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∫∫ $sqrt{frac{1-x^2-y^2 }{1+x^2+y^2} }$ dxdy
x $x^{2} + y^{2} leq 1$
$x geq 0$
$y geq 0$

$iint _{D} , sqrt{frac{1-x^2-y^2}{1+x^2+y^2} }dx, dy=[, x=rcdot cosvarphi ,; y=rsinvarphi ,; dx, dy=r, dr, dvarphi , ]==int _0^{frac{pi}{2}}dvarphi int _0^1sqrt{frac{1-r^2}{1+r^2}} cdot r, dr=Iintsqrt{frac{1-r^2}{1+r^2}}cdot r, dr=[t=r^2,dt=2rdr]=$

$=intsqrt{frac{1-t}{1+t}}cdot frac{1}{2}dt=$

$=frac{1}{2} int frac{sqrt{(1-t)(1-t)}}{sqrt{(1+t)(1-t)}}dt=frac{1}{2}int frac{1-t}{sqrt{1-t^2}}dt=frac{1}{2}int frac{dt}{sqrt{1-t^2}}-frac{1}{4}int frac{2tdt}{sqrt{1-t^2}}==frac{1}{2}arcsint+frac{1}{4}*2sqrt{1-t^2}+C=frac{1}{2}arcsin x^2+frac{1}{2}sqrt{1-x^4}+C$

$I=int _0^{frac{pi}{2}}(frac{1}{2}arcsinx^2|_0^1+frac{1}{2}sqrt{1-x^4}|_0^1)dvarphi ==int _0^{frac{pi}{2}}(frac{1}{2}*frac{pi}{2}-frac{1}{2})dvarphi=(frac{pi}{4}-frac{1}{2})varphi |_0^1=frac{pi}{4}-frac{1}{2}==$