1个回答
展开全部
∫∫D f(x,y)dxdy
=∫∫D1 x²dxdy+∫∫D2 1/∨(x²+y²)dxdy
=4∫(-1,0) x²dx∫(0,x+1)dy+4∫(0,π/2) dθ∫[1/(cosθ+sinθ),2/(cosθ+sinθ)] dr
=4∫(-1,0) (x³+x²)dx+4∫(0,π/2) 1/(sinθ+cosθ)dθ
=(x^4+4/3 x³)|(-1,0)+∨2 ln{[1-cos(θ+π/4)]/[1+cos(θ+π/4)]} |(0,π/2)
=1/3+∨2{ ln[(2+∨2)/(2-∨2)]-ln[(2-∨2)/(2+∨2)]}
=1/3 +∨2 ln[(6+4∨2)/(6-4∨2)]
=1/3+∨2 ln(17+12∨2)
=∫∫D1 x²dxdy+∫∫D2 1/∨(x²+y²)dxdy
=4∫(-1,0) x²dx∫(0,x+1)dy+4∫(0,π/2) dθ∫[1/(cosθ+sinθ),2/(cosθ+sinθ)] dr
=4∫(-1,0) (x³+x²)dx+4∫(0,π/2) 1/(sinθ+cosθ)dθ
=(x^4+4/3 x³)|(-1,0)+∨2 ln{[1-cos(θ+π/4)]/[1+cos(θ+π/4)]} |(0,π/2)
=1/3+∨2{ ln[(2+∨2)/(2-∨2)]-ln[(2-∨2)/(2+∨2)]}
=1/3 +∨2 ln[(6+4∨2)/(6-4∨2)]
=1/3+∨2 ln(17+12∨2)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询