A triangular laminate OAB of uniform area density,$\rho$, has vertices at (0,0), (0,2) and (1,0). The moment of inertia about an axis through O,perpendicular to the plane of the laminate is denoted by $I_O$ and the centre of mass is located G.
i)Calculate the coordinates of the centre of mass, and show that
$I_O$=$5\over 6$$\rho$
I've found the coordinates of the centre of mass to be MX=$2\over 3$$\rho$ and MY=$2\over 3$$\rho$ but I'm struggling to show $I_O$=$5\over 6$$\rho$
I know for the moment of inertia about a general point Q is $I_Q$=$\int$$\rho$$s^2$dV where s is the perpendicular distance of the point from the axis passing through Q.
I've tried doing the integral
$I_O$+=$\int_0^1$$\int_0^{2x}$($x^2$+$y^2$)dydx
but instead of getting the correct answer I get $7\over 6$$\rho$. I thought my limits might be wrong but I'm not sure and don't know what they'd be if they are.
You need $$ I_O = \rho \int_0^1 \int_0^{2-2x} (x^2 + y^2)\,{\rm d}y\,{\rm d}x = \frac{5}{6} \rho $$
The line between (0,2) and (0,1) is $y=2-2x$