1) calculate A_n = ∫∫_([0,n[^2 ) ((dxdy)/(√(x^2 +y^2 +4))) 2)find lim_(n→+∞) A


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Question Number 65399 by mathmax by abdo last updated on 29/Jul/19

$1) c a l c u l a t e A_{n} = \int \int_{[0, n [^{2}} \frac{d x d y}{\sqrt{x^{2} + y^{2} + 4}}$ $2) f i n d {l i m}_{n \to + \infty} A_{n}$
Commented by mathmax by abdo last updated on 31/Jul/19
$let consider the diffeomorphisme (r,θ)→(x,y) / x =r cosθ and y =rsinθ we have 0≤x<n and 0≤y<n ⇒ 0≤x^2 +y^2 <2n^2 ⇒0≤r^2 <2n^2 ⇒0≤r<n(√2) A_n =∫∫_(0≤r<n(√2) and 0≤θ≤(π/2)) ((rdr dθ)/(√(r^2 +4))) =∫_0 ^(n(√2)) ((rdr)/(√(r^2 +4))) ∫_0 ^(π/2) dθ =(π/2)[(√(r^2 +4))]_0 ^(n(√2)) =(π/2){(√(2n^2 +4))−2} 2)lim_(n→+∞) A_n =+∞$
$l e t c o n s i d e r t h e d i f f e o m o r p h i s m e (r, θ) \to (x, y) /$ $x = r c o s θ a n d y = r s i n θ w e h a v e 0 ⩽ x < n a n d 0 ⩽ y < n \Rightarrow$ $0 ⩽ x^{2} + y^{2} < 2 n^{2} \Rightarrow 0 ⩽ r^{2} < 2 n^{2} \Rightarrow 0 ⩽ r < n \sqrt{2}$ $A_{n} = \int \int_{0 ⩽ r < n \sqrt{2} a n d 0 ⩽ θ ⩽ \frac{π}{2}} \frac{r d r d θ}{\sqrt{r^{2} + 4}}$ $= \int_{0}^{n \sqrt{2}} \frac{r d r}{\sqrt{r^{2} + 4}} \int_{0}^{\frac{π}{2}} d θ = \frac{π}{2} {[\sqrt{r^{2} + 4}]}_{0}^{n \sqrt{2}} = \frac{π}{2} {\sqrt{2 n^{2} + 4} - 2}$ $2) {l i m}_{n \to + \infty} A_{n} = + \infty$
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