caculate ∫∫_D (x^2 −y^2 ) e^(−x^2 −y^2 ) dxdy with D ={(x,y)∈R^2 / x^2 +y^2 ≤4}


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Question Number 46848 by maxmathsup by imad last updated on 01/Nov/18
$caculate ∫∫_D (x^2 −y^2 ) e^(−x^2 −y^2 ) dxdy with D ={(x,y)∈R^2 / x^2 +y^2 ≤4}$
$c a c u l a t e \int \int_{D} (x^{2} - y^{2}) e^{- x^{2} - y^{2}} d x d y w i t h$ $D = {(x, y) \in R^{2} / x^{2} + y^{2} ⩽ 4}$
Commented by maxmathsup by imad last updated on 03/Nov/18
$let consider the diffeomrphisme (r,θ) →ϕ(r,θ)=(x,y) / x =r cosθ and y =rsinθ we have x^2 +y^2 ≤4 ⇒0<r≤2 and 0≤θ≤2π ∫∫_D (x^2 −y^2 )e^(−(x^2 +y^2 )) dx =∫∫_W foϕ(r,θ)rdrdθ =∫∫_(0<r≤2) r^2 {cos^2 θ −sin^2 θ}e^(−r^2 ) r dr dθ =∫_0 ^2 r^3 e^(−r^2 ) dr .∫_0 ^(2π) cos2θ dθ but ∫_0 ^(2π) cos(2θ)dθ =[((sin(2θ))/2)]_0 ^(2π) =0 ⇒ I =0 .$
$l e t c o n s i d e r t h e d i f f e o m r p h i s m e (r, θ) \to φ (r, θ) = (x, y) /$ $x = r c o s θ a n d y = r s i n θ w e h a v e x^{2} + y^{2} ⩽ 4 \Rightarrow 0 < r ⩽ 2 a n d 0 ⩽ θ ⩽ 2 π$ $\int \int_{D} (x^{2} - y^{2}) e^{- (x^{2} + y^{2})} d x = \int \int_{W} f o φ (r, θ) r d r d θ$ $= \int \int_{0 < r ⩽ 2} r^{2} {{c o s}^{2} θ - {s i n}^{2} θ} e^{- r^{2}} r d r d θ$ $= \int_{0}^{2} r^{3} e^{- r^{2}} d r . \int_{0}^{2 π} c o s 2 θ d θ b u t \int_{0}^{2 π} c o s (2 θ) d θ = {[\frac{s i n (2 θ)}{2}]}_{0}^{2 π} = 0 \Rightarrow$ $I = 0 .$
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