设D:x^2+y^2=0,f(x,y)为D上的连续函数,且f(x,y)=[1-(x^2+y^2)]^0.5-∏/8*∫∫f(x,y)dxdy,求f(x,y)
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设D:x^2+y^2=0,f(x,y)为D上的连续函数,且f(x,y)=[1-(x^2+y^2)]^0.5-∏/8*∫∫f(x,y)dxdy,求f(x,y)
设D:x^2+y^2=0,f(x,y)为D上的连续函数,且f(x,y)=[1-(x^2+y^2)]^0.5-∏/8*∫∫f(x,y)dxdy,求f(x,y)
设D:x^2+y^2=0,f(x,y)为D上的连续函数,且f(x,y)=[1-(x^2+y^2)]^0.5-∏/8*∫∫f(x,y)dxdy,求f(x,y)
注意：∫∫f(x,y)dxdy其实是一个常数,设a=∫∫f(x,y)dxdy
则：f(x,y)=[1-(x^2+y^2)]^0.5-πa/8
两边做二重积分得：
∫∫f(x,y)dxdy 积分区域为：x²+(y-1/2)²≤1/4,x≥0,圆的极坐标方程为：r=sinθ,θ：0--->π/2
=∫∫ {[1-(x^2+y^2)]^0.5-πa/8} dxdy
=∫∫ [1-(x^2+y^2)]^0.5dxdy-πa/8∫ dxdy
后一个被积函数为1,结果为区域面积,即半圆面积
=∫∫ [1-(x^2+y^2)]^0.5dxdy-(πa/8)*(1/2)*π(1/2)²
=∫∫ √(1-r²)*rdrdθ-(π²a/64)
=∫[0--->π/2]dθ∫[0--->sinθ] √(1-r²)*rdr-(π²a/64)
=1/2∫[0--->π/2]dθ∫[0--->sinθ] √(1-r²)d(r²)-(π²a/64)
=-(1/2)(2/3)∫[0--->π/2] (1-r²)^(3/2) |[0--->sinθ]dθ-(π²a/64)
=(1/3)∫[0--->π/2] (1-cos³θ) dθ-(π²a/64)
=(1/3)∫[0--->π/2] 1 dθ-(1/3)∫[0--->π/2] cos³θ dθ-(π²a/64)
=(1/3)(π/2)-(1/3)∫[0--->π/2] cos²θ d(sinθ)-(π²a/64)
=(1/3)(π/2)-(1/3)∫[0--->π/2] (1-sin²θ) d(sinθ)-(π²a/64)
=(1/3)(π/2)-(1/3)(sinθ-(1/3)sin³θ)-(π²a/64) |[0--->π/2]
=(π/6)-(1/3)(1-(1/3))-(π²a/64)
=(π/6)-(2/9)-(π²a/64)
因此：a=(π/6)-(2/9)-(π²a/64)
解得：a=[(π/6)-(2/9)]/[1+(π²/64)]=(32/9)*(3π-4)/(64+π²)
因此：f(x,y)=[1-(x^2+y^2)]^0.5-π/8*a
得：f(x,y)=[1-(x^2+y^2)]^0.5-(4π/9)*(3π-4)/(64+π²)
这么变态的结果,不知有没算错.
正确