Question

把I=∫∫(x^2+y^2<=2x)f(x^2+y^2,arctany/x)dxdy化为极坐标系下的二次积分

我已经被高数搞死了 大神来解救我吧~~o(>_<)o ~~

Answer 1

{ x = rcosθ
{ y = rsinθ
x² + y² ≤ 2x
(rcosθ)² + (rsinθ)² ≤ 2(rcosθ)
r² ≤ 2rcosθ
r ≤ 2cosθ、θ：- π/2→π/2
∫∫ f(x² + y²，arctan(y/x)) dxdy
= ∫(- π/2→π/2) ∫(0→2cosθ) f(r²，arctan(rsinθ/rcosθ) r drdθ
= ∫(- π/2→π/2) dθ ∫(0→2cosθ) f(r²，arctan(tanθ)) r dr
= ∫(- π/2→π/2) dθ ∫(0→2cosθ) f(r²，θ) r dr

Answer 2

I=∫∫(x^2+y^2<=2x)f(x^2+y^2,arctany/x)dxdy=∫(-π/2.π/2)dθ∫(0,2cosθ)f(r^2,θ)rdr