I'm trying to calculate the volume of a solid of revolution, given by $y=2$, $x=0$ and $y=x^{1/3}$, rotating about the $x$-axis. from $x=0$ to $x=8$.
One way I solved this was via an integral $\pi\displaystyle\int_0^8 \!(2-x^{1/3})^2 \, \mathrm{d}x$. This makes sense to me because I'm getting the radius of each disk via $2-x^{1/3}$ and then integrating the areas of the disks from $0$ to $8.$ this provides an answer $3.2\pi.$
Another way I approached this was to first find the volume of the entire cylinder, then subtracting the volume of the solid of revolution given by $y=x^{1/3}$. this gives an answer of $12.8\pi$
Both of these seem valid ways to calculate the volume, though both are clearly algebraically different and provide different values. Which one is the correct way to calculate the volume and why is the other one invalid?
I think the second way is correct, but the first way, whilst (I believe) being a nice attempt at applying Cavalieri's principle, is ultimately incorrect.
The red almost-rectangle on the far left has the same area as the blue almost-rectangle on the far left. But compare the volume of revolution formed when rotating each one of these around the $x-$ axis. Whilst similar, the red almost-rectangle will produce greater volume. The relative difference in the volumes of revolution formed between the red and blue almost-rectangles on the right is much greater. The red one will form a large ring, whereas the blue one will form a small ring (in fact, a ring with no hole). So I think we see that for volume, overall, red>blue.