# Derivation of the formula for the surface moment of inertia of an isosceles triangle

For a isosceles triangle with base $$b$$ and height $$h$$, the surface moment of inertia around the $$z$$-axis is $$\frac{bh^3}{36}$$ (considering that our coordinate system has $$z$$ in the horizontal and $$y$$ in the vertical axis, and has its origin on the triangle's center of mass (which is at $$\left\{\frac{b}{2},-\frac{h}{3}\right\}$$ if you put your coordinate system's origin at the bottom left corner if the triangle).

I know that the formula for the moment of inertia around the $$z$$-axis is $$I_z = \int_A{y^2 dA}$$, but I cannot figure out how to derive the formula from that. How is it done?

The infinitesimal area $\text{dA}$ is $2 z \text{dy}$.
The relationship between $z$ and $y$ can be got from the slope. $$\frac{z}{y-\frac{2 h}{3}}=\frac{0-\frac{b}{2}}{\frac{2 h}{3}+\frac{h}{3}}$$
Solving, we get $$z=\frac{b (2 h-3 y)}{6 h}$$
Thus $$I_{zz}=\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 dA =\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 2 z \, dy =\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 2 \frac{b (2 h-3 y)}{6 h} \, dy=\frac{b h^3}{36}$$