I need to find the moment of inertia about co-ordinate axes of a tetrahedron in the first octant bounded by $x+y+z=1$.
My attempt:
$x+y+z = 1$ forms the base of the tetrahedron and my limits for integration based on integrating first with respect to z are:
limits for $z: [0,1-x-y]$
limits for $y:[0,1-x]$
limits for $x:[0,1]$
The formula for moment of inertia in 3 dimensions is:
$ I_x = \iiint( y^2+ z^2 ) \delta dV $ where $\delta$ is the density function
$ I_y = \iiint( x^2+ z^2 ) \delta dV $
$ I_z = \iiint( x^2+ y^2 ) \delta dV $
then I found the moment of inertia about x axis,
$I_x = \int_0^1\int_0^{1-x}\int_0^{1-x-y}( y^2+ z^2 ) dzdydz$
$I_x = \int_0^1\int_0^{1-x}( y^2z+ \frac{z^3}{3} )|_0^{1-x-y} dydz$
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.
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$I_x = \frac{1}{30}$
Similarly for y and z.
For now, I just want to confirm whether I'm moving in the right direction and whether I've missed something or not. Thank you.
Edit: As far as I know, tetrahedron has to do with triple integration. However, the official method for this question is double integration for x and y moments and triple for z. Why is this so?