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Question Number 27182 by abdo imad last updated on 02/Jan/18

find the value of I_a = ∫∫_D_a e^(−((x^2 +y^2 )/2)) dxdy with D_a ={(x,y)∈R^2 / x^2 +y^2 ≤ a^2 }

findthevalueofIa=∫∫Dae−x2+y22dxdywithDa={(x,y)∈R2/x2+y2⩽a2}

Commented by abdo imad last updated on 05/Jan/18

we use the polar coordinates and the diffeomorphismeψ (r,θ)−>ψ(r,θ) =(x,y)=(rcosθ,rsinθ) I_(a ) =∫∫_(o≤r≤a) e^(−(r^2 /2)) rdr dθ = ∫_0 ^(2π) ( ∫_0 ^a r e^(−(r^2 /2)) dr )dθ= 2π ∫_0 ^a r e^(−(r^2 /2)) dr = 2π[ − e^(−(r^2 /2)) ]_0 ^a =2π(1− e^(−(a^2 /2)) )

weusethepolarcoordinatesandthediffeomorphismeψ(r,θ)−>ψ(r,θ)=(x,y)=(rcosθ,rsinθ)Ia=∫∫o⩽r⩽ae−r22rdrdθ=∫02π(∫0are−r22dr)dθ=2π∫0are−r22dr=2π[−e−r22]0a=2π(1−e−a22)

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