Werden die drei Theile dieses Abdruckes über die Länge AB fortgeführt, so erzeugt der erste
L
S
⋅
S
J
L
D
{\displaystyle \scriptstyle {\frac {LS\cdot SJ}{LD}}}
eine hyperbolische Fläche.[ 1]
Der zweite Theil
½JS
erzeugt das Rechteck
½AB · JS;
der dritte Theil
L
A
⋅
L
B
⋅
S
J
2
L
D
2
{\displaystyle \scriptstyle {\frac {LA\cdot LB\cdot SJ}{2LD^{2}}}}
bringt die Fläche
A
L
⋅
L
B
⋅
J
S
2
⋅
L
A
−
A
L
⋅
L
B
⋅
J
S
2
L
B
{\displaystyle \scriptstyle {\frac {AL\cdot LB\cdot JS}{2\cdot LA}}-{\frac {AL\cdot LB\cdot JS}{2LB}}}
[ 2] hervor.
Von der ersten Fläche subtrahire man die Summe der zweiten und dritten Fläche, welche Summe
= AB · JS = 2 · AS · JS
Fig. 115. wird; alsdann stellt der Rest die gesuchte Flache ABNA dar. Hiernach ergiebt sich auch folgende Construction der Aufgabe. In den Punkten L, A, S, B errichte man die Perpendikel
Ll, Aa, Ss = JS, Bb
und ziehe durch s zu den Asymptoten LI und LB die Hyperbel asb, welche die Perpendikel Aa und Bb in a und b schneidet. Subtrahirt man nun von der hyperbolischen Fläche Aa sb B das Rechteck 2 · AS · JS, so bleibt die gesuchte Fläche ABNA übrig.
Drittes Beispiel . Nimmt die nach den einzelnen Theilen der Kugel gerichtete Centripetalkraft im vierfachen Verhältniss der Abstände ab, so setze man
V =
P
E
4
2
⋅
A
S
3
{\displaystyle \scriptstyle {\frac {PE^{4}}{2\cdot AS^{3}}}}
worauf man, indem
PE =
2
⋅
P
S
⋅
L
D
{\displaystyle \scriptstyle {\sqrt {2\cdot PS\cdot LD}}}
gesetzt wird, für DN das Verhältniss
L
S
⋅
J
S
3
2
2
⋅
L
D
3
2
−
J
S
3
2
2
2
⋅
L
D
1
2
−
L
A
⋅
L
B
⋅
J
S
3
2
2
2
⋅
L
D
3
2
{\displaystyle \scriptstyle {\frac {LS\cdot JS^{\frac {3}{2}}}{{\sqrt {2}}\cdot LD^{\frac {3}{2}}}}-{\frac {JS^{\frac {3}{2}}}{2{\sqrt {2}}\cdot LD^{\frac {1}{2}}}}-{\frac {LA\cdot LB\cdot JS^{\frac {3}{2}}}{2{\sqrt {2}}\cdot LD^{\frac {3}{2}}}}}
erhält. Führt man die drei Theile desselben über die Länge AB fort, so erhält man für sie respective folgende Flächenräume:
2
⋅
L
S
⋅
J
S
3
2
L
A
1
2
−
2
⋅
L
S
⋅
J
S
3
2
L
B
1
2
{\displaystyle \scriptstyle {\frac {{\sqrt {2}}\cdot LS\cdot JS^{\frac {3}{2}}}{LA^{\frac {1}{2}}}}-{\frac {{\sqrt {2}}\cdot LS\cdot JS^{\frac {3}{2}}}{LB^{\frac {1}{2}}}}}
L
B
1
2
⋅
J
S
3
2
−
L
A
1
2
⋅
J
S
3
2
2
{\displaystyle \scriptstyle {\frac {LB^{\frac {1}{2}}\cdot JS^{\frac {3}{2}}-LA^{\frac {1}{2}}\cdot JS^{\frac {3}{2}}}{\sqrt {2}}}}
L
A
⋅
S
B
⋅
J
S
3
2
3
2
L
A
3
2
{\displaystyle \scriptstyle {\frac {LA\cdot SB\cdot JS^{\frac {3}{2}}}{3{\sqrt {2}}LA^{\frac {3}{2}}}}}
−
L
A
⋅
S
B
⋅
J
S
3
2
3
2
L
B
3
2
{\displaystyle \scriptstyle -{\frac {LA\cdot SB\cdot JS^{\frac {3}{2}}}{3{\sqrt {2}}LB^{\frac {3}{2}}}}}
[ 3]
↑ [589 No. 59 . S. 206 . Dies ergiebt sich unmittelbar wie im ersten Beispiel.↑ [589 No. 60 . S. 206 . Setzt man LD = x, so wird die zu findende Fläche bestimmt durch
A
L
⋅
L
B
⋅
J
S
2
∫
L
A
L
B
d
x
x
2
=
A
L
⋅
L
B
⋅
J
S
2
[
−
1
L
B
−
(
−
1
L
A
)
]
=
A
L
⋅
L
B
⋅
J
S
2
⋅
A
L
−
A
L
⋅
L
B
⋅
J
S
2
⋅
L
B
{\displaystyle \scriptstyle {\frac {AL\cdot LB\cdot JS}{2}}\int \limits _{LA}^{LB}{\frac {dx}{x^{2}}}={\frac {AL\cdot LB\cdot JS}{2}}\left[-{\frac {1}{LB}}-\left(-{\frac {1}{LA}}\right)\right]={\frac {AL\cdot LB\cdot JS}{2\cdot AL}}-{\frac {AL\cdot LB\cdot JS}{2\cdot LB}}}
↑ [589 No. 61 . S. 206 . Setzt man nämlich wieder LD = x, so erhält man nach der Reihe:
L
S
⋅
J
S
3
2
2
∫
L
A
L
B
d
x
x
3
2
=
L
S
⋅
J
S
3
2
2
[
−
2
⋅
L
B
−
1
2
+
2
⋅
L
A
−
1
2
]
{\displaystyle \scriptstyle {\frac {LS\cdot JS^{\frac {3}{2}}}{\sqrt {2}}}\int \limits _{LA}^{LB}{\frac {dx}{x^{\frac {3}{2}}}}={\frac {LS\cdot JS^{\frac {3}{2}}}{\sqrt {2}}}\left[-2\cdot LB^{-{\frac {1}{2}}}+2\cdot LA^{-{\frac {1}{2}}}\right]}
=
L
S
⋅
J
S
3
2
2
[
1
L
A
1
2
−
1
L
B
1
2
]
{\displaystyle \scriptstyle =LS\cdot JS^{\frac {3}{2}}{\sqrt {2}}\left[{\frac {1}{LA^{\frac {1}{2}}}}-{\frac {1}{LB^{\frac {1}{2}}}}\right]}
J
S
3
2
2
2
∫
L
A
L
B
d
x
x
1
2
=
J
S
3
2
2
2
2
[
L
B
1
2
−
L
A
1
2
]
{\displaystyle \scriptstyle {\frac {JS^{\frac {3}{2}}}{2{\sqrt {2}}}}\int \limits _{LA}^{LB}{\frac {dx}{x^{\frac {1}{2}}}}={\frac {JS^{\frac {3}{2}}}{2{\sqrt {2}}}}2\left[LB^{\frac {1}{2}}-LA^{\frac {1}{2}}\right]}
=
L
B
1
2
⋅
J
S
3
2
−
L
A
1
2
⋅
J
S
3
2
2
{\displaystyle \scriptstyle ={\frac {LB^{\frac {1}{2}}\cdot JS^{\frac {3}{2}}-LA^{\frac {1}{2}}\cdot JS^{\frac {3}{2}}}{\sqrt {2}}}}
L
A
⋅
L
B
⋅
J
S
3
2
2
2
∫
L
A
L
B
d
x
x
3
2
{\displaystyle \scriptstyle {\sqrt {{\frac {LA\cdot LB\cdot JS^{\frac {3}{2}}}{2{\sqrt {2}}}}\int \limits _{LA}^{LB}{\frac {dx}{x^{\frac {3}{2}}}}}}}
=
L
A
⋅
L
B
⋅
J
S
3
2
2
2
⋅
2
3
[
1
L
A
3
2
−
1
L
B
3
2
]
=
L
A
⋅
L
B
⋅
J
S
3
2
3
2
[
1
L
A
3
2
−
1
L
B
3
2
]
{\displaystyle \scriptstyle ={\frac {LA\cdot LB\cdot JS^{\frac {3}{2}}}{2{\sqrt {2}}}}\cdot {\frac {2}{3}}\left[{\frac {1}{LA^{\frac {3}{2}}}}-{\frac {1}{LB^{\frac {3}{2}}}}\right]={\frac {LA\cdot LB\cdot JS^{\frac {3}{2}}}{3{\sqrt {2}}}}\left[{\frac {1}{LA^{\frac {3}{2}}}}-{\frac {1}{LB^{\frac {3}{2}}}}\right]}
![{\displaystyle \scriptstyle {\frac {AL\cdot LB\cdot JS}{2}}\int \limits _{LA}^{LB}{\frac {dx}{x^{2}}}={\frac {AL\cdot LB\cdot JS}{2}}\left[-{\frac {1}{LB}}-\left(-{\frac {1}{LA}}\right)\right]={\frac {AL\cdot LB\cdot JS}{2\cdot AL}}-{\frac {AL\cdot LB\cdot JS}{2\cdot LB}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a8845bea2e7b3f004dc71e52e730bf7ef5fb22e)
![{\displaystyle \scriptstyle {\frac {LS\cdot JS^{\frac {3}{2}}}{\sqrt {2}}}\int \limits _{LA}^{LB}{\frac {dx}{x^{\frac {3}{2}}}}={\frac {LS\cdot JS^{\frac {3}{2}}}{\sqrt {2}}}\left[-2\cdot LB^{-{\frac {1}{2}}}+2\cdot LA^{-{\frac {1}{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d81bc6aa089d21c0bdbb29a23fd6e3f22845def5)
![{\displaystyle \scriptstyle =LS\cdot JS^{\frac {3}{2}}{\sqrt {2}}\left[{\frac {1}{LA^{\frac {1}{2}}}}-{\frac {1}{LB^{\frac {1}{2}}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24c130adadbc887668900e354a05438e3de9ce70)
![{\displaystyle \scriptstyle {\frac {JS^{\frac {3}{2}}}{2{\sqrt {2}}}}\int \limits _{LA}^{LB}{\frac {dx}{x^{\frac {1}{2}}}}={\frac {JS^{\frac {3}{2}}}{2{\sqrt {2}}}}2\left[LB^{\frac {1}{2}}-LA^{\frac {1}{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c905929f852231e4d9e9547b4499df01bc6e9ed1)
![{\displaystyle \scriptstyle ={\frac {LA\cdot LB\cdot JS^{\frac {3}{2}}}{2{\sqrt {2}}}}\cdot {\frac {2}{3}}\left[{\frac {1}{LA^{\frac {3}{2}}}}-{\frac {1}{LB^{\frac {3}{2}}}}\right]={\frac {LA\cdot LB\cdot JS^{\frac {3}{2}}}{3{\sqrt {2}}}}\left[{\frac {1}{LA^{\frac {3}{2}}}}-{\frac {1}{LB^{\frac {3}{2}}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/359411a6a48710ce3cd1f0c8b984ae9f93e07c00)
