二重积分:计算∫∫e^(y/(x+y))dxdy,其中D:x+y≤1,x≥0,y≥0. 10
2个回答
展开全部
解:∫∫<D>e^(y/(x+y))dxdy=∫<0,π/2>dθ∫<0,1/(sinθ+cosθ)>e^(sinθ/(sinθ+cosθ))rdr (做极坐标变换)
=(1/2)∫<0,π/2>e^(sinθ/(sinθ+cosθ))dθ/(sinθ+cosθ)²
=(1/2)∫<0,π/2>e^(sinθ/(sinθ+cosθ))d(sinθ/(sinθ+cosθ))
=(1/2)e^(sinθ/(sinθ+cosθ))│<0,π/2>
=(1/2)(e^(1/(1+0))-e^(0/(0+1)))
=(e-1)/2。
=(1/2)∫<0,π/2>e^(sinθ/(sinθ+cosθ))dθ/(sinθ+cosθ)²
=(1/2)∫<0,π/2>e^(sinθ/(sinθ+cosθ))d(sinθ/(sinθ+cosθ))
=(1/2)e^(sinθ/(sinθ+cosθ))│<0,π/2>
=(1/2)(e^(1/(1+0))-e^(0/(0+1)))
=(e-1)/2。
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
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
广告 您可能关注的内容 |