Help calculating the surface area given by the polar curve: $r=2(1-\cos\theta)$ I want to calculate the surface area given by the curve:
$$ r = 2(1-\cos(\theta)) $$
using an integral.
I have thought about doing this:
$$ x = r\cos(\theta), \,  y = r\sin(\theta) $$
$$ \iint r \,dr \,d\theta
$$
But I can't figure my integral boundaries. And is it alright to multiply with the Jacobian $r$? Or would that be a mistake?
I know that the $dr$ should by limited between $0$ and $2(1-\cos(\theta))$, but I could decide what about the limits of $d\theta$.
Do I need to go from $0$ to $\pi$ or from $0$ to $2\pi$   ?
Could someone please clarify this for me?
 A: From 

http://tutorial.math.lamar.edu/Classes/CalcII/PolarSurfaceArea.aspx

The formula for surface area is  $$2\pi\int_{a}^{b}  r\cos(\theta)\sqrt{r^2 + (r')^2}d\theta$$
Around the y-axis and it's $$2\pi\int_{a}^{b} r\sin(\theta)\sqrt{r^2 + (r')^2}d\theta$$
When around the x-axis.
EDIT: I noticed you were asking about the bounds. The main method to find out the bounds that you want to use is by graphing the function. Also please consider renaming the question to the area of the polar graph and not the surface area, because that implies you're revolving the function around an axis.
From

http://www.wolframalpha.com/input/?i=polar+graph+r+%3D2*%281-cos%28theta%29%29

You can see that theta is from $0$ to $2\pi$, and r is from $0$ to $2(1-\cos(\theta))$. And you're supposed to use the jacobian as well, because you're using a double integral, and the jacobian for polar graph is $rdrd\theta$
A: Your jacobian is correct, you must use:
$$ \iint rdrd\theta $$
When determining your bounds for r and $\theta$, think it terms of radii and angles.
For r, you are correct in that it must be between 0 and $ r = 2(1-cos(\theta)) $
For $\theta$, since it repeats from 0 to $2\pi$, that would be the bounds for $\theta.$
A: Call $D$ the domain enclosed by your curve. Then, you're right,
$$A=\text{Area}(D)=\iint_D\mathrm{d}A,$$
where $\mathrm{d}A$ is the surface element. In polar coordinates, $D$ is represented by
$$\Delta=\bigl\{(r,\theta)\in\mathbb{R}^2\;\bigm\vert\;0\leq\theta\leq2\pi,\ 0\leq r\leq2(1-\cos\theta)\bigr\}.$$
Then, using the well-known surface element in polar coordinates, i.e., $\mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\theta$ we obtain
$$A=\iint_\Delta r\,\mathrm{d}r\,\mathrm{d}\theta.$$
Since $\Delta$ is a normal domain, we can use iterated integrals:
$$A=\int_0^{2\pi}\left(\int_0^{2(1-\cos\theta)}r\,\mathrm{d}r\right)\,\mathrm{d}\theta.$$
I'm sure you can take it from here.
A: $$\text{Area}=\frac{1}{2}\int_0^{2\pi}r^2d\theta=\frac{1}{2}\int_0^{2\pi}\bigg[2\{1-\cos(\theta)\}\bigg]^2d\theta.$$
If we expand this, we find that $$\text{Area=2}\int_0^{2\pi}\bigg[1-2\cos(\theta)+\cos^2(\theta)\bigg]d\theta.$$ 
But $\cos^2(\theta) \equiv \frac{1+\cos(2\theta)}{2}$so you should be able to integrate this easily.
