2}\cdot (-2)+{4 \over 3}\pi a^{2}\cdot (-2)={64 \over 3}\pi a^{2}.}