求大神解答这道二重积分题，非常谢谢
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积分域 D 是圆 (x-1)^2 + y^2 = 1 的右半圆部分。化为极坐标是
r = 2cost， -π/4 ≤ t ≤ π/4, sect ≤ t ≤ 2cost
I = ∫<-π/4, π/4>dt ∫<sect, 2cost>dr/r^3 =
= (-1/2)∫<-π/4, π/4>dt [1/r^2]<sect, 2cost>
= (-1/2)∫<-π/4, π/4> [(1/4)(sect)^2-(cost)^2]dt
= (-1/4)∫<0, π/4> [(sect)^2-4(cost)^2]dt
= (-1/4)∫<0, π/4> [(sect)^2-(2+2cos2t)]dt
= (-1/4)[tant - 2t - sin2t]<0, π/4>
= (-1/4)[1-π/2 -1 ] = π/8
r = 2cost， -π/4 ≤ t ≤ π/4, sect ≤ t ≤ 2cost
I = ∫<-π/4, π/4>dt ∫<sect, 2cost>dr/r^3 =
= (-1/2)∫<-π/4, π/4>dt [1/r^2]<sect, 2cost>
= (-1/2)∫<-π/4, π/4> [(1/4)(sect)^2-(cost)^2]dt
= (-1/4)∫<0, π/4> [(sect)^2-4(cost)^2]dt
= (-1/4)∫<0, π/4> [(sect)^2-(2+2cos2t)]dt
= (-1/4)[tant - 2t - sin2t]<0, π/4>
= (-1/4)[1-π/2 -1 ] = π/8
追问
谢谢大神，大神积分t的范围不是-π/2→π/2吗？
大神我懂了，给你比心，谢谢你😊
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