Find the moments of inertia with respect
Find the moments of inertia with respect to the coordinate axes of the triangular area bounded by $x=0,\ y=0$, and $\dfrac xa+\dfrac yb=1$.
Explanation of Method
Set up the double integrals $I_x=\iint y^2\,dA$ and $I_y=\iint x^2\,dA$ over the triangle $0\le x\le a,\ 0\le y\le b(1-x/a)$, and evaluate each.
Step 1: About the $x$-axis
\[I_x=\int_0^a\frac{b^3}3\left(1-\frac xa\right)^3dx\]
With $u=1-x/a$: \[I_x=\frac{ab^3}3\int_0^1u^3\,du=\frac{ab^3}{12}\]
Step 2: About the $y$-axis
\[I_y=\int_0^ax^2\cdot b\left(1-\frac xa\right)dx = b\int_0^a\left(x^2-\frac{x^3}a\right)dx = b\left(\frac{a^3}3-\frac{a^3}4\right)=\frac{a^3b}{12}\]
Step 3: Polar moment
\[I_O=I_x+I_y=\frac{ab(a^2+b^2)}{12}\]
Step 4: Comparison with the rectangle
Both $I_x$ and $I_y$ here are exactly $\frac14$ of the corresponding rectangle values from Problem 144 — consistent with the triangle having half the rectangle's area, with that area concentrated closer to the axes.
Result
Final answer:
\[\boxed{I_x=\dfrac{ab^3}{12},\quad I_y=\dfrac{a^3b}{12},\quad I_O=\dfrac{ab(a^2+b^2)}{12}}\]