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\lhead{10. November 2004}
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\rhead{\bfseries Vorlesung 7}

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\begin{document}


\begin{aussage}[Satz]{(Integralsatz von Green)}{}
Sei $M$$\subset$$\RR^2$ offene Menge, die sich in endl. viele
Standardbereiche zerlegen l"asst.

$ V = ( V_1, V_2 )$ stetig diffb. Vektorfeld in Umgebung von $\overline{M} = M \setminus \delta  M$
$$ \Rightarrow \int_{M} \left( \frac{\partial  V_1}{\partial  y} -
\frac{\partial  V_2}{\partial  x} \right) d(x, y) = - \int_{\delta  M} V_{1}dx +
V_{2}dy $$

Hierbei soll der Rand so durchlaufen werden, dass das Innere immer links liegt.
\\ $ $ \\
Beweis: \\
Beweise Satz f"ur jeden Standardbereich. Allg. Satz folgt durch Aufsummierung.
\\ Also o.B.d.A. $M$ ist Standardbereich. \\
W"ahle folgende explizite Parameterdarstellung von $\delta  M$
$$
\left\{
\begin{array}{c c c}
[0,4] & \leftarrow & \RR^2 \\
t & \mapsto & ( x(t), y(t) )
\end{array} \right .
$$
mit 
$$
x(t) = \left \{ \begin{array}{c c}
a+t(b-a) & 0 \leq t \leq 1 \\
b & 1\leq t \leq 2 \\
b-(t-2)(b-a) & 2\leq t \leq 3 \\
a & 3 \leq t \leq 4 
\end{array} \right .
$$
$$
y(t) = \left \{ \begin{array}{c c}
\varphi(a+t(b-a)) & 0 \leq t \leq 1 \\
\varphi(b)+(t-1)(\psi(b)-\varphi(b)) & 1\leq t \leq 2 \\
\varphi(b-(t-2)(b-a)) & 2\leq t \leq 3 \\
\varphi(a)-(t-3)(\psi(a)-\varphi(a)) & 3 \leq t \leq 4 
\end{array} \right .
$$

\begin{eqnarray*}
\int_M\frac{\partial  V_1}{\partial  y} d(x,y) & \stackrel{Satz 1}{=} &
 \int_a^b(\int_{\varphi(x)}^{\psi(x)}\frac{\partial  V_1}{\partial  y}dy)dx \\
& = & \int_a^b(V_1(x,\psi(x))-V_1(x,\varphi(x)))dx \\
& = & - \int_b^aV_1(x,\psi(x))dx - \int_a^bV_1(x, \varphi(x))dx \\
& = & - \int_2^3V_1(x(t),y(t))x'(t)dt - \int_0^1V_1(x(t),y(t))x'(t)dt \\
& = & - \int_0^4V_1(x(t),y(t))x'(t)dt \qquad (\hbox{ denn } \int_1^2V_1(x(t),y(t))x'(t)=0) \\
& = & - \int_{\delta S}V_1(x,y)dx
\end{eqnarray*}

Analog ist zu zeigen: $ - \int_M \frac{\partial V_2}{\partial  x}d(x,y) = - \int_{\delta S}V_2dy $
\phantom{hallo}\hfill$\square$
\end{aussage}
\begin{aussage}[Beispiel]{}{}
    $ M = \{ ( x, y) \in \RR^2 \mid 0 \leq x^2 + y^2 < 1\} $ \\
    $ V_{1}= 2xy^3  \quad  V_{2}= 3x^2y^2 $

    $$\int_{\delta M} 2xy^3dx + 3x^2y^2dy=
    -\int_{M} (6xy^2 - 6xy^2) d( x, y) = 0 $$
\end{aussage}
\begin{aussage}[Beispiel]{}{} 
    $M$ wie oben 

    $V_{1}=x^{4} - y^3 \quad   V_{2}=x^3 - y^{4}$


    \begin{eqnarray*}
	 \int_{\delta M} (x^{4} - y^3)dx + (x^3 - y^{4})dy & = & - \int_{M} (-3y^2 - x^2)d(x,y) \\ & = &
    3\int_M (x^2+y^2) d(x,y) \quad ( \hbox{ Polarkoord.: }    x= r cos\varphi \hbox{, } y= r sin\varphi ) \\
    & = & 3 \int_{0}^{1}(\int_{0}^{2\pi} r^2r dr)d\varphi
    = 6\pi \int_{0}^{1} r^3 dr = \frac{6\pi}{4}= \frac{3}{2}\pi
    \end{eqnarray*}
\end{aussage} 

\begin{aussage}[Satz]{(Gaußscher Integralsatz)}{}

    Sei $M \subset \RR^3$ Standardbereich
    $V=(V_{1}, V_{2}, V_{3})$ stetig dif\/f'bares Vektorfeld auf
    Umgebung von $\overline{M}$ \\
    Sei $N$ "au"serer Normalenvektor auf $\delta M$ \\
    $$ \Rightarrow \int_{M} div V \cdot d(x, y, z) = \int_{\delta M} (V\cdot N) d\sigma $$
    $ div V = \frac{\partial V_{1}}{\partial x}+\frac{\partial V_{2}}{\partial y}+\frac{\partial V_{3}}{\partial z} $ \\
    $ M= \bigcup_{(x, y) \in M_{z}} \{(x, y, z) \in \RR^3 \mid \varphi(x, y) \leq z \leq \psi (x, y)\} $ \\
$$
    \begin{array}{c c c c c c c}
	 \delta M = & \Gamma_{\varphi} & \cup & \Gamma_{\psi} & \cup & \hbox{Teil parallel z-Achse} & \\
	    = & \delta M_1 & = & \delta M_2 & = & \delta M_3 &  
    \end{array}
$$
    F"ur $p = (x, y, z) \in \Gamma_{\psi}$ ist $N (p) = \frac{(-\psi_{x}, -\psi_{y}, 1)}{\sqrt{1+\psi_{x}^2+\psi_{y}^2}}$   \\

    F"ur $p = (x, y, z) \in \Gamma_{\varphi}$ ist $N(p) = \frac{(\varphi_{x}, \varphi_{y}, -1)}{\sqrt{1+\varphi_{x}^2+\varphi_{y}^2}}$ \\
\newpage
    Beweis: Setze $N (p) = (N_{1}, N_{2}, N_{3}) (p)$ \\

\begin{eqnarray*}
        \int_{M} \frac{\partial V_{3}}{\partial z} (x, y, z) d(x, y, z)
        & \stackrel{Satz 1}{=} & \int_{M_{z}} (\int_{\varphi(x, y)}^{\psi(x, y)}
        \frac{\partial V_{3}}{\partial z} (x, y, z)dz)d(x, y) \\ 
        & \stackrel{H.D.I.}{=} & \int_{M_{z}} [V_{3} (x, y, \psi(x, y)) - V_{3} (x, y, \varphi(x, y))] d(x, y) \\
        & = & \int_{M_{z}} V_{3} (x, y, \psi(x, y)) \frac{\sqrt{\psi_{x}^2+\psi_{y}^2+1}}{\sqrt{\psi_{x}^2+\psi_{y}^2+1}}d(x,y) \\
        & + & \int_{M_{z}} V_{3} (x, y, \varphi(x, y)) \frac{\sqrt{\varphi_{x}^2+\varphi_{y}^2+1}}{\sqrt{\varphi_{x}^2+\varphi_{y}^2+1}}d(x,y) \\
	& & \hbox{Auf dem oberen Rand } \delta M_{1} \hbox{ ist } N_{3}(p) = \frac{1}{\sqrt{\psi_{x}^2+\varphi_{y}^2+1}} \\
	& & \hbox{Auf dem unteren Rand } \delta M_{2} \hbox{ ist } N_{3}(p) = \frac{-1}{\sqrt{\psi_{x}^2+\varphi_{y}^2+1}} \\
	& & \hbox{Zur z-Achse paralleler Teil } \delta M_{3} \hbox{ ist } N_{3}(p) = 0 \\
        & = & \int_{M_{z}}V_{3}\cdot N_{3}\cdot\sqrt{\psi_{x}^2+\psi_{y}^2+1} \, d(x,y)\\
        & + & \int_{M_{z}} V_{3}\cdot N_{3}\cdot\sqrt{\varphi_{x}^2+\varphi_{y}^2+1} \, d(x,y) \\
        & = & \int_{\delta M_{1}} V_{3} N_{3} d\sigma + \int_{\delta M_{2}} V_{3} N_{3}
        d\sigma + \int_{\delta M_{3}} V_{3} N_{3} d\sigma \\
        & = & \int_{\delta M} V_{3}\cdot N_{3}\cdot d\sigma
\end{eqnarray*}
        Analog (wegen M x-projezierbar)

          $$  \int_{M} \frac{\partial V_{1}}{\partial x} d(x, y, z) =
            \int_{\delta M} V_{1} \cdot N_{1} d\sigma $$
          $$  \int_{M} \frac{\partial V_{2}}{\partial x} d(x, y, z) =
            \int_{\delta M} V_{2} \cdot N_{2} d\sigma $$
\begin{eqnarray*} 
        \Rightarrow \int_{M} div V d(x, y, z)& = & \int_{M}
        \left( \frac{\partial V_{1}}{\partial x}+ \frac{\partial V_{2}}{\partial y}
        + \frac{\partial V_{3}}{\partial z} \right) d(x, y, z) \\	
        & = & \int_{\delta M} (V_{1}N_{1} + V_{2}N_{2} + V_{3}N_{3}) d\sigma
        = \int_{\delta M} V\cdot N d\sigma 
\end{eqnarray*}

\end{aussage}

\begin{sonst}[Bemerkung]{}{} 
	$ \int_{\delta M} V_{i}N_{i} d\sigma$ ist f"ur
        parametrisierte Fl"achen definiert. \\
        Man hat f"ur die 3 Summanden verschiedene Parametriesierungen.

        $\Rightarrow$ Beweis, dass das Fl"achenintegral unabh"angig
        von der Parametrisierung ist kommt sp"ater.
\end{sonst}
\begin{aussage}[Beispiel]{}{}
            Sei $M =\{ (x, y, z) \in \RR^3 \mid 0 \leq x^2+y^2 < 1 \hbox{, } 0 < z < \sqrt{1 - (x^2 + y^2)} \} $ \\
            $V = (V_{1}, V_{2}, V_{3})$ mit $V_{1} = xz^2$, $ V_{2} = x^2y - z^3$, $V_{3} = 2xy + y^2z$
\begin{eqnarray*}
            \int_{\delta M} V\cdot N d\sigma & = & \int_{M} div V d(x, y, z) \\
            & = & \int_{M} (z^2 + x^2 + y^2) d(x, y, z) \\
            & = & \int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}
            r^2r^2 cos \theta d\theta dy dr \quad \hbox{ (mit Polarkoord.)} \\
            & = & \cdots 
\end{eqnarray*}
\end{aussage}

\begin{aussage}[Satz]{(Integralsatz von Stokes)}{}
Sei $D \subset \RR^3$ offen und $V = ( V_1, V_2, V_3 ): D \rightarrow
\RR^3$ zweimal stetig diff'bares Vektorfeld. \\
Sei $M$$\subset$$\RR^2$ offene Menge, die sich in endl. viele
Standardbereiche zerlegen l"asst. \\
Sei $ F:M \rightarrow F \subset \overline{F} \subset D \subset \RR^3$
eine parametrisierte Fl"ache
$$ \Rightarrow \int_{F}(rot{V} \cdot N)d\sigma = \int_{\partial F} ( V_1dx +
V_2dy + V_3dz ) $$
Bew.: \\
$$ \hbox{Sei }  \stackrel{\sim}{k}: \left \{ \begin{array}{l}
		[a,b] \rightarrow \partial M \\
		t \mapsto (\stackrel{\sim}{u}(t),\stackrel{\sim}{v}(t)) 
		\end{array} \right.
		\hbox{ parametrisierung von } \partial F 
$$
Sei $ f(u,v) = (x(u,v), y(u,v), z(u,v))$
\\ Setze zur Abk"urzung $\stackrel{\sim}{V_1}(u,v) = V_1(x(u,v), y(u,v),
z(u,v))$
\begin{eqnarray*}
\Rightarrow \int_{\partial F}V_1dx & = &
\intop_{a}^{b}\stackrel{\sim}{V_1}(\stackrel{\sim}{u}(t)\stackrel{\sim}{v}(t))\frac{d}{dt}
(x(\stackrel{\sim}{u}(t), \stackrel{\sim}{v}(t)))dt \\
& \stackrel{Kettenregel}{=} & \intop_{a}^{b}\stackrel{\sim}{V_1}(\stackrel{\sim}{u}(t)\stackrel{\sim}{v}(t))
\{\frac{\partial x}{\partial u}(\stackrel{\sim}{u}(t)\stackrel{\sim}{v}(t))\stackrel{\sim}{u}'(t)+
\frac{\partial x}{\partial v}(\stackrel{\sim}{u}(t) \stackrel{\sim}{v}(t))\stackrel{\sim}{v}'(t)\}dt \\
& \stackrel{Def. Kurvenint.}{=} & \int_{\partial F} \{ \stackrel{\sim}{V_1}(u,v) \frac{\partial x}{\partial u}(u,v)du
+\stackrel{\sim}{V_1}(u,v) \frac{\partial x}{\partial v}(u,v)dv\} \\
& \stackrel{Green}{=} & \int_{M}\{\frac{\partial}{\partial u}(\stackrel{\sim}{V_1}(u,v)\frac{\partial x}{\partial
v}(u,v)) - \frac{\partial}{\partial v}(\stackrel{\sim}{V_1}(u,v)\frac{\partial x}{\partial u}(u,v))\}d(u,v) \\
& = & \int_{M}\{\frac{\partial}{\partial u}V_1(x(u,v),y(u,v),z(u,v)) \frac{\partial x}{\partial v}(u,v)) - \\
& & \phantom{\int_{M}}\frac{\partial x}{\partial v}V_1(x(u,v),y(u,v),z(u,v)) \frac{\partial x}{\partial u}(u,v))\}d(u,v) \\
\end{eqnarray*}
\begin{eqnarray*}
& \stackrel{Produktregel}{=} & \int_{M} \{(\frac{\partial V_1}{\partial x}\frac{\partial x}{\partial u}
+ \frac{\partial V_1}{\partial y}\frac{\partial y}{\partial u} + 
\frac{\partial V_1}{\partial z}\frac{\partial z}{\partial u} ) \frac{\partial x}{\partial v} +
V_1\frac{\partial^2x}{\partial u \partial v } - \\ 
& & \phantom{\int_{M}} (\frac{\partial V_1}{\partial x}\frac{\partial x}{\partial v}
+ \frac{\partial V_1}{\partial y}\frac{\partial y}{\partial v}
+ \frac{\partial V_1}{\partial z}\frac{\partial z}{\partial v} \frac{\partial x}{\partial u}
- V_1\frac{\partial^2x}{\partial u \partial v } \} d(u,v) - \\
& & \phantom{\int_{M}} ( \frac{\partial x \partial y}{\partial u \partial v} - 
 \frac{\partial x \partial y}{\partial v \partial u} - \frac{\partial x \partial y}{\partial u \partial v})
 \frac{\partial V_1}{\partial y} + (\frac{\partial x \partial z}{\partial v \partial u} - 
 \frac{\partial x \partial z}{\partial u \partial v}) \frac{\partial V_1}{\partial z}\}d(u,v) = (\ast)
 \end{eqnarray*}
 Es war vergleiche $ \mathsection 5$:
 $$ 
 N(u,v)=\frac{(D_x, D_y, D_z)}{\sqrt{{D_x}^2+{D_y}^2+{D_z}^2}}(u,v) \hbox{ mit } 
 D_y = \left| \begin{array}{c c} \frac{\partial z}{\partial u} & \frac{\partial x}{ \partial u} \\
				 \frac{\partial z }{\partial v} & \frac{\partial x}{\partial v} \end{array}
				 \right |
				 \hbox{, }
 D_z = \left| \begin{array}{c c} \frac{\partial y}{\partial u} & \frac{\partial y}{ \partial u} \\
				 \frac{\partial x }{\partial v} & \frac{\partial y}{\partial v} \end{array}
				 \right |
				 \hbox{ und }
$$
$
				 D = \sqrt{{D_x}^2+{D_y}^2+{D_z}^2} = \|f_u \times f_v\|
$
\\
\begin{eqnarray*}
(\ast) & = & \int_{M}\{-D_z \frac{\partial V_1}{\partial y} + D_y \frac{\partial V_1}{\partial z}\}d(u,v) \\
	& = & \int_{M}(-\frac{\partial V_1}{\partial y}\frac{D_z}{D} + 
	\frac{\partial V_1}{\partial z}\frac{D_y}{D})\|f_u \times f_v\|d(u,v) \\
	& = & \int_{F}(-\frac{\partial V_1}{\partial y}\frac{D_z}{D} + 
	\frac{\partial V_1}{\partial z}\frac{D_y}{D})d\sigma
\end{eqnarray*}
Analog erh"alt man:
\begin{eqnarray*}
\int_{\partial F} V_2dy & = & \int_{F}(-\frac{\partial V_2}{\partial z}\frac{D_x}{D} +
					\frac{\partial V_2}{\partial x}\frac{D_z}{D})d\sigma \\
\int_{\partial F} V_3dy & = & \int_{F}(-\frac{\partial V_3}{\partial x}\frac{D_y}{D} +
					\frac{\partial V_3}{\partial y}\frac{D_x}{D})d\sigma
\end{eqnarray*}
Summation ergibt:
\begin{eqnarray*}
\int_{\partial F} V_1dx + V_2dy + V_dz & = & \int_{F}((\frac{\partial V_3}{\partial y} 
- \frac{\partial V_2}{\partial z})\frac{D_x}{D}+(\frac{\partial V_1}{\partial z} - \frac{\partial
V_3}{\partial x}) \frac{D_y}{D} + (\frac{\partial V_2}{\partial x } - \frac{\partial V_1}{\partial
y})\frac{D_x}{D})d \sigma \\ & = & \int_{F}rot{V} \cdot Nd \sigma
\end{eqnarray*}
\end{aussage}
\begin{sonst}[Beispiel]{}{}
$ D = \RR^3, \, V(x,y,z) = (-y,x,1)$ \\
$F=$ obere Halbkugel vom Radius $2$ \\
$M = \{(u,v) \in \RR^2 \, | \, u^2+v^2 \le 4 \} $
\\ $$ f : \left\{ \begin{array}{l}
								M \rightarrow F \subset \RR^3 \\
								(u,v) \mapsto (x(u,v),y(u,v),z(u,v))
						\end{array} \right. 
			\hbox{ mit } \left\{ \begin{array}{l l} 
								 		  x(u,v) = u \\
										  y(u,v) = v \\
										  z(u,v) = \sqrt{4-u^2-v^2} \end{array} \right.
	$$
\begin{eqnarray*}
rot{V} & = & (\frac{\partial V_3}{\partial y} - \frac{\partial V_2}{\partial z},
\frac{\partial V_1}{\partial z} - \frac{\partial V_3}{\partial x},
\frac{\partial V_2}{\partial x } - \frac{\partial V_1}{\partial y}) = (0,0,2) \\
N & = & \frac{(-\frac{\partial z }{\partial u},-\frac{\partial z }{\partial v},1)}
{\sqrt{1+(\frac{\partial z}{\partial u})^2 +(\frac{\partial z}{\partial v})^2}} \\
rot{V} \cdot N & = & \frac{2}{\sqrt{1+(\frac{\partial z}{\partial u})^2 +(\frac{\partial z}{\partial v})^2}} =
\frac{2}{\|f_u \times f_v \|} \\
\int_{F}rot{V} \cdot N d\sigma & = & \int_{M}\frac{2}{\|f_u \times f_v \|} \|f_u \times f_v \| d(u,v)
= \int_{M}2d(u,v)= 8\pi
\end{eqnarray*}
Satz von Stokes: \\
$$ \int_{\partial F} (-ydx+xdy+dz) = 8\pi $$

\end{sonst}
\end{document}
