设D={(x,y)/x^2+y^2≤x},求∫∫x^1/2dxdy
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设D={(x,y)/x^2+y^2≤x},求∫∫x^1/2dxdy
设D={(x,y)/x^2+y^2≤x},求∫∫x^1/2dxdy
设D={(x,y)/x^2+y^2≤x},求∫∫x^1/2dxdy
D={(x,y)|x^2+y^2≤x}={(x,y)|(x-1/2)^2+y^2≤1/2}
∫∫x^1/2dxdy=∫[0,1]x^1/2dx∫ [-(x-x^2)^1/2,(x-x^2)^1/2]dy
=∫[0,1]x^1/2*2(x-x^2)^1/2dx=∫[0,1]2x*2(1-x)^1/2dx
令t=(1-x)^1/2,x=1-t^2,dx=-2tdt
原式=∫[0,1]2(1-t^2)t*2tdt=4*∫[0,1](t^2-t^4)dt=4*1/3-4*1/5=8/15.
[ ,]表示[下限,上限]
可能运算时出错,方法应该是对的,仅供参考.
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