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\lhead{3. November 2004}
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\rhead{\bfseries Vorlesung 5}

\title{Analysis III}
\author{Johannes Kolb}
\date{3. 11. 2004}

\theoremstyle{definition}
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\newtheorem{lem}{Lemma}
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\begin{bsp}
\nl

Berechne $\int\limits_0^\infty{e^{-x^2}\ud x}$

\includegraphics*[width=5cm]{img5-1.pdf}

\begin{eqnarray*}
K_R & = & \{(x,y) \in \Rset^2 \prop x\geq 0,\, y\geq 0, \, x^2+y^2 \leq R^2 \} \\
Q_R & = & \{(x,y) \in \Rset^2 \prop 0 \leq x \leq R, \, 0 \leq y \leq R \} \\
K_{R\sqrt{2}} & = & \{(x,y) \in \Rset^2 \prop x \geq 0,\, y \geq 0,\, x^2+y^2 \leq 2R^2\}
\end{eqnarray*}
\begin{eqnarray*}
\int_{K_R}{e^{-(x^2+y^2)}\ud(x,y)} &=& \int_0^{\frac{\pi}{2}}{\left(\int_0^R{e^{-r^2}r\ud r}\right)\ud \phi} \\
&=& \int_0^{\frac{\pi}{2}}{\left[-\frac{1}{2}e^{-r^2}\right]_0^R\ud \phi} = \int_0^{\frac{\pi}{2}}{\left(-\frac{1}{2}e^{-R^2}+\frac{1}{2}\right)\ud \phi} = \frac{\pi}{4} - \frac{\pi}{4}e^{-R^2}
\end{eqnarray*}

Analog: $\int_{K_{R\sqrt{2}}}{e^{-(x^2+y^2)}\ud(x,y)} = \frac{\pi}{4}-\frac{\pi}{4}e^{-2R^2}$
\end{bsp}

\[
\int_{Q_R}{e^{-(x^2+y^2)}\ud(x,y)} = \int_0^R{\left(\int_0^R{e^{-x^2}e^{-y^2}\ud x}\right) \ud y} = \left(\int_0^R{e^{-x^2}\ud x}\right)^2
\]
\[ \implic \: \int_{K_R}{e^{-(x^2+y^2)}\ud(x,y)} \leq \int_{Q_R}{e^{-(x^2+y^2)}\ud(x,y)} \leq \int_{K_{\sqrt{2}R}}{e^{-(x^2+y^2)}\ud(x,y)} \]
\[ \equival \: \begin{array}{ccccc}
    \sqrt{\frac{\pi}{4}-\frac{\pi}{4}e^{-R^2}} & \leq & \int_0^R{e^{-x^2}\ud x} & \leq & \sqrt{\frac{\pi}{4}-\frac{\pi}{4}e^{-2R^2}} \\
    \downarrow R \to \infty & & & & \downarrow R \to \infty \\
    \sqrt{\frac{\pi}{4}} & \leq & \int_0^\infty{e^{-x^2}\ud x} & \leq & \sqrt{\frac{\pi}{4}}
\end{array} \]

\subsubsection*{Anwendung: Schwerpunkte}

\begin{Def}
Sei $M \subset \Rset^3$ me"sbar, keine Nullmenge
\[ (x_0,y_0,z_0) := \frac{1}{\mu(M)}\left(\int_M{x\ud(x,y,z)},\, \int_M{y\ud(x,y,z)},\, \int_M{z\ud(x,y,z)}\right)\]
hei"st Schwerpunkt der Menge $M$.
\end{Def}

\begin{satz}
F"ur den Schwerpunkt $(x_0,y_0,z_o)$ von $M$ gilt:
\[ \int_M{(x-x_0)\ud(x,y,z)} = \int_M{(y-y_0)\ud(x,y,z)} = \int_M{(z-z_0)\ud(x,y,z)} = 0 \]
\end{satz}
\begin{proof}
\begin{eqnarray*}
\int_M{(x-x_0)\ud(x,y,z)} & = & \int_M{x\ud(x,y,z)} - \int_M{x_0\ud(x,y,z)} \\
& = & x_0\mu(M) - x_0\mu(M) = 0
\end{eqnarray*}
\end{proof}

\begin{bsp}Schwerpunkt des Kugeloktanten

\[ K=\{(x,y,z) \prop x \geq 0,\, y \geq 0,\, z \geq 0,\, x^2+y^2+z^2 \leq 1 \} \]
Hatten: $\mu(K) = \frac{\pi}{6}$

Transformation in Polarkoordinaten:
\begin{eqnarray*}
    x & = & r\cdot\cos(\varphi)\cdot\cos(\theta) \\
    y & = & r\cdot\sin(\varphi)\cdot\cos(\theta) \\
    z & = & r\cdot\sin(\theta) \\
    \frac{\ud(x,y,z)}{\ud(r,\varphi,\theta)} = r^2\cos(\theta)
\end{eqnarray*}

\begin{eqnarray*}
    \int_K{x\ud(x,y,z)} & = & \int_0^\pihalb{\int_0^\pihalb{\int_0^1{r\cos\varphi\cos\theta\cdot r^2\cos\theta\ud{}r}\ud\varphi}\ud\theta} \\
    & = & \frac{1}{4}\int_0^\pihalb{\int_0^\pihalb{\cos\varphi\cdot\cos^2\theta\ud\varphi}\ud\theta} \\
    & = & \frac{1}{4}\int_0^\pihalb{\cos^2\theta\ud \theta} = \frac{1}{4}\left[\frac{1}{2}(\cos\theta\sin\theta+\theta)\right]_0^\pihalb = \frac{\pi}{16}
\end{eqnarray*}
denn $\frac{1}{2}(\cos\theta\sin\theta+\theta)'=\frac{1}{2}(\cos^2\theta-\sin^2\theta+1) = \cos^2\theta$
\[ \implic \: x_0 = \frac{1}{\mu(K)}\int_K{x\ud(x,y,z)} = \frac{6}{\pi}\frac{\pi}{16} = \frac{3}{8} \]
\[ \implic \textrm{Schwerpunkt }K=\left(\frac{3}{8},\frac{3}{8},\frac{3}{8}\right) \]
\end{bsp}

\subsubsection*{Anwendung: Tr"agheitsmoment}
\begin{Def}
Sei $M \subset \Rset^3$ me"sbar
\[I_x := \int_M{(y^2+z^2)\ud(x,y,z)} = \textrm{ Tr"agheitsmoment von $M$ bzgl. $x$-Achse } \]
\end{Def}

\begin{bsp}
\[ M_R := \{(x,y,z) \in \Rset^3 \prop x^2+y^2 \leq R,\, -\frac{h}{2}\leq z \leq \frac{h}{2}\} \]

\includegraphics*[width=5cm]{img5-2.pdf}

\textbf{Zylinderkoordinaten:}
\begin{eqnarray*}
    x & = & r\cos\varphi \\
    y & = & r\sin\varphi \\
    z & = & z \\
    \frac{\ud(x,y,z)}{\ud(r,\varphi,z)} & = & \det\left(\begin{array}{ccc}
	\cos\varphi & -r\sin\varphi & 0 \\
	\sin\varphi & r\cos\varphi  & 0 \\
	0 & 0 & 1 
    \end{array}\right) = r
\end{eqnarray*}

\begin{eqnarray*}
    I_x & = & \int_M{(x^2+y^2)\ud(x,y,z)} = \int_0^R{\int_0^{2\pi}{\int_{-\frac{h}{2}}^{\frac{h}{2}}{r^3\ud z}\ud \varphi} \ud r} \\
    & = & \frac{R^4}{4}\int_0^{2\pi}{\int_{-\frac{h}{2}}^{\frac{h}{2}}{\ud\varphi}\ud z} 
    = \frac{2\pi R^4}{4} \int_{-\frac{h}{2}}^{\frac{h}{2}}{\ud z} = \frac{\pi R^4}{2}h
\end{eqnarray*}
\end{bsp}

\section*{§5 Fl"achen im $\Rset^3$}
Sei $M \in \Rset^2$ offen und $f:\left\{\begin{array}{ccl}
    M & \to & \Rset^3 \\
    (u,v) & \mapsto & \left(x(u,v), y(u,v), z(u,v)\right)
\end{array}\right.$ einmal stetig differenzierbare Abbildung, s.d.
\[\rg\frac{\partial(x,y,z)}{\partial(u,v)} = \rg\left(\begin{array}{cc}
	                 \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
	\rule{0cm}{0.5cm}\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\
	\rule{0cm}{0.5cm}\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} 
    \end{array}\right) = 2 \]
Dann hei"st $F=\Im f$ eine (parametrisierte) Fl"ache in $\Rset^3$ \\
$f:M \to F \in \Rset^3$ Parameterdarstellung.

\begin{bsp}
\[M=\Rset^2\quad f:\Rset^2 \to \Rset^3 \textrm{ mit } \begin{array}{ccc}
    x(u,v) & = & u \\
    y(u,v) & = & v \\
    z(u,v) & = & 0 
\end{array}\]
\[ \frac{\partial(x,y,z)}{\partial(u,v)} = \left(\begin{array}{cc}
    1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right) \quad \rg=2 \quad F=\textrm{$x$-$y$-Ebene}
    \]
\end{bsp}

\begin{bsp}
\[ M=\{(u,v) \in \Rset^2 \prop 0 < u < \infty,\,0 < v < 2\pi\} \]
\[ f: M \to \Rset^3 \textrm{ mit } \begin{array}{ccl}
    x(u,v) & = & u\cos v \\
    y(u,v) & = & u\sin v \\
    z(u,v) & = & 0
    \end{array} \]
\[ \frac{\partial(x,y,z)}{\partial(u,v)} = \left(\begin{array}{cc}
	\cos v & -u\sin v \\
	\sin v & u\cos v \\
	0 & 0
    \end{array}\right) \quad \rg = 2 \]
$F$ ist l"angs der positiven $x$-Achse aufgeschlitzte $(x,y)$-Ebene
\end{bsp}

\begin{bsp}
\[ M = \{(u,v) \in \Rset^2 \prop u^2+v^2 \leq R \} \]
\[ f: M \to \Rset^3 \textrm{ mit }\begin{array}{ccl}
	x(u,v) & = & u \\
	y(u,v) & = & v \\
	z(u,v) & = & \sqrt{R^2-u^2-v^2}
    \end{array} \]
\[\frac{\partial(x,y,z)}{\partial(u,v)} = \left(\begin{array}{cc}
	1 & 0 \\
	0 & 1 \\
	\frac{-u}{\sqrt{R^2-u^2-v^2}} & \frac{-v}{\sqrt{R^2-u^2-v^2}}
    \end{array}\right) \quad \rg = 2 \]
$F$ ist obere Halbkugeloberfl"ache mit Mittelpunkt $(0,0,0)$ und Radius $R$.
\end{bsp}

\begin{satz}
Sei $M \subset \Rset^2$ offen und $\tilde{k}: [a,b] \to M$ parametrisierte ebene Kurve $\tilde{K}$ \\
Sei $f: M \to F \subset \Rset^3$ parametrisierte Fl"ache in $\Rset^3$ \\
\implic{} $k=f\circ \tilde{k}: [a,b] \to K \subset \Rset^3$ ist parametrisierte kurve im $\Rset^3$
\end{satz}
\begin{proof}
$k$ einmal stetig diffbar, denn $\tilde{k}$ und $f$ sind es.\\
zu zeigen: $k'(t) \neq 0 \: \forall t \in [a,b]$\\
\[ \textrm{Sei }\tilde{k}(t) = \left(u(t),v(t)\right),\quad f(u,v)=\left(x(u,v),y(u,v),z(u,v)\right) \]
\[ \implic f_u = \left(\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}\right) \quad f_v = \left(\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v}\right)\]
\[ \implic k'(t) = f_u(u,v)\cdot u'(t) + f_v(u,v)\cdot v'(t) \]
\[ \left. \begin{array}{rcl}
	\textrm{Da $\tilde{k}$ parametrisierte Kurve} & \implic & (u'(t),v'(t)) \neq (0,0) \\
	\textrm{Da $f$ parametrisierte Fl"ache}       & \implic & f_u, f_v \textrm{ linear unabh"angig} 
    \end{array}\right\} \: k'(t)\neq 0 \: \forall t \]


\end{proof}

Hatten gesehen, dass $f_u(u,v)$ und $f_v(u,v)$ linear unabh"angig sind $\forall(u_0,v_0)$
\begin{Def}
Die Ebene, die von $f_u(u_0,v_0)$ und $f_v(u_0,v_0)$ aufgespannt wird, hei"st Tagentialebene $T_{f,P}$ von $F$ im Punkt $f(u_0,v_0)$
\end{Def}

\textbf{Beh:}
Die Vektoren $f_u(u_0,v_0)$ $f_v(u_0,v_0)$ sind Tangentialvektoren von Kurven in $F$

\begin{proof}
Betrachten $\tilde{k}(t) = (t,v_0) \quad t \in [u_0-\varepsilon,u_0+\varepsilon]$
\[\implic k'(\mu_0) = f_u(u_0,v_0)\cdot u'(0) + f_v(u_0,v_0)\cdot v'(v_0) = f_u(u_0,v_0) \]
\end{proof}

\begin{Def}
Sei $f: M \to F \subset \Rset^3$ parametrisierte Fl"ache in $\Rset^3$.
\[N(u_0,v_0) := \frac{f_u(u_0,v_0) \times f_v(u_0,v_0)}{||f_u(u_0,v_0)\times f_v(u_0,v_0)||} \]
hei"st Normalenvektor von $f$ an der Stelle $p=f(u_0,v_0)$.

\[ \textrm{Hierbei: } 
    x=\left(\begin{array}{c} x_1\\x_2\\x_3\end{array}\right) \quad
    y=\left(\begin{array}{c} y_1\\y_2\\y_3\end{array}\right) \quad
    x\times y = \left(\begin{array}{c}
	x_2\cdot y_3 - x_3\cdot y_2 \\
	x_3\cdot y_1 - x_1\cdot y_3 \\
	x_1\cdot y_2 - x_2\cdot y_1
	\end{array}\right) \]

Ist $f(u,v) = (x(u,v),y(u,v),z(u,v))$
\[ \implic \: \frac{\partial(x,y,z)}{\partial(u,v)} = \left(\begin{array}{cc}
	\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
	\rule{0cm}{0.5cm}\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\
	\rule{0cm}{0.5cm}\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v}
    \end{array}\right) \]

\[\textrm{Setzen } D_x=\left|\begin{array}{cc}
	\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\
	\rule{0cm}{0.5cm}\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} 
    \end{array}\right|, \quad
    D_y=\left|\begin{array}{cc}
	\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} \\
	\rule{0cm}{0.5cm}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} 
    \end{array}\right|, \quad
    D_z=\left|\begin{array}{cc}
	\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
	\rule{0cm}{0.5cm}\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} 
    \end{array}\right| \]
\end{Def}
\[ \textrm{so ist } N(u_0,v_0) = \frac{\left(D_x(u_0,v_0), D_y(u_0,v_0), D_z(u_0,v_0)\right)}{\sqrt{\left(D_x^2+D_y^2+D_z^2\right)(u_0,v_0)}} \]

\textbf{Spezialfall:} Sei $g: M \to \Rset$ stetig differenzierbare Funktion
\[ \implic \Gamma_g=\{(x,y,z) \in \Rset^3 \prop x=u, y=v, z=g(u,v)\} \textrm{ ist parametrisierte Fl"ache} \]
\[ 
    D_x = \left|\begin{array}{cc} 0 & 1 \\ g_u & g_v \end{array}\right| = -g_u \quad
    D_y = \left|\begin{array}{cc} g_u & g_v\\ 1 & 0  \end{array}\right| = -g_v \quad
    D_z = \left|\begin{array}{cc} 1 & 0 \\ 0 & 1     \end{array}\right| = 1 
\]
\end{document}
