Question

计算二重积分∫∫√（x^2+y)dxdy,其中D：x^2+y^2≤2x请问极坐标θ角的取值范围是

Answer 1

设x=rcost y=rsint -π/2<=t<=π/2 所以r^2<=2rcost r<=2cost ∫∫√（x^2+y^2)dxdy =∫[-π/2,π/2] dt ∫[0,2cost] r^2dr =∫[-π/2,π/2] dt 1/3r^3 [0,2cost] =8/3 ∫[-π/2,π/2] cos^3t dt =8/3∫[-π/2,π/2] (1-sin^2t) d(sint) =8/3*(sint-1/3sin^3t) [-π/2,π/2] =32/9