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let-a-gt-0-calculate-D-a-xdxdy-a-2-x-2-y-2-and-D-a-x-y-R-2-x-2-y-2-a-2-and-x-gt-0-




Question Number 78703 by abdomathmax last updated on 20/Jan/20
let a>0  calculate ∫∫_D_a      ((xdxdy)/(a^2  +x^2  +y^2 ))  and D_a ={(x,y)∈R^2 / x^2  +y^2 ≤a^2   and x>0}
leta>0calculateDaxdxdya2+x2+y2andDa={(x,y)R2/x2+y2a2andx>0}
Answered by mind is power last updated on 20/Jan/20
x=rcos(θ),y=rsin(θ)  D_a ⇔{(r,θ)∈R^2 /r^2 <a^2 �cos(θ)>0}  r∈[0,a],θ∈[−(π/2),(π/2)[  ∫∫_D_a  ((xdxdy)/(a^2 +x^2 +y^2 ))=∫_((−π)/2) ^(π/2) ∫_0 ^a ((r^2 cos(θ)drdθ)/(a^2 +r^2 ))  =∫_(−(π/2)) ^(π/2) cos(θ)dθ.∫_0 ^a (r^2 /(a^2 +r^2 ))dr  =[sin(θ)]_(−(π/2)) ^(π/2) .[1−(a^2 /(a^2 +r^2 ))]_0 ^a   2[x−aarctan((r/a))]_0 ^a 2(a−a(π/4))=a(((4−π)/2))
x=rcos(θ),y=rsin(θ)Da{(r,θ)R2/r2<a2cos(θ)>0}r[0,a],θ[π2,π2[Daxdxdya2+x2+y2=π2π20ar2cos(θ)drdθa2+r2=π2π2cos(θ)dθ.0ar2a2+r2dr=[sin(θ)]π2π2.[1a2a2+r2]0a2[xaarctan(ra)]0a2(aaπ4)=a(4π2)

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