surface area of cylindrical spiral

Question

consider

$$\{(r\cos(\phi),r\sin(\phi),\phi)|r \in (0,1), \phi \in (0,2\pi)\}.$$

is the surface area the same as that of the unit circle, that is $\pi$? Intuitively yes, maybe not :S

Answer 1

If the surface is that described by a radius having its endpoints at $P=(\cos(\phi),\sin(\phi),\phi)$ and $Q=(0,0,\phi)$, then its area is given by $$ \int \sqrt{dz^2+r^2d\phi^2}dr= \int_0^{2\pi}\,d\phi\int_0^{1}\sqrt{1+r^2}\,dr=\pi(\sqrt{2}+\sinh^{-1}1). $$ Intuitively it is quite obvious that the area cannot be that of a circle of unit radius: just think that by "stretching" your spiral by a different amount along $z$ you are changing the surface and its area.