Question

数学分析，这个怎么求

数学分析，这个怎么求帮帮忙，非常感谢

Answer 1

因为积分区域D关于x轴对称，且y+xyf(x^2m+y^2m)是关于y的奇函数，所以
原式=∫∫(D) xdσ
将D换算成极坐标形式
r^2<=2mrcosθ
r<=2mcosθ，θ∈[-π/2,π/2)
所以原式=∫(-π/2,π/2)dθ∫(0,2mcosθ) rcosθ*rdr
=∫(-π/2,π/2) cosθdθ*(r^3/3)|(0,2mcosθ)
=∫(-π/2,π/2) (8/3)*m^3*cos^4θdθ
=(4/3)*m^3*∫(0,π/2) (2cos^2θ)^2dθ
=(4/3)*m^3*∫(0,π/2) (1+cos2θ)^2dθ
=(4/3)*m^3*∫(0,π/2) (1+2cos2θ+cos^2(2θ))dθ
=(2/3)*m^3*∫(0,π/2) (2+4cos2θ+2cos^2(2θ))dθ
=(2/3)*m^3*∫(0,π/2) (3+4cos2θ+cos4θ)dθ
=(1/6)*m^3*(12θ+8sin2θ+sin4θ)|(0,π/2)
=(1/6)*m^3*6π
=πm^3