# Question about the graphical interplay between displacement and area

A question asked to show that the length of the polar curve $r(\theta) = e^\theta$ from $- \infty$ to $0$ is convergent. Well, using the formula $ds = \sqrt{r^2 + (\frac{dr}{d\theta})^2}$ one can easily find the length to be $\sqrt{2}$. However, I also wondered if this is a legitimate argument:

The curve from $- \infty$ to $0$ is enclosed within a circle with radius $1$ and area $\pi$. Thus, the length of the curve is less than or equal to pi, so it converges.

Is comparing repeated displacement to area like this legitimate or rigorous? My gut tells me no because the units aren't the same: For example, the line $y = x$ from $0 \to 1$ is completely enclosed in a square of sidelength $1$, but the length of that line, $\sqrt{2}$, is greater than the area of the square, $1$. If not, why does the argument seem to make sense in this case?

• You can get an awfully long curve completely contained in a circle of radius $1$. – André Nicolas Feb 24 '14 at 3:28

Consider the curve $r=\frac{1}{\theta}$, where $\theta$ goes say from $1$ to $\infty$. We get a spiral, which, like your spiral, is wholly contained within a circle of radius $1$. The arclength formula gives $$\int_1^\infty \sqrt{\frac{1}{\theta^2}+\frac{1}{\theta^4}}\,d\theta.$$ This integral diverges.