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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.

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Now, wouldn't x = (1/L) lim_T-->∞ ∫[0 to T] e^(-bθ) ds? But I don't know how to compute that integral!! – Mark T Apr 12 '13 at 16:52
What is the centroid of a curve? A region in the plane has a centroid. – Stefan Smith Apr 13 '13 at 1:16
up vote 1 down vote accepted

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 $$\xi={2b^2\over 1+4b^2},\qquad \eta={b\over1+4b^2}\ .$$

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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 $b$.

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I don't know how to do the centroid though! help! – Mark T Apr 12 '13 at 17:29
You presumably calculated the arclength using – RecklessReckoner Apr 12 '13 at 17:38
You presumably calculated the arclength using $\sqrt{1+b^2}\int^{\infty}_{0} e^{-b\theta} d\theta $ to get the answer you quoted. Now set the upper limit of the arclength integral to some unknown angle $\Theta$ ; you will get an arclength function which you set equal to $\frac{\sqrt{1+b^2}}{2b}$. You would then solve this for the angle $\Theta$; this is the angle at which the midpoint of the arclength occurs. You can put this into the polar curve equation to get $r$. With the polar coordinates in hand, you can convert these to rectangular coordinates. – RecklessReckoner Apr 12 '13 at 17:45

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