let A_n =∫∫_w_n e^(−x^2 −y^2 ) (√(x^2 +y^2 ))dxdy with w_n =[(1/n),n]×[(1/n),n] 1) calculate A_n interms of n 2) find lim_(n→+∞) A


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Question Number 57103 by turbo msup by abdo last updated on 30/Mar/19

$l e t A_{n} = \int \int_{w_{n}} e^{- x^{2} - y^{2}} \sqrt{x^{2} + y^{2}} d x d y$ $w i t h w_{n} = [\frac{1}{n}, n] \times [\frac{1}{n}, n]$ $1) c a l c u l a t e A_{n} i n t e r m s o f n$ $2) f i n d {l i m}_{n \to + \infty} A_{n}$
Commented by 121194 last updated on 30/Mar/19

$x = r \cos θ$ $y = r \sin θ$ $J (r, θ) = r$ $\int \int_{w_{n}^{}} e^{- r^{2}} r \cdot r d r d θ = \int \int_{w_{n}^{}} e^{- r^{2}} r^{2} d r d θ$
Commented by maxmathsup by imad last updated on 30/Mar/19
$let use the changement x=rcosθ and y=rsinθ we have (1/n)≤x≤n and (1/n) ≤y≤n ⇒(2/n^2 ) ≤x^2 +y^2 ≤2n^2 ⇒(2/n^2 ) ≤r^2 ≤2n^2 ⇒((√2)/n) ≤r≤n(√2) also we have x≥0 and y ≥0 ⇒ 0≤θ ≤(π/2) ⇒ A_n =∫∫_(((√2)/n)≤r≤n(√2)and 0≤θ≤(π/2)) e^(−r^2 ) r rdrdθ =∫∫_(w^′ n) r^2 e^(−r^2 ) drdθ =(∫_((√2)/n) ^(n(√2)) r^2 e^(−r^2 ) dr)∫_0 ^(π/2) dθ =(π/2) ∫_((√2)/n) ^(n(√2)) r^2 e^(−r^2 ) dr by parts u=r and v^′ =r e^(−r^2 ) ∫_((√2)/n) ^(n(√2)) r(r e^(−r^2 ) )dr =[−(1/2) r e^(−r^2 ) ]_((√2)/n) ^(n(√2)) +(1/2) ∫_((√2)/n) ^(n(√2)) e^(−r^2 ) dr =(1/2){((√2)/n) e^(−(2/n^2 )) −n(√2)e^(−2n^2 ) } +(1/2) ∫_((√2)/n) ^(n(√2)) e^(−r^2 ) dr ⇒ A_n =((π(√2))/4){ (1/n) e^(−(2/n^2 )) −n e^(−2n^2 ) } +(π/4) ∫_((√2)/n) ^(n(√2)) e^(−r^2 ) dr... 2) we have lim_(n→+∞) (1/n) e^(−(2/n^2 )) −n e^(−2n^2 ) =0 and lim_(n→+∞) ∫_((√2)/n) ^(n(√2)) e^(−r^2 ) dr =∫_0 ^∞ e^(−r^2 ) dr =((√π)/2) ⇒ lim_(n→+∞) A_n =((π(√π))/8) .$
$l e t u s e t h e c h a n g e m e n t x = r c o s θ a n d y = r s i n θ w e h a v e$ $\frac{1}{n} ⩽ x ⩽ n a n d \frac{1}{n} ⩽ y ⩽ n \Rightarrow \frac{2}{n^{2}} ⩽ x^{2} + y^{2} ⩽ 2 n^{2} \Rightarrow \frac{2}{n^{2}} ⩽ r^{2} ⩽ 2 n^{2} \Rightarrow \frac{\sqrt{2}}{n} ⩽ r ⩽ n \sqrt{2}$ $a l s o w e h a v e x ⩾ 0 a n d y ⩾ 0 \Rightarrow 0 ⩽ θ ⩽ \frac{π}{2} \Rightarrow$ $A_{n} = \int \int_{\frac{\sqrt{2}}{n} ⩽ r ⩽ n \sqrt{2} a n d 0 ⩽ θ ⩽ \frac{π}{2}} e^{- r^{2}} r r d r d θ = \int \int_{w^{^{'}} n} r^{2} e^{- r^{2}} d r d θ$ $= (\int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} r^{2} e^{- r^{2}} d r) \int_{0}^{\frac{π}{2}} d θ = \frac{π}{2} \int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} r^{2} e^{- r^{2}} d r b y p a r t s u = r a n d v^{^{'}} = r e^{- r^{2}}$ $\int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} r (r e^{- r^{2}}) d r = {[- \frac{1}{2} r e^{- r^{2}}]}_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} + \frac{1}{2} \int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} e^{- r^{2}} d r$ $= \frac{1}{2} {\frac{\sqrt{2}}{n} e^{- \frac{2}{n^{2}}} - n \sqrt{2} e^{- 2 n^{2}}} + \frac{1}{2} \int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} e^{- r^{2}} d r \Rightarrow$ $A_{n} = \frac{π \sqrt{2}}{4} {\frac{1}{n} e^{- \frac{2}{n^{2}}} - n e^{- 2 n^{2}}} + \frac{π}{4} \int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} e^{- r^{2}} d r . . .$ $2) w e h a v e {l i m}_{n \to + \infty} \frac{1}{n} e^{- \frac{2}{n^{2}}} - n e^{- 2 n^{2}} = 0 a n d {l i m}_{n \to + \infty} \int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} e^{- r^{2}} d r$ $= \int_{0}^{\infty} e^{- r^{2}} d r = \frac{\sqrt{π}}{2} \Rightarrow {l i m}_{n \to + \infty} A_{n} = \frac{π \sqrt{π}}{8} .$
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