※ 引述《RCSQQ ()》之铭言： : 请问 : Find the volume of the region the first octant (x,y,z> 0) above the parabolic : = : 2 2 2 cylinder z=y and be:low the paraboloid z=16-4x -3y above means the lower bound of integral below means the upper bound of integral 2 2 ∫∫∫dv = ∫∫∫dzdxdy = ∫∫ 16-4x - 4y (r)drdδ 2 2 2 2 2 boundary condition: 16-4x - 3y = y gives 16 = 4x +4y 2 4 = r r =0~2 2 2 2 2 so 2∫∫ 4 - r dr dδ = ∫(4 - r ) dδ where δ is 0~1/2π ∵ first octant answer = 8π : 2. : Calculate ∫∫ F ds, where F=(0,0,x) and S is the parametrized surface : s : 2 3 2 : Φ(u,v)=(u ,v ,u - v ), 0<u ,v<1 : = = : 这两题怎算呢? : 谢谢!!! 这题是曲面积分 且曲面是参数式 可以参考课本surface integral ∫∫ds = ∫∫|ru×rv|dudv 有这条式子 Φ表曲面 大多课本用 r表示 ru 表对u偏微 rv 则表对v偏微 3 4 ∫∫(0,0,u^2) dot n dudv = ∫∫ 2 u dudv = 1/2 (u ) = 1/2 2 ru = (2u , 0 , 3u ) rv = (0 , 1 , -2v ) 2 n = ru ×rv = (-3u , 4uv ,2u) --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 114.34.122.244 ※ 编辑: rygb 来自: 114.34.122.244 (06/18 22:25) ※ 编辑: rygb 来自: 114.34.122.244 (06/18 22:40)