Finding the centroid of a polar curve The curve is $r = e^{-b\theta}$ where $b > 0$ and  $θ \in [0, \infty)$.
I got that the arc length is $\frac{\sqrt{b^2 + 1}}{b}$ (is this correct?), but computing the centroid $(x, y)$ looks awful.  I'm not sure where to start.
 A: Your curve has the parametric representation
$$\gamma:\quad \theta\mapsto(e^{-b\theta}\cos\theta, e^{-b\theta}\sin\theta)\qquad(0\leq\theta<\infty)\ .$$
It follows that
$$ds=\sqrt{(x'(\theta))^2+(y'(\theta))^2}\ d\theta=\sqrt{1+b^2}\ e^{-b\theta}\ d\theta\ .$$
The centroid $(\xi,\eta)$ of $\gamma$ is characterized by the so-called moment equations
$$\xi \ L(\gamma)=\int_\gamma x\ ds,\qquad \eta\ L(\gamma)=\int_\gamma y\ ds\ .$$
You already have computed $L(\gamma)={\sqrt{1+b^2}\over b}$. In addition we need
$$\int_\gamma x\ ds=\sqrt{1+b^2}\int_0^\infty e^{-2b\theta}\ \cos\theta\ d\theta=\ldots={2b\sqrt{1+b^2}\over 1+4b^2}$$
and
$$\int_\gamma y\ ds=\sqrt{1+b^2}\int_0^\infty e^{-2b\theta}\ \sin\theta\ d\theta=\ldots={\sqrt{1+b^2}\over 1+4b^2}\ .$$
It follows that
$$\xi={2b^2\over 1+4b^2},\qquad \eta={b\over1+4b^2}\ .$$
A: I must admit I've never seen it asked that a student find the centroid of a polar curve, so I'm not sure of the context for your problem.  I would guess, since the curve is a one-dimensional object in a two-dimensional space (the plane), that you are being asked to find the point at which one-half of the total arclength (which is finite) is reached, passing inward from $\theta = 0$.  (And I'm leaving this as an answer, rather than a comment, since I don't have enough rep yet...)
EDIT: I concur with your arclength result, so you want to find the value of $\theta$ at which you reach an arclength of $\frac{\sqrt{1+b^2}}{2b}$, then evaluate the radius of the curve there.*  If you're asked to give the Cartesian coordinates, you can then also carry out the requisite transformation.
*Interestingly, this result is independent of  $b$.
