Question

微积分二重积分问题！！在线等！

D是封闭区域。0<=y<=sinx,0<=x<=pai

Answer 1

解；
∫∫D (x^2-y^2)ds
=∫[0,π] dx {∫[0,sinx] (x^2-y^2)dy}
=∫[0,π] {x^2*y - (y^3)/3}[0,sinx] dx
=∫[0,π] {x^2*sinx-(sinx)^3/3} dx
先求:x^2*sinx-(sinx)^3/3的不定积分
∫ x^2*sinx-(sinx)^3/3 dx
=∫ x^2*sinx dx - ∫ - (1-(cos)^2) d(cosx)
={-x^2*cosx + ∫ 2x*cosx dx} + {-sinx-(cosx)^3/3}
={-x^2*cosx + 2x*sinx - ∫ 2sinx dx}+{-sinx-(cosx)^3/3}
= -x^2*cosx + 2x*sinx + 2cosx-sinx-(cosx)^3/3 + C
再求：
∫[0,π] {x^2*sinx-(sinx)^3/3} dx
={-x^2*cosx + 2x*sinx + 2cosx-sinx-(cosx)^3/3 + C}[0,π]
=(π^2-2+1/3)-(2-1/3)
=π^2-4/3
即:∫∫D (x^2-y^2)ds=π^2-4/3
希望能帮助你哈

Answer 2

见图

Answer 3

没有被积函数怎么求啊？