Evaluate : Ω=∫_0 ^( 1) ∫_0 ^( 1) (1/(2−x^2 − y^2 )) dxdy=??? ★★♣♣★★


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Question Number 108954 by mnjuly1970 last updated on 20/Aug/20

$E v a l u a t e :$ $Ω = \int_{0}^{1} \int_{0}^{1} \frac{1}{2 - x^{2} - y^{2}} d x d y = ? ? ?$ $★ ★ ♣ ♣ ★ ★$
Commented by kaivan.ahmadi last updated on 20/Aug/20

$x = r c o s θ, y = r s i n θ \Rightarrow - x^{2} - y^{2} = - r^{2}$ $0 ⩽ y ⩽, 0 ⩽ x ⩽ 1 \Rightarrow 0 ⩽ θ ⩽ \frac{π}{2}, 0 ⩽ r ⩽ 1$ $Ω = \int_{0}^{\frac{π}{2}} \int_{0}^{1} \frac{1}{2 - r^{2}} r d r d θ$ $f i r s t f i n d \int_{0}^{1} \frac{r}{2 - r^{2}} d r$ $u = 2 - r^{2} \Rightarrow d u = - 2 r d r$ $\Rightarrow \frac{- 1}{2} \int \frac{d u}{u} = \frac{- 1}{2} l n u = \frac{- 1}{2} l n (2 - r^{2}) ∣_{0}^{1} = \frac{- 1}{2} (l n 1 - l n 2) =$ $\frac{l n 2}{2}$ $\Rightarrow Ω = \int_{0}^{\frac{π}{2}} \frac{l n 2}{2} d θ = \frac{l n 2}{2} \times \frac{π}{2} = \frac{π}{4} l n 2$
Commented by mnjuly1970 last updated on 20/Aug/20

$0 ⩽ r ⩽ s e c (θ) & 0 ⩽ θ ⩽ \frac{π}{4} b e c a u s e$ $i n t e g r a n d i s s y m m e t r i c$ $f (x, y) = f (y, x)$ $f i n a l a n s w e r i s G : c a t a l a n c o n s t a n t . . .$
Commented by kaivan.ahmadi last updated on 20/Aug/20

$H i s i r, w h y d o w e u s e a d i f i c u l t w a y w h e n w e h a v e a$ $h a v e a s i m p l e w a y ?$
Commented by mathmax by abdo last updated on 20/Aug/20

$0 ⩽ x ⩽ 1 and 0 ⩽ y ⩽ 1 \Rightarrow 0 ⩽ x^{2} + y^{2} ⩽ 2 \Rightarrow 0 ⩽ r ⩽ {\sqrt{2}}^{'} . .!$
Commented by kaivan.ahmadi last updated on 21/Aug/20

$s t u d y p o l a r c o o r d i n a t e s y s t e m p l e a s e .$
	Answered by mathmax by abdo last updated on 20/Aug/20
	$Ω =∫_0 ^1 ∫_0 ^1 ((dxdy)/(2−x^2 −y^2 )) we considere the diffeomorphism { ((x =rcosθ _(⇒ ) Ω =∫∫ _(o≤r≤(√2)and 0≤θ≤(π/2)) (1/(2−r^2 ))r dr dθ)),((y =rsinθ)) :} =(π/2)∫_0 ^(√2) ((rdr)/((2−r^2 ))) =−(π/4) ∫_0 ^(√2) ((−2r)/(2−r^2 ))dr =−(π/4)[ln∣2−r^2 ∣]_0 ^(√2) =∞ this integral is divergent...$
	$Ω = \int_{0}^{1} \int_{0}^{1} \frac{dxdy}{2 - x^{2} - y^{2}} we considere the diffeomorphism$ ${\begin{cases} x = rcos θ_{\Rightarrow} Ω = \int \int_{o ⩽ r ⩽ \sqrt{2} and 0 ⩽ θ ⩽ \frac{π}{2}} \frac{1}{2 - r^{2}} r dr d θ \\ y = rsin θ \end{cases}$ $= \frac{π}{2} \int_{0}^{\sqrt{2}} \frac{rdr}{(2 - r^{2})} = - \frac{π}{4} \int_{0}^{\sqrt{2}} \frac{- 2 r}{2 - r^{2}} dr = - \frac{π}{4} {[\ln ∣ 2 - r^{2} ∣]}_{0}^{\sqrt{2}} = \infty$ $this integral is divergent . . .$
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