Find the center of mass of an isosceles right triangle if
Find the center of mass of an isosceles right triangle if at each point the surface density is proportional to its distance from the hypotenuse.
Explanation of Method
With the right angle at the origin and legs of length $a$ along the axes, the hypotenuse is $x+y=a$, so the density $\mu=k(a-x-y)/\sqrt2$ is a simple linear function inside the triangle. Each of the three integrals needed — mass, and the two first moments — reduces to an elementary polynomial integral after integrating out $y$ first.
- Set up the density function from the distance to the hypotenuse.
- Compute the total mass $m$.
- Compute the moments $M_x,M_y$ and the center of mass.
Step 1 · Density function
With vertices $O=(0,0),\,A=(a,0),\,B=(0,a)$, the hypotenuse is $x+y=a$, and inside the triangle $x+y\le a$, so the distance is $d=\dfrac{a-x-y}{\sqrt2}$, giving $\mu(x,y)=\dfrac{k(a-x-y)}{\sqrt2}$.
Step 2 · Total mass
$$ m = \frac k{\sqrt2}\int_0^a\int_0^{a-x}(a-x-y)\,dy\,dx = \frac k{2\sqrt2}\int_0^a(a-x)^2\,dx = \frac{ka^3}{6\sqrt2}. $$
Step 3 · Moments and center of mass
$$ M_y = \frac k{2\sqrt2}\int_0^ax(a-x)^2\,dx = \frac k{2\sqrt2}\cdot\frac{a^4}{12} = \frac{ka^4}{24\sqrt2}. $$
By the $x\leftrightarrow y$ symmetry of both the triangle and $\mu$: $M_x=M_y=\dfrac{ka^4}{24\sqrt2}$. Therefore:
$$ \bar x = \frac{M_y}m = \frac{ka^4/(24\sqrt2)}{ka^3/(6\sqrt2)} = \frac a4, \qquad \bar y = \frac a4. $$
Additional Notes
- This derivation matches the source exactly. The center of mass $\left(\frac a4,\frac a4\right)$ lies on the symmetry line $y=x$, pulled toward the right-angle vertex (where the density, and hence distance from the hypotenuse, is greatest) rather than toward the centroid $\left(\frac a3,\frac a3\right)$ of the uniform triangle.
- Verified numerically (e.g. $a=2.3$): $m\approx1.4339$, $\bar x=\bar y\approx0.575=a/4$, matching exactly.
Result
The center of mass is at: