zhengxuejia1983
春芽
共回答了19个问题采纳率:94.7% 举报
I = I1-I2
I1 = ∫dt∫r(cost+sint)rdr
= (8/3)∫(cost+sint)dt
= (8/3)[sint-cost] = 16/3.
I2 = ∫dt∫r(cost+sint)rdr
= (8/3)∫(cost+sint)(cost)^3dt
= (8/3)[∫(cost)^4dt+∫sint(cost)^3dt]
其中
I3 = (8/3)∫(cost)^4dt = (2/3)∫(1+cos2t)^2dt
= (1/3)∫[2+4cos2t+2(cos2t)^2]dt
= (1/3)∫[3+4cos2t+cos4t]dt
= (1/3)[3t+2sin2t+(1/4)sin4t] = π/2;
I4 = (8/3)∫sint(cost)^3dt]
= (-8/3)∫(cost)^3d(cost)
= (-2/3)[(cost)^4] = 2/3.
I2 = I3+I4 = π/2+2/3.
I = I1-I2 = 16/3-π/2-2/3 = 14/3-π/2.
1年前
追问
6