第5个回答 2020-08-28
可以啊。
I = ∫<0, 2>y^2 dy ∫<-2, -√(2y-y^2)> dx
= ∫<0, 2>y^2[2-√(2y-y^2)]dy
= 2∫<0, 2>y^2dy - ∫<0, 2>y^2√(2y-y^2)dy
= (2/3)[y^3]<0, 2> - I1 = 16/3 - I1
对于 I1, √(2y-y^2) = √[1-(y-1)^2], 令 y-1 = sint,
则 √[1-(y-1)^2] = cost
I2 = ∫<-π/2, π/2>(1+sint)^2 (cost)^2 dt
= ∫<-π/2, π/2>[1+2sint+(sint)^2](cost)^2 dt
= ∫<0, π/2>[2(cost)^2+2(sint)^2(cost)^2]dt
= ∫<0, π/2>[1+cos2t+(1/2)(sin2t)^2]dt
= ∫<0, π/2>[1+cos2t+1/4+(1/4)cos4t]dt
= [5t/4+(1/2)sin2t+(1/16)sin4t]<0, π/2> = 5π/8
I = 16/3 - 5π/8本回答被提问者采纳