发布网友
共2个回答
热心网友
如图所示:
热心网友
联立解 z^2 = x^2+y^2/4, 2z = x^2+y^2/4 , 得 z = 0, z = 2
则两曲面交线在 xOy 平面的投影是曲线 x^2+y^2/4 = 4,
即 D : x^2/4+y^2/16 = 1.
作广义极坐标变换 x = 2rcost,y = 4rsint, D : r = 1
z^2 = x^2+y^2/4 即 z = 2r ; 2z = x^2+y^2/4 即 z = 2r^2
V = ∫<0, 2π>dt∫<0, 1>(2r-2r^2) 2· 4rdr
= 32π[r^3/3-r^4/4]<0, 1> = 32π/12 = 8π/3
热心网友
如图所示:
热心网友
联立解 z^2 = x^2+y^2/4, 2z = x^2+y^2/4 , 得 z = 0, z = 2
则两曲面交线在 xOy 平面的投影是曲线 x^2+y^2/4 = 4,
即 D : x^2/4+y^2/16 = 1.
作广义极坐标变换 x = 2rcost,y = 4rsint, D : r = 1
z^2 = x^2+y^2/4 即 z = 2r ; 2z = x^2+y^2/4 即 z = 2r^2
V = ∫<0, 2π>dt∫<0, 1>(2r-2r^2) 2· 4rdr
= 32π[r^3/3-r^4/4]<0, 1> = 32π/12 = 8π/3