# Moment of Inertia of a Tetrahedron about the X-axis and its Centroid.

My task is to compute the moment of inertia and the radius of gyration of a constant-density tetrahedron defined by $$x,y,z\ge0$$ and $$\frac xa + \frac yb + \frac zc \leq 1$$ about the x-axis.

I know that the moment of inertia, $$I$$, of an object is defined by: $$I=\int_RdI$$ over some region of integration R. Here, I'd let R be the volume of the tetrahedron. Then, $$dI = r^2dM$$ for mass element $$dM=ρ(x,y,z)dV$$ for some density constant density $$ρ(x,y,z) = k$$. So, we should have:

$$I = \int_0^c \int_0^{b- \frac bcz} \int_0^{a- \frac aby - \frac acz} ρ(x,y,z)r^2dxdydz$$

$$r^2$$ should represent the distance from any point in the tetrahedron to the x-axis, so we can (I think) let $$r^2=y^2+z^2$$. Hence:

$$I = k\int_0^c \int_0^{b- \frac bcz} \int_0^{a- \frac aby - \frac acz}(y^2+z^2)dxdydz$$.

From here, I would compute $$I$$ and then I would be able to calculate the radius of gyration, $$r_g$$, where $$r_g=\sqrt{\frac IM}$$.

This integral looks unnecessarily complicated and I'm not sure if I set it up correctly. Any idea where I might have gone wrong?

• Should that be $z/c$? – Frpzzd Oct 25 '18 at 19:16
• yeah that's a typo. – Jackson Joffe Oct 25 '18 at 19:17
• The integral isn't really that complicated. Split it into two integrals and then integrate $y^2$ and $z^2$ separately. – Frpzzd Oct 25 '18 at 19:19
• Why "unnecessarily"? The reasoning is correct and all formulas are in their place. Do the integral and set $M=abc/3$ – Rafa Budría Oct 25 '18 at 19:21
• Actually, the density is $k=M/(\frac{abc}{3})$. The moment of inertia is a tensor. The one you're calculating is $I_{xx}$. See en.wikipedia.org/wiki/Moment_of_inertia – minmax Oct 25 '18 at 19:47

$$I = k\int_0^c dz \int_0^{b- \frac bcz} dx\int_0^{a- \frac aby - \frac acz}(y^2+z^2)\,dy$$