Moment of inertia tensor bounds Can someone explain to me how did they determine the bounds in the following problem I did this in cylindrical coordinates but I would like to also understand this in order to fall back on cartesian coordinate in case if there is nothing else to rely on.
http://scienceworld.wolfram.com/physics/MomentofInertiaCone.html
 A: To determine bounds, you start on the outside and work your way in.
Since they've chosen to integrate in the order $dx\,dy\,dz$, we start with the $z$-coordinate. Since this is a cone of height $h$, we integrate from $0$ to $h$ because these are the minimum and maximum possible $z$-values in the cone. Easy enough.
Now, take a particular value of $z$, say $z_0$, and consider the cross section of the cone satisfying $z = z_0$. What we get, of course, is a circle of some radius. Using a bit of geometry, we can determine that the radius at height $z_0$ is $R(h-z_0)/h$. Since we integrate with respect to $y$ next, we need to determine the minimum and maximum possible values of $y$ in this cross section. The answer is easy though: since the circle is centered at $(0, 0, z_0)$, $y$ runs from $-R(h-z_0)/h$ to $R(h-z_0)/h$ in this cross section. The webpage didn't use this notation though; they simply used $z$ where I used $z_0$ here, hence the bounds of the second integral.
Finally, pick a particular value of $z$, say $z_0$ again, and a particular value of $y$, say $y_0$. What is the cross section of the cone satisfying $z = z_0$ and $y = y_0$? It is the same as the cross section of our circle of radius $R(h-z_0)/h$ satisfying $y=y_0$. Thus, it is simply a line segment cutting through our circle parallel to the $x$-axis. Using the Pythagorean theorem, it is easy to see that it is the line segment running from $x = -\sqrt{(R(h-z_0)/h))^2 - y_0^2}$ to $x = \sqrt{(R(h-z_0)/h))^2 - y_0^2}$. Again, they simply used $z$ and $y$ instead of $z_0$ and $y_0$, hence the bounds on the third integral.
