Question

计算二重积分∫∫√（x^2+y^2)dxdy,其中D：x^2+y^2≤2x。 D

Answer 1

化成极坐标，x^2+y^2≤2x，变成r=2cosθ
积分区域；0≤r≤2cosθ，
π/2≤θ≤π/2，
区域以X轴为上下对称，只求第一象限区域，明大毁再2倍即可，
I=2∫[0,π/2] dθ∫仿亮[0,2cosθ] r*rdr
=2∫[0,π/2] dθ (r^3/3)[0,2cosθ]
=(2/3)∫[0,π/2] *8(cosθ)^3 dθ
=（16/3）∫[0,π/2] [1-(sinθ)^2]d(sinθ)
=(16/3)[sinθ-(sinθ)^3/3] [0,π/2]
=(16/3)[1/2-1/激备8)
=32/9.

Answer 2

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 ∫或凯仿[0,2cost] r^2dr
=∫[-π/2,π/2] dt 1/孙迅3r^3 [0,2cost]
=8/3 ∫[-π/衫纤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