There are two solid figures (I don't know what they're called just see the picture) and their radius on top ($r_1$) and bottom ($r_2$) same but height different ($h_1,h_2$).
When we use calculus to compute their volume we get the strange conclusion that they have equal volume:$$V_1=V_2=\int_{r_1}^{r_2}\pi r^2\mbox{d}r.$$Of course $V_1<\pi r_2^2h_1\quad V_2>\pi r_1^2h_2$ so when $h_2$ is much larger than $h_1$ there's $V_1<V_2$ so here's a contradiction.
Since calculus is like a sum, I suppose the problem is how was the summing process done to $r$ form $r_1$ to $r_2$. Normally we would say each time increase a bit until $r$ got to $r_2$ but how much that bit is matters.
So how to explain this and get the volume correctly?
For what it's worth, these are called frustrums.
The issue is because you're not actually calculating a volume.
Think geometrically about what $\pi r^2 \, \mathrm{d} r$ means: you've got the area of a circle, which is good -- but then a small change in the radius? That would just net you some sort of area. After all, $r$ and $\mathrm{d} r$ are in the same "plane", in a sense: they're both radii, just one being a differential change in one. You need something that goes up a dimension, in some way.
What you really would want, in this framing, is some sort of differential change in height; this would generate a volume. However, the radius clearly depends on the height, so your integral would really be
$$\int_0^h \pi r(h)^2 \, \mathrm{d}h$$
If you could find out what $r(h)$ is, the integral might be feasible, but there's a simpler way. Imagine taking a right trapezoid, of top length $r_2/2$ and bottom length $r_1/2$, and height $h$ (be it $h_1$ or $h_2$). Place the vertical side against the $y$-axis, and rotate it about said axis. You should immediately see how this generates a frustrum with the desired parameters.
Then all you need to do is apply the techniques used in finding the volume of a solid of revolution. Paul's Online Math Notes is a good resource for that.