Question

求这个二重积分怎么算？

Answer 1

二重积分的计算方法

Answer 2

x = (1-y^2)/2 即 y^2 + 2x - 1 = 0,
化为极坐标 r^2(sint)^2 + 2rcost -1 = 0，
r ≥ 0, 则 r = (1-cost)/(sint)^2 = 1/(1+cost) , 0 ≤ t ≤ π/2
I = ∫<下0, 上π/2>dt ∫<下0, 上1/(1+cost)> r rdr
= (1/3)∫<下0, 上π/2>dt [r^3]<下0, 上1/(1+cost)>
= (1/3)∫<下0, 上π/2>dt/(1+cost)^3
= (1/3)(1/8)∫<下0, 上π/2>dt/[cos(t/2)]^6
= (1/12)∫<下0, 上π/2>[sec(t/2)]^6 d(t/2)
= (1/12)∫<下0, 上π/2>[sec(t/2)]^4 dtan(t/2)
= (1/12)∫<下0, 上π/2>{[tan(t/2)]^2+1}^2 dtan(t/2)
= (1/12)∫<下0, 上π/2>{[tan(t/2)]^4+2[tan(t/2)]^2+1}dtan(t/2)
= (1/12){(1/5)[tan(t/2)]^5+(2/3)[tan(t/2)]^3+tan(t/2)}<下0, 上π/2>
= (1/12)(1/5 + 2/3 + 1) = 7/45