Using calculus to compute volume 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?
 A: 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.
