Question Number 132261 by muneer0o0 last updated on 12/Feb/21
Answered by Olaf last updated on 13/Feb/21
![Ω = ∫_(−1) ^(+1) ∫_(−(√(1−y^2 ))) ^(+(√(1−y^2 ))) sec(x^2 +y^2 )dxdy Ω = ∫_0 ^1 ∫_0 ^(2π) sec(r^2 )rdθdr Ω = ∫_0 ^1 ((2πr)/(cosr^2 ))dr Ω = 2π[(1/2)ln(1+sin(1))−(1/2)ln(cos(1))] Ω = πln(((1+sin(1))/(cos(1))))](https://www.tinkutara.com/question/Q132290.png)
$$\Omega\:=\:\int_{−\mathrm{1}} ^{+\mathrm{1}} \int_{−\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} ^{+\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \mathrm{sec}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dxdy} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{sec}\left({r}^{\mathrm{2}} \right){rd}\theta{dr} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}\pi{r}}{\mathrm{cos}{r}^{\mathrm{2}} }{dr} \\ $$$$\Omega\:=\:\mathrm{2}\pi\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{sin}\left(\mathrm{1}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{1}\right)\right)\right] \\ $$$$\Omega\:=\:\pi\mathrm{ln}\left(\frac{\mathrm{1}+\mathrm{sin}\left(\mathrm{1}\right)}{\mathrm{cos}\left(\mathrm{1}\right)}\right) \\ $$