Help in finding moments of inertia in planar laminas Hello I'm a student how to figure out how to solve these types of problems and I haven't found any good resources on it.  A specific problem I'm working on is this:
Let $R$ be the lamina with density $1/4$ bounded by $y\geq0$, $x\geq0$, and $y=2-7x^2$.  Set up the integral for Moment $M_y$, using slices parallel to the $y$ axis. 
The answer it gives is $$\int_{0}^{\sqrt{2/7}}\frac{1}{4}x(2-7x^2)dx$$ but I have no idea how they got this solution. Any help would be much appreciated. 
 A: Imagine a thin vertical strip, going from $x$ to $x+dx$. The height of the strip is everywhere about $2-7x^2$. 
The strip is almost a rectangle. So the area of the strip is about $(2-7x^2)\,dx$. The mass  of the strip is therefore about $(1/4)(2-7x^2)\,dx$.
Every point in the strip is at about a perpendicular distance $x$ from the $y$-axis. So the moment of the strip about the $y$-axis is about  $x(1/4)(2-7x^2)\,dx$.
"Add up" over all such strips (integrate). It is not hard to see that the curve $y=2-7x^2$ meets the $x$-axis at $x=\sqrt{2/7}$. So our moment is
$$\int_0^{\sqrt{2/7}}x(1/4)(2-7x^2)\,dx.$$
Remark: The above is an informal way to find the appropriate integral. If you want to be more technical, divide the part of the $x$-axis from $x=0$ to $x=\sqrt{2/7}$ into a large number $n$ of equal subintervals. Let the division points be $0=x_0,x_1,\dots,x_n=\sqrt{2/7}$. Let $\delta x$ be the length of these subintervals. (Here $\Delta x=\sqrt{2/7}/n$.) The moment of the lamina can be approximated by
$$\sum_{i=0}^{n-1} x_i(1/4)(2-x_i^2)\Delta x.$$
This is a Riemann sum. As $n\to\infty$, the approximation approaches the true moment. It also approaches the integral of the answer. 
