Question

高数题，在线等

密度为p的均匀物体占空间区域由旋转单双曲面x^2+y^2-z^2=1及平面z=0和z=1围成，求对z轴的转动惯量。
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!谁会用∫p(x^2+y^2)dv转化为柱面坐标做,要具体过程.

Answer 1

∫p(x^2+y^2)dv=∫p(r^2*(cosa)^2+r^2*(sina)^2)*r*drdadz
=∫(0,1)dz∫(0,根下(1+z^2))dr∫(0,2π)p*r^3da
=∫(0,1)dz∫(0,根下(1+z^2))p2πr^3dr
=(πp/2)∫(0,1)(1+z^2)^2dz=14π/15

Answer 2

竖直方向为z轴，当距离为Z时，r=根号下1+Z^2

对每个dz，从0积到r，求出每个dz的转动惯量I=∫ p2πxdxdz•x^2= ∫p2πdz•x^3dx=(πp（1+z^2）^2)•dz/2

对上式的dz从0积到1，答案是14πp/15

Answer 3

设竖直方向为z轴
设质点和转轴的垂直距离=半径R=(x^2+y^2)^(1/2)
x^2+y^2-z^2=1
R^2=x^2+y^2=1+z^2---------------(1)
质量=密度*体积=p*2πrdrdz
转动惯量=质量*(距离^2)=(p*2πrdrdz)*(r^2)
总转动惯量=∫(0,1)dz ∫(0,R)r^2*p*2πrdr
=2πp∫(0,1)dz∫(0,R)r^3dr
=2πp∫(0,1)dz(R^4)/4=(1/2)*πp∫(0,1)dz(1+z^2)^2 (从(1))
=(1/2)*πp∫(0,1)dz(1+2z^2+z^4)=(1/2)*πp[1+2/3+1/5)
=14πp/15

Answer 4

二重积分应用