Find the moment of inertia, with respect
Find the moment of inertia, with respect to the hypotenuse, of the right isosceles triangle from the previous problem: legs of length $a$, surface density $\mu(x,y)=\dfrac{k(a-x-y)}{\sqrt2}$ proportional to the distance from the hypotenuse.
Explanation of Method
The moment of inertia about the hypotenuse is $\iint d^2\mu\,dA$, where $d=\frac{a-x-y}{\sqrt2}$ is again the distance to the hypotenuse. Since $\mu\propto d$, the integrand is proportional to $d^3=(a-x-y)^3/(2\sqrt2)$, and the same $y$-then-$x$ integration order used for the mass in the previous problem reduces this to a single elementary power integral.
- Set up the moment-of-inertia integral $I=\iint d^2\mu\,dA$.
- Evaluate the resulting $(a-x-y)^3$ double integral.
- Express $I$ in terms of the total mass $m$.
Step 1 · Setup
$$ I = \iint_D d^2\mu\,dA = \iint_D\frac{(a-x-y)^2}2\cdot\frac{k(a-x-y)}{\sqrt2}\,dx\,dy = \frac k{2\sqrt2}\iint_D(a-x-y)^3\,dx\,dy. $$
Step 2 · Evaluate
Integrating $y$ first ($c=a-x$): $\displaystyle\int_0^c(c-y)^3\,dy=\frac{c^4}4=\frac{(a-x)^4}4$, so:
$$ I = \frac k{8\sqrt2}\int_0^a(a-x)^4\,dx = \frac k{8\sqrt2}\cdot\frac{a^5}5 = \frac{ka^5}{40\sqrt2}. $$
Step 3 · In terms of the mass
From the previous problem, $m=\dfrac{ka^3}{6\sqrt2}\Rightarrow k=\dfrac{6\sqrt2m}{a^3}$. Substituting:
$$ I = \frac{a^5}{40\sqrt2}\cdot\frac{6\sqrt2m}{a^3} = \frac{3ma^2}{20}. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $a=2.3$): $I\approx1.1378$, matching both $\dfrac{ka^5}{40\sqrt2}$ (with $k=1$) and $\dfrac{3ma^2}{20}$ exactly.
Result
The moment of inertia about the hypotenuse is: