Finding the centroid of a polar curve I have absolutely no idea how to find the area centroid of this problem. I have been working at this one for ages but can't seem to get anywhere. 
Any first steps? How would one go about solving this? 
Find the centroid of the area by direct integration. 

 A: Here we go. Now we want to calculate the centroid $(\overline{r}, \overline{\theta})$ of the area that was defined by a polar function $r = r(\theta), (\alpha \leqslant \theta \leqslant \beta)$.
We know the general formula for centroid:
$$
\left\{
\begin{aligned}
\overline{x} = \dfrac{1}{A}\int_A x \,\mathrm{d}A\\
\overline{y} = \dfrac{1}{A}\int_A y \,\mathrm{d}A
\end{aligned}
\right.
$$
For each polar point $(r(\theta), \theta)$ on the curve, we can take a fan-shaped surface element just like the following figure.

The centroid of this surface element is $\left(\frac{2}{3}r(\theta), \theta\right)$(Why there is a $\frac{2}{3}$ just like you ask? We can consider the surface element as an triangle, and the centroid of this triangle is obviously at here.) The cartesian coordinate of it's centroid is $\left(\frac{2}{3}r(\theta)\cos\theta, \frac{2}{3}r(\theta)\sin\theta\right)$. And the area of this surface element $\mathrm{d}A = \frac{1}{2}r^2(\theta)\mathrm{d}\theta$. So we have
$$
\left\{
\begin{aligned}
\overline{x} = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\cos\theta \,\mathrm{d}\theta\\
\overline{y} = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\sin\theta \,\mathrm{d}\theta
\end{aligned}
\right.\\
\mbox{with}~A = \dfrac{1}{2}\int_{\alpha}^{\beta}r^2(\theta)\,\mathrm{d}\theta
$$
Now we get the cartesian coordinate $(\overline{x}, \overline{y})$ of the centroid of the hole form. But we want the polar coordinate $(\overline{r}, \overline{\theta})$. Notice that
$$
\left\{
\begin{aligned}
& \overline{r} = \left|\overline{x}+\mathrm{i}\,\overline{y}\right|\\
& \overline{\theta} = \arg(\overline{x}+\mathrm{i}\,\overline{y})
\end{aligned}
\right.
$$
with
$$
\begin{aligned}
\overline{x}+\mathrm{i}\,\overline{y} & = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\cos\theta \,\mathrm{d}\theta + \mathrm{i}\,\left(\dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\sin\theta \,\mathrm{d}\theta\right)\\
& = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)(\cos\theta + \mathrm{i}\,\sin\theta) \,\mathrm{d}\theta\\
& = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\,\mathrm{e}^{\mathrm{i}\theta} \,\mathrm{d}\theta
\end{aligned}
$$
Finally we have
$$
\left\{
\begin{aligned}
& \overline{r} = \dfrac{1}{3A}\left|\underline{Z}\right|\\
& \overline{\theta} = \arg\left(\underline{Z}\right)
\end{aligned}
\right.\\
~\\
\mbox{with}~A = \dfrac{1}{2}\int_{\alpha}^{\beta}r^2(\theta)\,\mathrm{d}\theta\\
\mbox{and}~\underline{Z} = \int_{\alpha}^{\beta} r^3(\theta)\,\mathrm{e}^{\mathrm{i}\theta} \,\mathrm{d}\theta
$$
Now apply this conclusion in this problem. Here, $r(\theta) = a\mathrm{e}^{\theta},(0 \leqslant \theta \leqslant \pi)$. So
$$
\begin{aligned}
A & = \dfrac{1}{2}\int_{0}^{\pi}a^2\mathrm{e}^{2\theta}\,\mathrm{d}\theta\\
& = \dfrac{1}{4}a^2(\mathrm{e}^{2\pi}-1)
\end{aligned}
$$
$$
\begin{aligned}
\underline{Z} & = \int_{0}^{\pi} a^3\mathrm{e}^{(3+\mathrm{i})\theta}\,\mathrm{d}\theta\\
& = -\dfrac{a^3(\mathrm{e}^{3\pi}+1)}{3+\mathrm{i}}
\end{aligned}
$$
So
$$
\begin{aligned}
\overline{r} & = \dfrac{1}{3A}\left|\underline{Z}\right|\\
& = \dfrac{2\sqrt{10}}{15}\dfrac{\mathrm{e}^{3\pi}+1}{\mathrm{e}^{2\pi}-1}\,a \approx 9.78\,a
\end{aligned}
$$
$$
\begin{aligned}
\overline{\theta} & = \arg(\underline{Z})\\
& = \pi - \arctan \dfrac{1}{3} \approx 161.57^\circ
\end{aligned}
$$
Plot the centroid $C = (\overline{r}, \overline{\theta})$ on the figure:

A: So the center of mass is given by $$\int \vec{r} dm$$
Your mass density is uniform. Your limits of integration will be $\theta: (0,\pi)$ and $r: (0,a e^{\theta})$. The Jacobian will be simply $r$. Remember that $\vec{r}=r \cos(\theta) \hat{i} + r\sin(\theta) \hat{j}$. You'll be left with a tricky enough integration, I would advise integrating with respect to $dr$ first. 
Can you handle it from here?
A: You have already seen the solution to the problem and I see that they have made recourse to complex variables. I would like to demonstrate that the solution might be somewhat simpler if one were to fully embrace the complex plane.
To begin with, for $r=ae^{\theta}$ the curve can be expressed in the complex plane simply as
$$z=ae^{(1+i)\theta}$$
This is seen to be a logarithmic spiral with a flair coefficient of unity, which means that the radius increases by factor $e$ per radian. You'll notice that at $\theta=\pi$ the radius is $r(\pi)=e^{\pi}$. Now, we know that the area and centroid are given by (see reference)
$$
A=\frac{1}{2}\int\Im\{z^*\dot z\}d\theta\\
Z_c=\frac{1}{3A}\int z\ \Im\{z^*\dot z\}d\theta
$$
where $()^*$ is the conjugate and $(^\dot \,)$ is the derivative with respect to $\theta$. Therefore we can calculate the centroid as follows:
$$
\dot z=a(1+i)e^{(1+i)\theta}\\
z^*=ae^{(1-i)\theta}\\
z^*\dot z=a^2(1+i)e^{2\theta}\\
\Im\{z^*\dot z\}=a^2e^{2\theta}\\
$$
Then
$$A=\frac{a^2}{2}\int_0^{\pi}e^{2\theta}\,d\theta=\frac{a^2}{4}(e^{2\pi}-1)\\
\begin{align}
Z_c
&=\frac{a^2}{3A}\int_0^{\pi}e^{(1+i)\theta}e^{2\theta}\,d\theta=\frac{a^2}{3A}\frac{(e^{(3+i)\pi}-1)}{(3+i)}\\
&=\frac{4a}{3}\frac{1}{(3+i)}\frac{(e^{(3+i)\pi}-1)}{(e^{2\pi}-1)}\approx-9.274a+3.091ai
\end{align}
$$
An finally, we can the centroid in polar coordinates from
$$
r_c=|Z_c|\approx9.776a\\
\theta_c=\arg Z_c\approx161.87^{\circ}
$$
It's been my experience that it's much simpler to solve problems like this in the complex plane.
The reference for this work is: Zwikker, C. (1968). The Advanced Geometry of Plane Curves and Their Applications, Dover Press.
