积分区域如图,是正方形,并不适合用极坐标求积分,很繁琐。由图可见,ρ的长度是角度的函数。以π/4分界,2个区域ρ 的表达式不同,所以2个区域要分别积分,再求和;变换为极坐标,x^2 + y^2 =ρ^2,x =ρcosα;dxdy =ρdρdα;上半部分积分:∫( π/4,π/2 ) cosα dα ∫( 0,2/sinα )ρ *ρ^2 * ρ dρ=∫( π/4,π/2 ) cosα dα [ρ^5/5 ]( 0,2/sinα )= (32/5)∫( π/4,π/2 ) cosα/(sinα)^5dα= (32/5)∫( π/4,π/2 ) d(sinα)/(sinα)^5= (32/5) * (-1/4)[ 1/(sinα)^4]( π/4,π/2 )= (-8/5)[ 1 - 4 ]= 24/5;下半部分积分:∫( 0,π/4 ) cosα dα ∫( 0,2/cosα ) ρ * ρ^2 * ρ dρ= ∫( 0,π/4 ) cosα dα [ ρ^5/5 ]( 0,2/cosα )= (32/5)∫( 0,π/4 ) 1/(cosα)^4 dα= (32/5) [ 2tanα/3 + sinα/3(cosα)^3 ]( 0,π/4 )= (32/5) [ 2/3 + 2/3 ]= 128/15;两部分积分之和 =24/5 + 128/15 = 200/15 = 40/3 。