This problem requires you to compute components of the moment of inertia tensor for several simple shapes. In each case, the object will be spinning about its centre of mass with angular speed $\omega$, and we ask you to compute the total angular momentum. In each case, you may assume the mass density is constant.
(i) A ring of radius $R$ and mass $M$ spinning about an axis perpendicular to the plane of the ring and through the centre of the ring.
So I know that, provided the appropriate coordinates are chosen, the inertia tensor is $\mathbb{I}=\mathrm{diag}(I_{xx}, I_{yy}, I_{zz})$, where $I_{xx} = \sum_i m_i (y²_i +z²_i)$ and likewise for the others.
The issue I'm having is trying to justify what the integral is. If we say the ring is embedded in $x=0$ with equation $y^2+z^2=R^2$, then I would expect there to be a double integral (in polar, for simplicity), as we're working in two dimensions ($y$ and $z$).
But, $dy \ dz = |J(r, \theta)| \ dr \ d\theta$, so the integral would be $$\iint \rho (y^2+z^2) \ dy \ dz = \iint \rho (y^2+z^2) r \ dr \ d\theta .$$
Using the fact that $y^2+z^2=R^2, \ r=R$, the integrand is actually the constant $\rho R^3$. Thus $$I_{xx} = \rho R^3 \iint dr \ d\theta. $$
The trouble is that $r$ is a constant, so the integral vanishes.
I realise that you actually want to use a single integral, i.e. $\int \rho (y^2+z^2) r \ d\theta$, but then you're saying $dy \ dz = r \ d\theta$, which isn't what you would get by computing the determinant of the Jacobian.
So what's the correct line of reasoning?