Moment of Inertia of Rectangular Prism about one of its edges

Question: What is the moment of inertia of a rectangular prism with dimenions $l\times w\times h$ represented by $a\times b\times c$ about one of its edges? Are my bounds correct, and what is $r$?
Motivation: Imagine a book standing upright on a table. A force is applied to the top of the book so it begins to rotate about one of it's axes without slipping. How much energy must be applied, knowing the dimensions & mass of the book, to tip the book over (i.e. to move the center of mass over the pivoting axis)? With moment of inertia I think I know how to do this - if there's a better way, answer the question anyway because I'm curious :)
Attempt: Here's what I've got. I know the equation for moment of inertia is: $$I=\iint_D \rho r^2\,dA\\OR\\I=\iiint_R \rho r^2 \,dV$$ I think the triple integral is more appropriate here, and filling in the bounds gives $$I=\int_0^a\int_0^b\int_0^c\rho r^2\,dz\,dy\,dz.$$ Let's assume constant density (1). Is $r^2$ simply $x^2+y^2+z^2$? This doesn't seem quite right. If it is, that gives $$I=\int_0^a\int_0^b\int_0^c x^2+y^2+z^2\,dz\,dy\,dz.$$ Which is easily solvable. Which I don't feel it should be.

Solution 1:

If you want the moment on inertia with respect to - say the the $x$ axis, then the correct formula is $$ I_x=\iiint_R \rho (y^2+z^2)\; dV, $$ with $$ R=\{(x,y,z)\;|\; 0 \le x \le a, 0\le y \le b,0\le z \le c \} $$