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

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).