过程最好有图
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过程最好有图过程最好有图过程最好有图(2)I=∫∫(x^2-y^2)dxdy=∫dx∫(x^2-y^2)dy=∫dx[yx^2-y^3/3]=∫[x^2sinx-(sinx)^3/3]dx.I1=∫x
过程最好有图
过程最好有图
过程最好有图
(2) I =∫∫(x^2-y^2)dxdy = ∫dx∫(x^2-y^2)dy
= ∫dx[yx^2-y^3/3]
= ∫[x^2sinx-(sinx)^3/3]dx.
I1 =∫x^2sinxdx = -∫x^2dcosx
= -[x^2cosx]+2∫xcosxdx = π^2+2∫xdsinx
= π^2+2[xsinx]-2∫sindx
= π^2+0+2[cosx] = π^2-2.
I2 = (1/3)∫(sinx)^3dx
= (-1/3)∫[1-(cosx)^2]dcosx
= (-1/3)[cosx-(cosx)^3/3] = 4/9.
则 I = I1-I2 = π^2-22/9.
(3) 化极坐标,D:r=4cost,0≤t≤π/2.
I = ∫∫√(6-x^2-y^2)dxdy
=∫dt∫√(6-r^2)rdr
= (-1/2)∫dt∫√(6-r^2)d(6-r^2)
= (-1/3)∫dt[(6-r^2)^(3/2)]
= (1/3)∫{6√6-[6-4(cost)^2]^(3/2)}dt
= π√6-(1/3)∫[2+4(sint)^2]^(3/2)dt
下面可作代换 u=√[2+4(sint)^2] 不胜其烦 !
请先核题 !
给个好评吧，我教你，谢谢！！！
我貌似不会做这种题