Question Number 190419 by horsebrand11 last updated on 02/Apr/23
$$\:\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\underset{−\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} {\overset{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} {\int}}\:\mathrm{ln}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right){dx}\:{dy}\:=? \\ $$
Answered by witcher3 last updated on 04/Apr/23
![x=rcos(a) y=rsin(a) −1≤rsin(a)≤1,⇒a∈[−(π/2),(π/2)] ∣x∣≤(√(1−y^2 )) ⇔0≤x^2 +y^2 ≤1⇔0≤r≤1 −(1/r)≤sin(a)≤(1/r) ∫_0 ^1 ∫_(−(π/2)) ^(π/2) rln(1+r^2 )dadr=(π/2)∫_0 ^1 ln(1+r^2 )dr^2 =(π/2)∫_0 ^1 ln(1+x)dx=(π/2).(2ln(2)−1)](https://www.tinkutara.com/question/Q190505.png)
$$\mathrm{x}=\mathrm{rcos}\left(\mathrm{a}\right) \\ $$$$\mathrm{y}=\mathrm{rsin}\left(\mathrm{a}\right) \\ $$$$−\mathrm{1}\leqslant\mathrm{rsin}\left(\mathrm{a}\right)\leqslant\mathrm{1},\Rightarrow\mathrm{a}\in\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\mid\mathrm{x}\mid\leqslant\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }\:\:\:\Leftrightarrow\mathrm{0}\leqslant\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{1}\Leftrightarrow\mathrm{0}\leqslant\mathrm{r}\leqslant\mathrm{1} \\ $$$$−\frac{\mathrm{1}}{\mathrm{r}}\leqslant\mathrm{sin}\left(\mathrm{a}\right)\leqslant\frac{\mathrm{1}}{\mathrm{r}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{rln}\left(\mathrm{1}+\mathrm{r}^{\mathrm{2}} \right)\mathrm{dadr}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{r}^{\mathrm{2}} \right)\mathrm{dr}^{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}=\frac{\pi}{\mathrm{2}}.\left(\mathrm{2ln}\left(\mathrm{2}\right)−\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$