The moment-of-inertia (MOI) tensor is real (no imaginary terms), symmetric, and positive-definite. Linear algebra tells us that for any (3x3) matrix that has those three properties, there's always a set of three perpendicular axes such that the MOI tensor can be expressed as a diagonal tensor in the basis of those axes. These are called the principal axes (or eigenvectors) of rotation, and the physical meaning behind them is that if you rotate the object around one of those axes, the angular momentum will lie along the axis. So one important thing to realize is that there is nothing fundamentally meaningful about off-diagonal elements; you can always rotate your coordinates to get rid of them. If the object has a symmetry axis, then that will be a principal axis. (Though, having a principal axis does not imply any symmetry of the object.)
On the other hand, what if the body is rotating about an axis that isn't one of the principal axes? This is equivalent to writing your MOI in a basis where the rotation axis is one of the basis vectors, in which case there are off-diagonal elements, which is what your question is asking about. So, off-diagonal elements in your MOI are equivalent to having a rotation axis that is not aligned with any of the principal axes. Again, this only happens when your body is not symmetric about the rotation axis.
And what does this mean physically? For one thing, the angular momentum is not aligned with the angular velocity. For example, imagine your object spinning inside a nicely symmetric little satellite in space. You can see its rotation axis, but if the satellite grabs onto the object, it will absorb the angular momentum, and you'll find the satellite spinning on a different axis.
Alternatively, you can think of the expression relating torque and angular acceleration $\vec{\tau} = I \cdot \vec{\alpha}$. An off-diagonal element in the MOI means that if I apply a torque about a certain axis, the object will accelerate its rotation about a different axis.
That looks lie a homework style question. So here the necessary hints.
The tensors for the base shapes are listed here. It should be clear how to get the tensors if you put two objects together. The same applies if subtracting the inner cone or cylinder for a hollow object. For additional details check the parallel axis theorem
Edit
While I think this is not a calculus help site, I will try to give an almost complete solution.
The dimensions are defined in the image. Note that this is a general solution that covers basically all your cases. For $A=B=0$ it is solid and hollow otherwise. For $B=D$ and $A=C$ it is a (hollow) cylinder. It is simple to verify that for those cases the following results coincide with the special case formulae.
The volume is given by
$$V= 2 \pi \int_0^h \mathrm d z \int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r$$
where
$$r_\mathrm i(z)= A+(C-A)\frac z h \quad r_\mathrm o(z)= B+(D-B)\frac z h$$
This results in
$$V= \frac {\pi}{ 3} \left[(B^2+BC+D^2)-(A^2+AC+C^2)\right]$$
One can see easily that for $A=B=C=0$ and $D=R$ this gives the volume of a simple cone. The centre of mass is at:
$$\begin{eqnarray}
z_\mathrm{cm}&=&\frac 1 V 2 \pi \int_0^h \mathrm d z \int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r z\\
&=&\frac h 4 \frac{(B^2+2BD+3D^2)-(A^2+2AC+3C^2)}{(B^2+BD+D^2)-(A^2+AC+C^2)}
\end{eqnarray}$$Again we see that in the cone limit this gives the known result of $z_\mathrm{cm}=3 h/4$. Also note that this general formula also allows to put the tip at the top, i.e. only $B\neq 0$ in which case one sees easily that the centre of mass is at $h/4$.
Now for $I_{xx}=I_{yy}$ and $I_{zz}$ (Note, I do this in units of the body's mass, so I divide by the volume, assuming constant density $\rho$):
$$\begin{eqnarray}
I_{zz}&=&\frac 1 V 2 \pi \int_0^h \mathrm d z \int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r (x^2+y^2)\\
&=& \frac 3 {10} \frac{(B^4+B^3D+ B^2D^2+BD^3+D^4)-(A^4+A^3C+A^2C^2+AC^3+C^4)}{(B^2+BD+D^2)-(A^2+AC+C^2)}
\end{eqnarray}$$
which is in the above mentioned limit $3/10 R^2$. Finally, we have
$$\begin{eqnarray}
I_{xx}&=&\frac 1 V \int_0^h \mathrm d z \int_0^{2 \pi} \mathrm d \varphi\int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r (z^2+y^2)\\
&=& \frac 3 {30} h^2 \frac{(B^2+3BD+6D^2)-(A^2+3AC+6C^2)}{(B^2+BD+D^2)-(A^2+AC+C^2)}+\\
&&+\frac 3 {20} \frac{(B^4+B^3D+ B^2D^2+BD^3+D^4)-(A^4+A^3C+A^2C^2+AC^3+C^4)}{(B^2+BD+D^2)-(A^2+AC+C^2)}
\end{eqnarray}$$
which again gives the well known factors $3/5 h^2$ and $3/20 R^2$ in case of $A=B=C=0$ and $D=R$.
Eventually, we need to check the moment in the centre of mass.
Now this I will leave open for the OP. Just consider this:
This calculation is obviously not in the centre of mass.$I_{zz}$ should not change as we shift along this axes. We also know that having the centre of mass solution, going to the coordinate system we have, is done by using the parallel axis theorem, i.e.
$$I_{xx}=I_{xx,\mathrm{cm}}+z_\mathrm{shift}^2$$
(again in units of $m$).
You may, as an alternative, reformulate the last integral.
Best Answer
Proposition: Given an arbitrary rigid body (and wrt. to an arbitrary choice of pivotal point for the rigid body, and wrt. to an arbitrary choice of Cartesian coordinates $x$, $y$, and $z$), then the diagonal elements $I_{xx}$, $I_{yy}$, and $I_{zz}$ of the inertia tensor satisfy the triangle inequality, $$ I_{xx} +I_{yy} ~\geq~ I_{zz}, \qquad I_{yy} +I_{zz} ~\geq~ I_{xx}, \qquad I_{zz} +I_{xx} ~\geq~ I_{yy}. \tag{1} $$ Sketched proof: Write down the definition of moment of inertia.$\Box$
Observation: It follows from the triangle inequality (1) alone that $$I_{xx}, I_{yy},I_{zz}~\geq~0 \tag{2}$$ are non-negative. (The ineq. (2) of course also follows from the definition of moment of inertia.)
Corollary of Proposition: Given an arbitrary rigid body (and wrt. to an arbitrary choice of pivotal point for the rigid body), then the three moments of inertia $I_x$, $I_y$, and $I_z$, around the three principal axes (which we will call $x$, $y$, and $z$) satisfy the triangle inequality, $$ I_x +I_y ~\geq~ I_z, \qquad I_y +I_z ~\geq~ I_x, \qquad I_z +I_x ~\geq~ I_y. \tag{3} $$
In other words, if a semi-positive definite symmetric real $3\times 3$ matrix with non-negative eigenvalues $I_x$, $I_y$, and $I_z$ does not satisfy the triangle inequality (3), it doesn't represent a physically possible distribution of mass.
Conversely, one may show that given three eigenvalues $I_x$, $I_y$, and $I_z$ that satisfy (3), they may be reproduced by a solid ellipsoid with a unique choice of non-negative semi-axes $a$, $b$, and $c$ (unique up to the scaling of the total mass $m$). $$ \frac{2}{5}m a^2~=~I_y +I_z -I_x~\geq~0, $$ $$ \frac{2}{5}m b^2~=~I_z +I_x -I_y~\geq~0, $$ $$ \frac{2}{5}m c^2~=~I_x +I_y -I_z~\geq~0.\tag{4} $$