Oberflächenintegral < mehrere Veränderl. < reell < Analysis < Hochschule < Mathe < Vorhilfe
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| Status: |
(Frage) beantwortet  | | Datum: | 17:19 Sa 28.11.2009 | | Autor: | babapapa |
| Aufgabe | | Bestimme die Oberfläche des Körpers, der von der xy - Ebene, der xz - Ebene der yz - Ebene und der Fläche [mm] x^2 [/mm] + [mm] y^2 [/mm] + z = 4 begrenzt wird. |
Hallo!
Folgende Überlegungen.
z = f(x,y) = 4 - [mm] x^2 [/mm] - [mm] y^2
[/mm]
[mm] x^2 [/mm] + [mm] y^2 [/mm] = [mm] r^2
[/mm]
außerdem
[mm] x^2 [/mm] + [mm] y^2 [/mm] = 4 - z =>
[mm] r^2 [/mm] = 4 - z =>
r = [mm] \wurzel{4 - z}
[/mm]
Nun weiß ich dass der Bereich folgendermaßen begrenzt ist:
0 [mm] \le [/mm] z [mm] \le [/mm] 4, r = [mm] \wurzel{4 - z}, [/mm] und 0 [mm] \le \phi \le \pi [/mm] (weil erster Quadrant)
nun habe ich eine Frage zur Koordinatendarstellung - bei den Parametern. (wenn ich mir den Plot so ansehe - sieht aus wie ein Elliptisches Paraboloid)
ich habe also:
[mm] x(r,\phi) [/mm] = r [mm] \cdot{} [/mm] cos [mm] \phi [/mm]
[mm] y(r,\phi) [/mm] = r [mm] \cdot{} [/mm] sin [mm] \phi
[/mm]
[mm] z(r,\phi) [/mm] = 4 - [mm] x^2 [/mm] - [mm] y^2
[/mm]
=>
[mm] z(r,\phi) [/mm] = 4 - (r [mm] \cdot{} [/mm] cos [mm] \phi )^2 [/mm] - (r [mm] \cdot{} [/mm] sin [mm] \phi )^2
[/mm]
[mm] z(r,\phi) [/mm] = 4 - [mm] r^2 (cos^2 \phi \cdot{} sin^2 \phi [/mm] )
[mm] z(r,\phi) [/mm] = 4 - [mm] r^2 [/mm]
Nun laut Buch:
[mm] \bruch{\partial x(r,\phi)}{\partial r} [/mm] = cos [mm] \phi
[/mm]
[mm] \bruch{\partial x(r,\phi)}{\partial \phi} [/mm] = -sin [mm] \phi [/mm] * r
[mm] \bruch{\partial y(r,\phi)}{\partial r} [/mm] = sin [mm] \phi
[/mm]
[mm] \bruch{\partial y(r,\phi)}{\partial \phi} [/mm] = cos [mm] \phi [/mm] * r
[mm] \bruch{\partial z(r,\phi)}{\partial r} [/mm] = -2r
[mm] \bruch{\partial z(r,\phi)}{\partial \phi} [/mm] = 0
E = [mm] (\bruch{\partial x(r,\phi)}{\partial r})^2 [/mm] + [mm] (\bruch{\partial y(r,\phi)}{\partial r})^2 [/mm] +
[mm] (\bruch{\partial z(r,\phi)}{\partial r})^2 [/mm] = [mm] cos^2 \phi [/mm] + [mm] sin^2 \phi [/mm] + [mm] 4r^2
[/mm]
= 1 + [mm] 4r^2
[/mm]
G = [mm] (\bruch{\partial x(r,\phi)}{\partial \phi})^2 [/mm] + [mm] (\bruch{\partial y(r,\phi)}{\partial \phi})^2 [/mm] + [mm] (\bruch{\partial z(r,\phi)}{\partial \phi})^2 [/mm] = [mm] sin^2 \phi [/mm] * [mm] r^2 [/mm] + [mm] cos^2 \phi [/mm] * [mm] r^2 [/mm] + 0 = [mm] r^2
[/mm]
F = [mm] \bruch{\partial x(r,\phi)}{\partial r} [/mm] * [mm] \bruch{\partial x(r,\phi)}{\partial \phi} [/mm] + [mm] \bruch{\partial y(r,\phi)}{\partial r} [/mm] * [mm] \bruch{\partial y(r,\phi)}{\partial \phi} [/mm] + [mm] \bruch{\partial z(r,\phi)}{\partial r} [/mm] * [mm] \bruch{\partial z(r,\phi)}{\partial \phi} [/mm] = - cos [mm] \phi [/mm] * sin [mm] \phi [/mm] * r + sin [mm] \phi [/mm] * cos [mm] \phi [/mm] * r = 0
dS = [mm] \wurzel{EG - F^2} [/mm] dr [mm] d\phi [/mm] = [mm] \wurzel{1 + 4*r^2 + r^2} [/mm] = [mm] \wurzel{1 + 5*r^2}
[/mm]
Also
[mm] \integral_{0}^{4}{\integral_{0}^{\pi}{\wurzel{1 + 5*r^2}
d\phi} dr}
[/mm]
Stimmt meine Überlegung soweit?
lg
Babapapa
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Hallo babapapa,
> Bestimme die Oberfläche des Körpers, der von der xy -
> Ebene, der xz - Ebene der yz - Ebene und der Fläche [mm]x^2[/mm] +
> [mm]y^2[/mm] + z = 4 begrenzt wird.
> Hallo!
>
> Folgende Überlegungen.
>
> z = f(x,y) = 4 - [mm]x^2[/mm] - [mm]y^2[/mm]
>
> [mm]x^2[/mm] + [mm]y^2[/mm] = [mm]r^2[/mm]
> außerdem
> [mm]x^2[/mm] + [mm]y^2[/mm] = 4 - z =>
> [mm]r^2[/mm] = 4 - z =>
> r = [mm]\wurzel{4 - z}[/mm]
>
> Nun weiß ich dass der Bereich folgendermaßen begrenzt
> ist:
> 0 [mm]\le[/mm] z [mm]\le[/mm] 4, r = [mm]\wurzel{4 - z},[/mm] und 0 [mm]\le \phi \le \pi[/mm]
> (weil erster Quadrant)
>
Für den ersten Quadraten gilt aber: [mm] 0 \le \phi \le \bruch{\pi}{2}[/mm]
(da [mm]x \ge 0[/mm] und [mm]y \ge 0[/mm])
>
> nun habe ich eine Frage zur Koordinatendarstellung - bei
> den Parametern. (wenn ich mir den Plot so ansehe - sieht
> aus wie ein Elliptisches Paraboloid)
>
> ich habe also:
>
> [mm]x(r,\phi)[/mm] = r [mm]\cdot{}[/mm] cos [mm]\phi[/mm]
> [mm]y(r,\phi)[/mm] = r [mm]\cdot{}[/mm] sin [mm]\phi[/mm]
> [mm]z(r,\phi)[/mm] = 4 - [mm]x^2[/mm] - [mm]y^2[/mm]
> =>
> [mm]z(r,\phi)[/mm] = 4 - (r [mm]\cdot{}[/mm] cos [mm]\phi )^2[/mm] - (r [mm]\cdot{}[/mm] sin
> [mm]\phi )^2[/mm]
> [mm]z(r,\phi)[/mm] = 4 - [mm]r^2 (cos^2 \phi \cdot{} sin^2 \phi[/mm]
> )
> [mm]z(r,\phi)[/mm] = 4 - [mm]r^2[/mm]
>
>
>
> Nun laut Buch:
> [mm]\bruch{\partial x(r,\phi)}{\partial r}[/mm] = cos [mm]\phi[/mm]
> [mm]\bruch{\partial x(r,\phi)}{\partial \phi}[/mm] = -sin [mm]\phi[/mm] * r
>
>
> [mm]\bruch{\partial y(r,\phi)}{\partial r}[/mm] = sin [mm]\phi[/mm]
> [mm]\bruch{\partial y(r,\phi)}{\partial \phi}[/mm] = cos [mm]\phi[/mm] * r
>
> [mm]\bruch{\partial z(r,\phi)}{\partial r}[/mm] = -2r
> [mm]\bruch{\partial z(r,\phi)}{\partial \phi}[/mm] = 0
>
> E = [mm](\bruch{\partial x(r,\phi)}{\partial r})^2[/mm] +
> [mm](\bruch{\partial y(r,\phi)}{\partial r})^2[/mm] +
> [mm](\bruch{\partial z(r,\phi)}{\partial r})^2[/mm] = [mm]cos^2 \phi[/mm] +
> [mm]sin^2 \phi[/mm] + [mm]4r^2[/mm]
> = 1 + [mm]4r^2[/mm]
>
> G = [mm](\bruch{\partial x(r,\phi)}{\partial \phi})^2[/mm] +
> [mm](\bruch{\partial y(r,\phi)}{\partial \phi})^2[/mm] +
> [mm](\bruch{\partial z(r,\phi)}{\partial \phi})^2[/mm] = [mm]sin^2 \phi[/mm]
> * [mm]r^2[/mm] + [mm]cos^2 \phi[/mm] * [mm]r^2[/mm] + 0 = [mm]r^2[/mm]
>
> F = [mm]\bruch{\partial x(r,\phi)}{\partial r}[/mm] *
> [mm]\bruch{\partial x(r,\phi)}{\partial \phi}[/mm] + [mm]\bruch{\partial y(r,\phi)}{\partial r}[/mm]
> * [mm]\bruch{\partial y(r,\phi)}{\partial \phi}[/mm] +
> [mm]\bruch{\partial z(r,\phi)}{\partial r}[/mm] * [mm]\bruch{\partial z(r,\phi)}{\partial \phi}[/mm]
> = - cos [mm]\phi[/mm] * sin [mm]\phi[/mm] * r + sin [mm]\phi[/mm] * cos [mm]\phi[/mm] * r = 0
>
> dS = [mm]\wurzel{EG - F^2}[/mm] dr [mm]d\phi[/mm] = [mm]\wurzel{1 + 4*r^2 + r^2}[/mm]
> = [mm]\wurzel{1 + 5*r^2}[/mm]
>
>
> Also
> [mm]\integral_{0}^{4}{\integral_{0}^{\pi}{\wurzel{1 + 5*r^2}
d\phi} dr}[/mm]
>
> Stimmt meine Überlegung soweit?
Die Integrationsgrenze von [mm]\phi[/mm] mußt nochmal überdenken.
>
> lg
> Babapapa
>
Gruss
MathePower
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| Status: |
(Frage) beantwortet | | Datum: | 18:41 Sa 28.11.2009 | | Autor: | babapapa |
Danke für die schnelle Antwort!
Die Parameter scheinen zu stimmen - das war meine größte Sorge.
Ja du hast recht mit [mm] \phi!
[/mm]
Vielen Dank!
lg
Babapapa
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