Question Number 65399 by mathmax by abdo last updated on 29/Jul/19
$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} =\:\int\int_{\left[\mathrm{0},{n}\left[^{\mathrm{2}} \right.\right.} \:\:\:\frac{{dxdy}}{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$\left.\mathrm{2}\right){find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 31/Jul/19
![let consider the diffeomorphisme (r,θ)→(x,y) / x =r cosθ and y =rsinθ we have 0≤x<n and 0≤y<n ⇒ 0≤x^2 +y^2 <2n^2 ⇒0≤r^2 <2n^2 ⇒0≤r<n(√2) A_n =∫∫_(0≤r<n(√2) and 0≤θ≤(π/2)) ((rdr dθ)/(√(r^2 +4))) =∫_0 ^(n(√2)) ((rdr)/(√(r^2 +4))) ∫_0 ^(π/2) dθ =(π/2)[(√(r^2 +4))]_0 ^(n(√2)) =(π/2){(√(2n^2 +4))−2} 2)lim_(n→+∞) A_n =+∞](Q65574.png)
$${let}\:{consider}\:{the}\:{diffeomorphisme}\:\left({r},\theta\right)\rightarrow\left({x},{y}\right)\:/ \\ $$$${x}\:={r}\:{cos}\theta\:\:{and}\:{y}\:={rsin}\theta\:\:{we}\:{have}\:\mathrm{0}\leqslant{x}<{n}\:{and}\:\mathrm{0}\leqslant{y}<{n}\:\Rightarrow \\ $$$$\mathrm{0}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} <\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow\mathrm{0}\leqslant{r}^{\mathrm{2}} <\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow\mathrm{0}\leqslant{r}<{n}\sqrt{\mathrm{2}} \\ $$$${A}_{{n}} =\int\int_{\mathrm{0}\leqslant{r}<{n}\sqrt{\mathrm{2}}\:{and}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{rdr}\:{d}\theta}{\sqrt{{r}^{\mathrm{2}} \:+\mathrm{4}}} \\ $$$$=\int_{\mathrm{0}} ^{{n}\sqrt{\mathrm{2}}} \:\:\frac{{rdr}}{\sqrt{{r}^{\mathrm{2}} \:+\mathrm{4}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{d}\theta\:=\frac{\pi}{\mathrm{2}}\left[\sqrt{{r}^{\mathrm{2}} \:+\mathrm{4}}\right]_{\mathrm{0}} ^{{n}\sqrt{\mathrm{2}}} \:\:=\frac{\pi}{\mathrm{2}}\left\{\sqrt{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{4}}−\mathrm{2}\right\} \\ $$$$\left.\mathrm{2}\right){lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =+\infty \\ $$