# Moment of inertia of a cone

## Homework Statement

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

The cone: $x^2+y^2<=z^2, 0<=z<=h$

$I_z = \int\int\int_T(x^2+y^2)dxdydz$

## Homework Equations

Representing the cone in cylindrical coords:

$x=zcos\theta$
$y=zsin\theta$
$z=z$

## The Attempt at a Solution

The integral in cylindrical coords is:

$I_z = \int_0^h \int_0^{2\pi} \int_0^z (z^2)rdrd\theta dz$

Evaluating the triple integral gives:
$\pi h^5/5$

But the answer in the book is:
$\pi h^5/10$

I don't see what I did wrong

LCKurtz
Homework Helper
Gold Member

## Homework Statement

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

The cone: $x^2+y^2<=z^2, 0<=z<=h$

$I_z = \int\int\int_T(x^2+y^2)dxdydz$

## Homework Equations

Representing the cone in cylindrical coords:

$x=zcos\theta$
$y=zsin\theta$
$z=z$

## The Attempt at a Solution

The integral in cylindrical coords is:

$I_z = \int_0^h \int_0^{2\pi} \int_0^z (z^2)rdrd\theta dz$

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

Evaluating the triple integral gives:
$\pi h^5/5$

But the answer in the book is:
$\pi h^5/10$

I don't see what I did wrong

I see:

$x=rcos\theta, y=rsin\theta$ where $0<=r<=z$

Thanks alot