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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}
caculate∫∫D(x2−y2)e−x2−y2dxdywithD={(x,y)∈R2/x2+y2⩽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 .
letconsiderthediffeomrphisme(r,θ)→φ(r,θ)=(x,y)/x=rcosθandy=rsinθwehavex2+y2⩽4⇒0<r⩽2and0⩽θ⩽2π∫∫D(x2−y2)e−(x2+y2)dx=∫∫Wfoφ(r,θ)rdrdθ=∫∫0<r⩽2r2{cos2θ−sin2θ}e−r2rdrdθ=∫02r3e−r2dr.∫02πcos2θdθbut∫02πcos(2θ)dθ=[sin(2θ)2]02π=0⇒I=0.

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