设平面区域满足0<y<根号下2x-x^2,0<x<1则∫∫f(x,y)dxdy在极坐标下的二重积分为什么?答案为[∫0到π/4dθ ∫0到secθ f(rcosθ,rsinθ)rdr]+[∫π/4到π/2dθ ∫0到2cosθ f(rcosθ,rsinθ)rdr] 分从0到π/4 和π
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![设平面区域满足0<y<根号下2x-x^2,0<x<1则∫∫f(x,y)dxdy在极坐标下的二重积分为什么?答案为[∫0到π/4dθ ∫0到secθ f(rcosθ,rsinθ)rdr]+[∫π/4到π/2dθ ∫0到2cosθ f(rcosθ,rsinθ)rdr] 分从0到π/4 和π](/uploads/image/z/5060520-0-0.jpg?t=%E8%AE%BE%E5%B9%B3%E9%9D%A2%E5%8C%BA%E5%9F%9F%E6%BB%A1%E8%B6%B30%26lt%3By%26lt%3B%E6%A0%B9%E5%8F%B7%E4%B8%8B2x-x%5E2%2C0%26lt%3Bx%26lt%3B1%E5%88%99%E2%88%AB%E2%88%ABf%28x%2Cy%29dxdy%E5%9C%A8%E6%9E%81%E5%9D%90%E6%A0%87%E4%B8%8B%E7%9A%84%E4%BA%8C%E9%87%8D%E7%A7%AF%E5%88%86%E4%B8%BA%E4%BB%80%E4%B9%88%3F%E7%AD%94%E6%A1%88%E4%B8%BA%5B%E2%88%AB0%E5%88%B0%CF%80%2F4d%CE%B8+%E2%88%AB0%E5%88%B0sec%CE%B8+f%28rcos%CE%B8%2Crsin%CE%B8%29rdr%5D%2B%5B%E2%88%AB%CF%80%2F4%E5%88%B0%CF%80%2F2d%CE%B8+%E2%88%AB0%E5%88%B02cos%CE%B8+f%28rcos%CE%B8%2Crsin%CE%B8%29rdr%5D+%E5%88%86%E4%BB%8E0%E5%88%B0%CF%80%2F4+%E5%92%8C%CF%80)
设平面区域满足0<y<根号下2x-x^2,0<x<1则∫∫f(x,y)dxdy在极坐标下的二重积分为什么?答案为[∫0到π/4dθ ∫0到secθ f(rcosθ,rsinθ)rdr]+[∫π/4到π/2dθ ∫0到2cosθ f(rcosθ,rsinθ)rdr] 分从0到π/4 和π
设平面区域满足0<y<根号下2x-x^2,0<x<1则∫∫f(x,y)dxdy在极坐标下的二重积分为什么?
答案为[∫0到π/4dθ ∫0到secθ f(rcosθ,rsinθ)rdr]+[∫π/4到π/2dθ ∫0到2cosθ f(rcosθ,rsinθ)rdr] 分从0到π/4 和π/4到π/2 而且r的区域为什么不一样啊
设平面区域满足0<y<根号下2x-x^2,0<x<1则∫∫f(x,y)dxdy在极坐标下的二重积分为什么?答案为[∫0到π/4dθ ∫0到secθ f(rcosθ,rsinθ)rdr]+[∫π/4到π/2dθ ∫0到2cosθ f(rcosθ,rsinθ)rdr] 分从0到π/4 和π
会画图就是了
用极坐标,积分区域被y = x分开为两部分
D₁是个等腰三角形:y = 0、x = 1、y = x
D₂是个弓形:y = x,y = √(2x - x²)
化为极坐标,
D₁:θ:0→π/4,x = 1 ==> rcosθ = 1 ==> r = secθ
D₂:θ:π/4→π/2,y = √(2x - x²) ==> r² = 2rcosθ ==> r = 2cosθ
所以∫∫D f(x,y) dxdy
= ∫∫D₁ f(x,y) dxdy + ∫∫D₂ f(x,y) dxdy
= ∫(0→π/4) ∫(0→secθ) f(rcosθ,rsinθ) rdrdθ + ∫(π/4→π/2) ∫(0→2cosθ) f(rcosθ,rsinθ) rdrdθ