Evaluate-0-1-0-1-1-2-x-2-y-2-dxdy-

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 \dots

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 = \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 \dots