Question

高数二重积分求解D={(x,y)|x^2+y^2<=y,x>=0}

f(x,y)=根号(1-x^2-y^2) - 8/pi *二重积分f(u,v)dudv 求f(x,y)

Answer 1

简单计算一下即可，答案如图所示

Answer 2

积分域
D={(x,y)|x^2+y^2<=y,
x>=0}
，是以P(0,1/2)为圆心，半径为1/2,
在第一象限的半圆。
则其面积为∫<D>∫dxdy
=
π/8.
∫<D>∫f(u,v)dudv
是一个常数，设为
C
=∫<D>∫f(u,v)dudv，则
f(x,y)
=
√(1-x^2-y^2)
-
(8/π)∫<D>∫f(u,v)dudv
=
√(1-x^2-y^2)
-
(8/π)C,
上式两边同时在D上作二重积分，得
C=∫<D>∫f(x,y)dxdy
=
∫<D>∫√(1-x^2-y^2)dxdy
-
(8/π)C∫<D>∫dxdy
C=∫<D>∫f(x,y)dxdy
=
∫<D>∫√(1-x^2-y^2)dxdy
-
(8/π)C(π/8),
2C
=∫<D>∫√(1-x^2-y^2)dxdy
,
则
C
=
∫<D>∫f(x,y)dxdy
=
(1/2)∫<D>∫√(1-x^2-y^2)dxdy,
得
f(x,y)=(1/2)√(1-x^2-y^2).