Find the moment of inertia around the z axis of the region bounded by x=0, y=0, z=0, x+y+z = 1.
My attempt at a solution:
I believe that this should require a triple integral of the form $\int_0^a\int_0^{y(x)}\int_0^{z(x,y)}z^2dzdydx$.
I can get the upper limit of z by solving as z = 1-x-y, giving the integral $\int_0^{1-x-y}dz$. However, for the life of me, I can't figure out how to get y in terms of x, as I've already used up my only equation which involves all 3 variables, or even two of them. What's my next step here? I can solve the final integral, I just need a hand setting it up.
First off, you need to integrate the squared distance of any point in the region to the $z$ axis, which is given by $(x^2+y^2)$. Next, for any fixed $z$ coordinate, note that the cross-section of the region is a right-angled triangle bounded by $x=0$, $y=0$ and $x+y=1-z$. Finally fix a $y$ too, then $x$ varies between $0$ and $1-y-z$, which results in the integral
$$ \int_0^1 \int_0^{1-z} \int_0^{1-y-z} (x^2+y^2)dx dy dz $$
which is easy to solve.