题目

计算∫∫(x^2-y^2)dσ,其中D是闭区域0≤y≤sinx,0≤x≤π

提问时间：2020-10-31

答案

∫∫(x^2-y^2)dσ
=∫∫(x^2-y^2)dxdy
=∫[0,π]dx∫[0,sinx](x^2-y^2)dy
=∫[0,π]dx*(y*x^2-y^3/3)|[0,sinx]
=∫[0,π][x^2*sinx-(sinx)^3/3]dx
=∫[0,π]x^2*sinx*dx-1/3*∫[0,π](sinx)^3*dx
=(-x^2*cosx+2xsinx+2cosx)|[0,π]-2/3*∫[0,π/2](sinx)^3*dx
=(π^2-2)-2-2/3*2/3
=π^2-4-4/9
=π^2-40/9