题目

∫∫(D)arctan y/x dxdy. D:1≤x^2+y^2≤4,y≥0,y≤x
x=rcosθ y=rsinθ ∫∫(D)arctan y/x dxdy=∫∫(D')arctan(sinθ/cosθ)rdrdθ 其中D':1<=r<=2,0<=θ<=π/4 那么 ∫∫(D)arctan y/x dxdy=∫∫(D')arctan(sinθ/cosθ)rdrdθ= ∫(0->π/4)∫(1->2)θr dr dθ= ∫(0->π/4) θ/2*r^2|(1->2) dθ= ∫(0->π/4) θ/2*(4-1) dθ= 3/4*θ^2|(0->π/4)=3π^2/64 其中∫∫(D')arctan(sinθ/cosθ)rdrdθ= ∫(0->π/4)∫(1->2)θr dr dθ是怎么化简的

提问时间：2021-02-04

答案

∫∫(D')arctan (sinθ/cosθ)rdrdθ= ∫∫(D')arctan(tan θ)rdrdθ = ∫∫(D') θrdrdθ