What is the Moment of Inertia of a Cone about its Longitudinal Axis?

[c0der]

Homework Statement

Find the moment of inertia of a solid cone about its longitudinal axis (z-axis)

The cone: [itex]x^2+y^2<=z^2, 0<=z<=h[/itex]

[itex]I_z = \int\int\int_T(x^2+y^2)dxdydz[/itex]

Homework Equations

Representing the cone in cylindrical coords:

[itex] x=zcos\theta [/itex]
[itex] y=zsin\theta [/itex]
[itex] z=z [/itex]

The Attempt at a Solution

The integral in cylindrical coords is:

[itex]I_z = \int_0^h \int_0^{2\pi} \int_0^z (z^2)rdrd\theta dz[/itex]

Evaluating the triple integral gives:
[itex]\pi h^5/5[/itex]

But the answer in the book is:
[itex]\pi h^5/10[/itex]

I don't see what I did wrong

 

[LCKurtz]

Homework Statement

Find the moment of inertia of a solid cone about its longitudinal axis (z-axis)

The cone: [itex]x^2+y^2<=z^2, 0<=z<=h[/itex]

[itex]I_z = \int\int\int_T(x^2+y^2)dxdydz[/itex]

Homework Equations

Representing the cone in cylindrical coords:

[itex] x=zcos\theta [/itex]
[itex] y=zsin\theta [/itex]
[itex] z=z [/itex]

The Attempt at a Solution

The integral in cylindrical coords is:

[itex]I_z = \int_0^h \int_0^{2\pi} \int_0^z (z^2)rdrd\theta dz[/itex]

The moment arm for the second moment from the z axis is ##r^2##, not ##z^2##.

Evaluating the triple integral gives:
[itex]\pi h^5/5[/itex]

But the answer in the book is:
[itex]\pi h^5/10[/itex]

I don't see what I did wrong

 

[c0der]

I see:

[itex] x=rcos\theta, y=rsin\theta[/itex] where [itex]0<=r<=z[/itex]

Thanks alot