Question

均匀圆锥面z=√(x^2+y^2)被平面z=1截下的有限部分对y轴的转动惯量Iy用第一类曲面积分表示为什么?密度为1.

Answer 1

曲面求偏导得到

z'x=x/√x^2+y^2

z'y=y/√x^2+y^2

Iy

=∫∫(x^2+z^2)dS

=∫∫(x^2+x^2+y^2)dS

=∫∫(2x^2+y^2)dS

=(3/2)∫∫(x^2+y^2)dS

=(3/2)∫∫(x^2+y^2)√[1+(z'x)^2+(z'y)^2] dxdy

=(3√2/2)∫∫(x^2+y^2)dxdy

=(3√2/2)∫∫r^2 rdrdθ

=(3√2/2)∫(0->2π)dθ ∫(0->1)r^3dr

=3√2π/4