∫∫(D)arctan y/x dxdy.D:1≤x^2+y^2≤4,y≥0,y≤x

题目

∫∫(D)arctan y/x dxdy.D:1≤x^2+y^2≤4,y≥0,y≤x

答案

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