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1-find-W-xdx-a-2-x-2-y-2-with-W-a-x-2-y-2-a-2-and-x-gt-0-a-gt-0-2-calculate-W-1-xdx-x-2-y-2-1-




Question Number 82872 by abdomathmax last updated on 25/Feb/20
1)find ∫∫_W  ((xdx)/(a^2  +x^2  +y^2 )) with  W_a →x^2  +y^2  ≤a^2  and x>0     (a>0)  2)calculate ∫∫_W_1     ((xdx)/(x^2 +y^2  +1))
1)findWxdxa2+x2+y2withWax2+y2a2andx>0(a>0)2)calculateW1xdxx2+y2+1
Commented by mathmax by abdo last updated on 25/Feb/20
1)  we use the diffeomorphism  (r,θ)→(x,y)/x=rcosθ  and y=rsinθ  wehave  x^2  +y^2 ≤a^2  ⇒0≤r≤a   x>0 ⇒θ∈]−(π/2),(π/2)[  ⇒∫∫_W_a   ((xdxdy)/(a^2  +r^2 )) =∫_0 ^a  ∫_(−(π/2)) ^(π/2)   ((rcosθ rdr dθ)/(a^2  +r^2 ))   ∫_0 ^a  (r^2 /(r^2  +a^2 )) dr ×∫_(−(π/2)) ^(π/2)  cosθ dθ  =2 ∫_0 ^a  (r^2 /(r^2  +a^2 ))dr  =2 ∫_0 ^a  ((r^2  +a^2 −a^2 )/(r^2  +a^2 ))dr =2a −2a^2  ∫_0 ^a  (dr/(r^2  +a^2 ))  (ch. r=ax)  =2a−2a^2  ∫_0 ^1  ((adx)/(a^2 (1+x^2 ))) =2a−2a [arctanx]_0 ^1   =2a−2a×(π/4) =(2−(π/2))a   2) ∫∫_W_1      ((xdx)/(x^2  +y^2  +1)) =_(a=1)    2−(π/2)
1)weusethediffeomorphism(r,θ)(x,y)/x=rcosθandy=rsinθwehavex2+y2a20rax>0θ]π2,π2[Waxdxdya2+r2=0aπ2π2rcosθrdrdθa2+r20ar2r2+a2dr×π2π2cosθdθ=20ar2r2+a2dr=20ar2+a2a2r2+a2dr=2a2a20adrr2+a2(ch.r=ax)=2a2a201adxa2(1+x2)=2a2a[arctanx]01=2a2a×π4=(2π2)a2)W1xdxx2+y2+1=a=12π2

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