Consider a right circular cone of radius 10 meters and height 10 meters with a uniform density of 1(kg/m^3). Find the moment of inertia of this cone about its axis and use this to compute the kinetic energy of the cone spinning about its axis at a rate of 1 rev. per second.
My attempt:
$\delta=density$
$w=velocity (radians/sec.)$
Kinetic Energy of Mass $= \frac{1}{2}(mass)({velocity^2})=\frac{1}{2}\delta{w^2}{r^2} dV$
Total Kinetic Energy $=\iiint_S \frac{1}{2}{w^2}\delta{r^2} dV $
Moment of Inertia: $$I_S=\iiint_S (\delta){r^2} dV$$
Kinetic Energy $=\frac{1}{2}I_S{w^2}$
Volume of Cone $=\frac{1}{3}\pi{r^2}h=\frac{1}{3}\pi{10^2}(10)$
Mass of Cone $=(density)(volume)=(\delta)(volume)=1(\frac{1}{3}\pi{10^2}(10))$
Moment of Inertia: $$I_S=\iiint_S \frac{3}{10}(mass of cone){r^2}= ...$$
Kinetic Energy $$k.e.= \frac{1}{2}I_S{w^2}= ...$$
Not sure if this is at all correct... please help!
You were okay clear up to here $$I_S=\iiint_S \frac{3}{10}(mass of cone){r^2}= ...$$ But that line makes no sense. One integrates differentials, and suddenly you have no differential in your integral. You seem to be mistaking $\delta\,dV$ with $\delta V$. The latter is the density times the volume of the cone, which gives the mass of the cone as you've calculated.
But $dV$ is not $V$. It is the differential of volume, and it only provides an actual number when you integrate it. You can't substitute it out with the mass of the entire cone.
Face the facts: you are going to have work out what $\iiint_S r^2\,dV$ by using actual calculus.
As a hint: use cylindrical coordinates $r, \theta, z$, so $dV = rdrd\theta dz$. Note that the $r$ here is the same $r$ you were using above - the distance from the point of integration to the $z$ axis.