# On volume change when paper is folded to give hollow cylinder

I have a question whose answer should be seemingly simple. If a paper of length $l$, breadth $b$ and thickness $t$ were folded to form a cylinder, the cylinder would be hollow with the same thickness $t$. Also, the outer circle would have radius $\frac {b}{2\pi}$ and the inner circle radius $\frac {b}{2\pi}-t$. On calculation I find the volume of the paper to be different from the volume of the outer cylinder minus the volume of the inner cylinder. Why is this the case? Thanks in advance.

• Why do you take the radii to be $b/2\pi$ and $b/2\pi-t$ rather than $b/2\pi+t$ and $b/2\pi$ -- or even $b/2\pi\pm t/2$? – Henning Makholm Aug 21 '18 at 15:22
• Rohpa if you fold it as a cylinder you will get $2\pi r=b$ also $2\pi (r+t)=b$. Won't it contradict? It means that the paper will deform. – prog_SAHIL Aug 21 '18 at 15:25

The paper of a thickness $t$ will never be able to form a perfect cylinder.

As breadth of the paper is $b$, you can write that,

$$2\pi r=b$$

where $r$ is the inner radius of the cylinder.

But the circumference of outer cylinder should also be $b$.

So,

$$2\pi (r+t)=b$$

which is a contradiction, proving that the paper must deform.

• I thought as much, thanks for your answer. Just a small temporary confusion. – Rohpa Aug 21 '18 at 16:11