 | 積分範囲が分かればあとはできますね。
>① D={ay>x^2,ax>y^2}
∬D xydxdy=∫[x:0→a]∫[y:x^2/a→√(ax)] xydxdy=…
>② D={x^2+y^2<ax,x>y>0}の時
D={x^2+y^2<ax,x>y>0}={y<x<√{(a/2)^2-y^2}+a/2,0<y<a/2}
∬D xdxdy=…
③
∬y>0 1/(x^2+y+1)^2dxdy=∫[x:-∞→∞]∫[y:0→∞]∫1/(x^2+y+1)^2dxdy=
∫[x:-∞→∞][-1/(x^2+y+1)][y:0→∞]dx=∫[x:-∞→∞]1/(x^2+1)dx
=[Tan^(-1)x][x:-∞→∞]=π
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