作业帮求解一道三重积分∫[0→π/2] cosθsinθ dθ∫[0→π/4] sin³φdφ∫[0→2] r^4 dr
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![求解一道三重积分∫[0→π/2] cosθsinθ dθ∫[0→π/4] sin³φdφ∫[0→2] r^4 dr](/uploads/image/z/8126780-68-0.jpg?t=%E6%B1%82%E8%A7%A3%E4%B8%80%E9%81%93%E4%B8%89%E9%87%8D%E7%A7%AF%E5%88%86%E2%88%AB%5B0%E2%86%92%CF%80%2F2%5D+cos%CE%B8sin%CE%B8+d%CE%B8%E2%88%AB%5B0%E2%86%92%CF%80%2F4%5D+sin%26%23179%3B%CF%86d%CF%86%E2%88%AB%5B0%E2%86%922%5D+r%5E4+dr)
求解一道三重积分∫[0→π/2] cosθsinθ dθ∫[0→π/4] sin³φdφ∫[0→2] r^4 dr
求解一道三重积分
∫[0→π/2] cosθsinθ dθ∫[0→π/4] sin³φdφ∫[0→2] r^4 dr
求解一道三重积分∫[0→π/2] cosθsinθ dθ∫[0→π/4] sin³φdφ∫[0→2] r^4 dr
这个直接三个分开积啊
=∫[0→π/2] (1/2)sin(2θ)dθ∫[0→π/4] sin^2 φ d(-cosφ) (r^5/5)|[0→2]
=(1/4)(-cos(2θ))|[0→π/2]
*∫[0→π/4] (1-cos^2 φ) d(-cosφ)
*(2^5/5)
=(1/4)*(-1-1)*-(cosφ-(cos^3 φ)/3)|[0→π/4]
*32/5
=16/5*[(1-根号2/2)-(1-(根号/2)^3)/3]
=16/5*(2/3-5根号2/12)
=32/15-(4根号2)/3