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db -> differenzierbar

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Martin Thoma 2012-12-06 22:21:33 +01:00
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documents/Analysis III/Kapitel-13.tex
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@ -41,7 +41,7 @@ und $\gamma(t) = (R\cos t,R\sin t)$, für $t\in[0,2\pi]$, dann gilt:
\begin{beweis}
Wir beweisen nur (1). ((2) beweist man analog und (3) folgt aus (1) und (2))\\
O.B.d.A: $(x_0,y_0) = (0,0)$ und $R$ stetig db. Also $\gamma = (\gamma_1,\gamma_2)$, $\gamma (t) = (\underbrace{R(t)\cos t}_{= \gamma_1(t)},\underbrace{R(t)\sin t)}_{=\gamma_2(t)}$. $R$ stetig differenzierbar. $A:= \int_B u_x(x,y)d(x,y)$\\
O.B.d.A: $(x_0,y_0) = (0,0)$ und $R$ stetig differenzierbar. Also $\gamma = (\gamma_1,\gamma_2)$, $\gamma (t) = (\underbrace{R(t)\cos t}_{= \gamma_1(t)},\underbrace{R(t)\sin t)}_{=\gamma_2(t)}$. $R$ stetig differenzierbar. $A:= \int_B u_x(x,y)d(x,y)$\\
Zu zeigen: $A=\int_0^{2\pi} u(\gamma (t))\cdot \gamma_2'(t) dt$.\\
Mit Polarkoordinaten, Transformations-Satz und Fubini:
\begin{displaymath}