Allgemeines zur Variablentransformation

Wir betrachten das Bereichsintegral

$$\mathit{I}=\underset{\mathit{B}}{\iint }\mathit{f}(\mathit{x},\mathit{y})\text{d}\mathit{x}\text{d}\mathit{y},$$

und wandeln die Koordinaten folgendermaßen um:

$$\begin{array}{rcl}\mathit{u}& =& {\phi }_1(\mathit{x},\mathit{y})\text{ ,}\\ \mathit{v}& =& {\phi }_2(\mathit{x},\mathit{y}).\end{array}$$

Der Integrationsbereich $\mathit{B}$ auf der $\mathit{x},\mathit{y}$-Ebene wurde in den entsprechenden Bereich $\mathit{B}\text{'}$ auf der $\mathit{u},\mathit{v}$-Ebene umgewandelt. Wir setzen voraus, dass diese Transformation umkehrbar ist:

$$\begin{array}{rcl}\mathit{x}& =& {\psi }_1(\mathit{u},\mathit{v})\text{ ,}\\ \mathit{y}& =& {\psi }_2(\mathit{u},\mathit{v}).\end{array}$$

Gemäß der Definition des zweidimesionalen Bereichsintegrals

$$\mathit{I}=\underset{\mathit{M}\to \infty }{lim}\underset{\mathit{N}\to \infty }{lim}\sum _{\mathit{m}=0}^{\mathit{M}-1}\sum _{\mathit{n}=0}^{\mathit{N}-1}\mathit{f}({\mathit{x}}_{\mathit{m}},{\mathit{y}}_{\mathit{n}})\Delta \mathit{x}\Delta \mathit{y}$$

ist die Doppelsumme in den transformierten Koordinaten gegeben durch

$$\sum _{\mathit{m}=0}^{\mathit{M}-1}\sum _{\mathit{n}=0}^{\mathit{N}-1}\mathit{f}({\psi }_1({\mathit{u}}_{\mathit{m}},{\mathit{v}}_{\mathit{n}}),{\psi }_2({\mathit{u}}_{\mathit{m}},{\mathit{v}}_{\mathit{n}}))\frac{\Delta \mathit{x}\Delta \mathit{y}}{\Delta \mathit{u}\Delta \mathit{v}}\Delta \mathit{u}\Delta \mathit{v}$$

Die Größe $\frac{\Delta \mathit{x}\Delta \mathit{y}}{\Delta \mathit{u}\Delta \mathit{v}}$ gibt das Verhältnis des Flächeninhaltes $\Delta \mathit{A}$ eines Flächenelementes in der $\mathit{x},\mathit{y}$-Ebene zum Inhalt des entsprechenden Elementes $\Delta \mathit{A}\text{'}$ in der $\mathit{u},\mathit{v}$-Ebene an. Im Grenzfall $\Delta \mathit{x},\Delta \mathit{y},\Delta \mathit{u},\Delta \mathit{v}\to 0$ ist dieses Verhältnis durch den Betrag der Funktionaldeterminante $\left|\frac{\partial (\mathit{x},\mathit{y})}{\partial (\mathit{u},\mathit{v})}\right|$ gegeben

$$\begin{array}{rcl}\mathit{I}& =& \underset{\mathit{M}\to \infty }{lim}\underset{\mathit{N}\to \infty }{lim}\sum _{\mathit{m}=0}^{\mathit{M}-1}\sum _{\mathit{n}=0}^{\mathit{N}-1}\mathit{f}({\psi }_1({\mathit{u}}_{\mathit{m}},{\mathit{v}}_{\mathit{n}}),{\psi }_2({\mathit{u}}_{\mathit{m}},{\mathit{v}}_{\mathit{n}}))\frac{\Delta \mathit{x}\Delta \mathit{y}}{\Delta \mathit{u}\Delta \mathit{v}}\Delta \mathit{u}\Delta \mathit{v}\\ & =& \underset{\mathit{B}\text{'}}{\iint }\mathit{f}({\psi }_1(\mathit{u},\mathit{v}),{\psi }_2(\mathit{u},\mathit{v}))\left|\frac{\partial (\mathit{x},\mathit{y})}{\partial (\mathit{u},\mathit{v})}\right|\text{d}\mathit{u}\text{d}\mathit{v}.\end{array}$$

Zur Veranschaulichung des Ergebnisses betrachten wir drei Vektoren ${\mathit{a}}_0,{\mathit{a}}_1,{\mathit{a}}_2$, die ein infinitesimales Flächenelement in der Umgebung des Punktes $({\mathit{u}}_0,{\mathit{v}}_0)$ definieren (Abb. 1)

$$\begin{array}{rcl}{\mathit{a}}_0& =& {\psi }_1({\mathit{u}}_0,{\mathit{v}}_0)\mathit{i}+{\psi }_2({\mathit{u}}_0,{\mathit{v}}_0)\mathit{j}\\ {\mathit{a}}_1& =& {\psi }_1({\mathit{u}}_0+\text{d}\mathit{u},{\mathit{v}}_0)\mathit{i}+{\psi }_2({\mathit{u}}_0+\text{d}\mathit{u},{\mathit{v}}_0)\mathit{j}\\ {\mathit{a}}_2& =& {\psi }_1({\mathit{u}}_0,{\mathit{v}}_0+\text{d}\mathit{v})\mathit{i}+{\psi }_2({\mathit{u}}_0,{\mathit{v}}_0+\text{d}\mathit{v})\mathit{j}.\end{array}$$

Mit Hilfe der Taylor-Reihe für Funktionen zweier Veränderlicher ergibt sich

$$\begin{array}{rcl}{\psi }_{1,2}({\mathit{u}}_0+\text{d}\mathit{u},{\mathit{v}}_0)& =& {\psi }_{1,2}({\mathit{u}}_0,{\mathit{v}}_0)+\frac{\partial {\psi }_{1,2}}{\partial \mathit{u}}\text{d}\mathit{u}+\dots \\ {\psi }_{1,2}({\mathit{u}}_0,{\mathit{v}}_0+\text{d}\mathit{v})& =& {\psi }_{1,2}({\mathit{u}}_0,{\mathit{v}}_0)+\frac{\partial {\psi }_{1,2}}{\partial \mathit{v}}\text{d}\mathit{v}+\dots .\end{array}$$

Die Vektoren ${\mathit{a}}_1-{\mathit{a}}_0$ und ${\mathit{a}}_2-{\mathit{a}}_0$ spannen ein Parallelogramm auf, dessen Flächeninhalt in kartesischen Koordinaten durch den Betrag des Kreuzprodukts

$$\text{d}\mathit{A}=\text{d}\mathit{x}\text{d}\mathit{y}=|({\mathit{a}}_1-{\mathit{a}}_0)\times ({\mathit{a}}_2-{\mathit{a}}_0)|$$

gegeben ist. Mit Hilfe von und ergibt sich

$$\begin{array}{rcl}{\mathit{a}}_1-{\mathit{a}}_0& =& \frac{\partial {\psi }_1}{\partial \mathit{u}}\text{d}\mathit{u}\mathit{i}+\frac{\partial {\psi }_2}{\partial \mathit{u}}\text{d}\mathit{u}\mathit{j}\\ {\mathit{a}}_2-{\mathit{a}}_0& =& \frac{\partial {\psi }_1}{\partial \mathit{v}}\text{d}\mathit{v}\mathit{i}+\frac{\partial {\psi }_2}{\partial \mathit{v}}\text{d}\mathit{v}\mathit{j}.\end{array}$$

Das Kreuzprodukt ist nun

$$({\mathit{a}}_1-{\mathit{a}}_0)\times ({\mathit{a}}_2-{\mathit{a}}_0)=\left|\begin{array}{rrr}\mathit{i}& \mathit{j}& \mathit{k}\\ \frac{\partial {\psi }_1}{\partial \mathit{u}}\text{d}\mathit{u}& \frac{\partial {\psi }_2}{\partial \mathit{u}}\text{d}\mathit{u}& 0\\ \frac{\partial {\psi }_1}{\partial \mathit{v}}\text{d}\mathit{v}& \frac{\partial {\psi }_2}{\partial \mathit{v}}\text{d}\mathit{v}& 0\end{array}\right|=\left|\begin{array}{rr}\frac{\partial {\psi }_1}{\partial \mathit{u}}& \frac{\partial {\psi }_2}{\partial \mathit{u}}\\ \frac{\partial {\psi }_1}{\partial \mathit{v}}& \frac{\partial {\psi }_2}{\partial \mathit{v}}\end{array}\right|\text{d}\mathit{u}\text{d}\mathit{v}\mathit{k}.$$

(siehe Rechenregeln für Determinanten) und somit ist

$$\text{d}\mathit{A}=\left|\begin{array}{rr}\frac{\partial {\psi }_1}{\partial \mathit{u}}& \frac{\partial {\psi }_2}{\partial \mathit{u}}\\ \frac{\partial {\psi }_1}{\partial \mathit{v}}& \frac{\partial {\psi }_2}{\partial \mathit{v}}\end{array}\right|\text{d}\mathit{u}\text{d}\mathit{v}=\frac{\partial (\mathit{x},\mathit{y})}{\partial (\mathit{u},\mathit{v})}\text{d}\mathit{A}\text{'}$$

wegen

$$\left|\begin{array}{rr}\frac{\partial {\psi }_1}{\partial \mathit{u}}& \frac{\partial {\psi }_2}{\partial \mathit{u}}\\ \frac{\partial {\psi }_1}{\partial \mathit{v}}& \frac{\partial {\psi }_2}{\partial \mathit{v}}\end{array}\right|=\left|\begin{array}{rr}\frac{\partial {\psi }_1}{\partial \mathit{u}}& \frac{\partial {\psi }_1}{\partial \mathit{v}}\\ \frac{\partial {\psi }_2}{\partial \mathit{u}}& \frac{\partial {\psi }_2}{\partial \mathit{v}}\end{array}\right|=\frac{\partial ({\psi }_1,{\psi }_2)}{\partial (\mathit{u},\mathit{v})}=\frac{\partial (\mathit{x},\mathit{y})}{\partial (\mathit{u},\mathit{v})}\text{und}\text{d}\mathit{A}\text{'}=\text{d}\mathit{u}\text{d}\mathit{v}.$$