Question

五设+D:x^2+y^2Rx,+其中+R>0+求+I=f((k^

Answer 1

问：五设+D:x^2+y^2Rx,+其中+R>0+求+I=f((k^

答：亲你好很高兴为你服务注意：∫∫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-cos

答：=(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)

问：好像题没拍清楚

答：嗯嗯

问：就是这个sin要写绝对值吗

答：是的