二重积分计算∫∫(x^2-y^2)dxdy D是闭区域0<=y<=sinx 0<=x<=π

2024-11-08 20:43:08
推荐回答(1个)
回答1:

使用直角坐标,
∫∫(x^2-y^2)dxdy
=∫[0,π]dx∫[0,sinx](x^2-y^2)dy
=∫[0,π](x^2y-1/3y^3)|[0,sinx]dx
=∫[0,π](x^2sinx-1/3(sinx)^3)dx

=∫(x^2sinx-1/3(sinx)^3)dx
=-x^2cosx+2xsinx+2cosx-∫1/3(sinx)^3dx
=-x^2cosx+2xsinx+2cosx-∫1/3(sinx)(1-(cosx)^2)dx
=-x^2cosx+2xsinx+2cosx-∫1/3(sinx)+1/3(sinx)((cosx)^2)dx
=-x^2cosx+2xsinx+2cosx+1/3cosx-1/9(cosx)^3dx
代入积分区间(0,π)
π^2-2-1/3+1/9-2-1/3+1/9
=π^2-40/9