设f(x)连续,F(t)=∫∫∫(k)[x^2+f(x^2+y^2)]dxdydz,其中k：0

Answer 1

用柱坐标
F(t)=∫ ∫ ∫ (k)[x^2+f(x^2+y^2)]dxdydz
=∫ ∫ ∫ [r²cos²θ+f(r²)]rdzdrdθ
=∫[0--->2π]dθ ∫ [0--->t]dr∫[0--->h] [r²cos²θ+f(r²)]rdz
=h∫[0--->2π]dθ ∫[0--->t] [r²cos²θ+f(r²)]rdr
=h∫[0--->2π]dθ ∫[0--->t] r³cos²θdr+h∫[0--->2π]dθ ∫[0--->t] rf(r²)dr
=(h/2)∫[0--->2π] (1+cos2θ)dθ ∫[0--->t] r³dr+h∫[0--->2π]dθ ∫[0--->t] rf(r²)dr
=(h/2)(θ+(1/2)sin2θ) |[0--->2π] ∫[0--->t] r³dr+h∫[0--->2π]dθ ∫[0--->t] rf(r²)dr
=πh∫[0--->t] r³dr+h∫[0--->2π]dθ ∫[0--->t] rf(r²)dr
因此：F'(t)=πht³+2πh*t f(t²)