Berechnen Sie das Integral:

Question

Aufgabe:

Berechnen Sie das Integral:

\( \iiint_{R} \frac{z}{(x+y)^{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z, \quad R=[1,2] \times[1,2] \times[0,2] \)

Answer 1

\( \iiint_{R} \frac{z}{(x+y)^{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z, \quad R=[1,2] \times[1,2] \times[0,2] \)

\( \int \limits_{0}^{2}\int \limits_{1}^{2}\int \limits_{1}^{2}\frac{z}{(x+y)^{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z \)

\(= \int \limits_{0}^{2}\int \limits_{1}^{2}\frac{z}{y+1}-\frac{z}{y+2}  \mathrm{~d} y \mathrm{~d} z \)

\(= \int \limits_{0}^{2}ln(\frac{9}{8})*z \mathrm{~d} z =2*ln(\frac{9}{8}) \)

Answer 2

Aloha :)

Da hier die Integrationsgrenzen für alle Variablen konstant sind, können wir in beliebiger Reihenfolge integrieren:

$$I=\iiint\limits_R\frac{z}{(x+y)^2}dV=\int\limits_1^2dx\int\limits_1^2dy\int\limits_0^2dz\,\frac{z}{(x+y)^2}=\int\limits_1^2dx\int\limits_1^2dy\left[\frac{z^2/2}{(x+y)^2}\right]_{z=0}^2$$$$\phantom{I}=\int\limits_1^2dx\int\limits_1^2dy\frac{2}{(x+y)^2}=\int\limits_1^2dx\left[\frac{-2}{x+y}\right]_{y=1}^2=\int\limits_1^2dx\left(\frac{-2}{x+2}+\frac{2}{x+1}\right)$$$$\phantom{I}=\left[-2\ln|x+2|+2\ln|x+1|\right]_1^2=(-2\ln4+2\ln3)-(-2\ln3+2\ln2)$$$$\phantom{I}=-2\ln4+4\ln3-2\ln2=\ln(3^4)-2(\ln4+\ln2)=\ln81-2\ln8=\ln81-\ln64$$$$\phantom{I}=\ln\left(\frac{81}{64}\right)\approx0,2356$$