Area between two circles as a double integral in polar coordinates Find the area between the circles $x^{2} + y^{2} = 4$ and  $x^{2} + y^{2} = 6x$ using polar coordinates.

I have found that the equation of the first circle, call it $C_1$, is $r=2$ on the other hand, for $C_2$, I get that its equation is $r = 6cos{\theta}$. Then, to find the bounds of integration, I have found that their angle of intersection should be $\theta = \arccos(1/3)$ and $\theta = -\arccos(1/3)$. Then, to set up the double integral:
$A= \displaystyle\int_{-\arccos(1/3)}^{\arccos(1/3)} \displaystyle\int_{6\cos{\theta}}^2  \mathrm{r}\,\mathrm{d}r\, \mathrm{d}{\theta}$
However, when evaluating this integral with the calculator, I get a negative value. What would be the problem in this case? Thanks in advance for your help.
 A: You should have two separate integrals, since there is a change in the boundaries of integration, as measured from the origin.  You can also apply symmetry about the $ \ x-$ axis and write
$$ A \ = \ 2 \ \left[ \int_0^{\arccos(1/3)}  \int_0^{2} \ r dr \ d \theta \ + \ \int_{\arccos(1/3)}^{\pi / 2} \   \int_0^{6 \cos \theta} \ r dr \ d \theta \ \right] . $$
The "radial arm" extends to the $ \ r \ = \ 2 \ $ circle up to the angle $ \ \theta \ = \ \arccos(\frac{1}{3}) \ , $ but then "switches" to the other circle until $ \ \theta \ = \ \frac{\pi}{2} \ . $  Here is a graph of the situation:

The point is that for the region "between" these two circles, your "radius" out from the origin is not bounded anywhere by one circle as the "outer circle" and the other as the "inner circle"; you only change over from one circle to the second one.
EDIT:  I looked at your problem statement again, and then my graph again, and realized that the statement is ambiguous.  I interpreted this as "the area inside both $ \ r \ = \ 6 \cos \theta \ \ \text{and} \ \ r \ = \ 2 \ $ " [which I've filled with blue] .
The integral you wrote could be applied to "the area inside $ \ r \ = \ 6 \cos \theta \ , $ but outside $  \ r \ = \ 2 \  $ " [which I've now filled in orange].  In that case, your approach is correct, except that $ \ r \ = \ 6 \cos \theta \ $ is the "outer curve" and $ \ r \ = \ 2 \ $ , the "inner curve", so you should have written
$$ A \ = \ \int_{-\arccos(1/3)}^{\arccos(1/3)} \displaystyle\int^{6\cos{\theta}}_2  \mathrm{r}\,\mathrm{d}r\, \mathrm{d}{\theta} \ \ . $$
Was this in fact the area you meant to cover?  (Sorry if I misunderstood the intended problem.)  You have the boundaries swapped in the integral, which certainly explains the negative result.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\!\!\!\!\!{\cal A}&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\Theta\pars{4 - x^{2} - y^{2}}\Theta\pars{6x - x^{2} - y^{2}}\,\dd x\,\dd y
\\[3mm]&=\int_{0}^{\infty}\dd r\,r\int_{0}^{2\pi}\Theta\pars{4 - r^{2}}
\Theta\pars{6r\cos\pars{\theta} - r^{2}}
\\[3mm]&=\int_{0}^{\infty}\dd r\,r\int_{0}^{2\pi}\dd\theta\,\Theta\pars{2 - r}
\Theta\pars{6\cos\pars{\theta} - r}
=\int_{0}^{2}\dd r\,r\int_{0}^{2\pi}\dd\theta\,
\Theta\pars{\cos\pars{\theta} - {r \over 6}}\tag{1}
\end{align}
where $\Theta\pars{z}$ is the Heaviside Step Function.

\begin{align}
&\left.\int_{0}^{2\pi}
\Theta\pars{\cos\pars{\theta} - {r \over 6}}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
=\left. 2\int_{0}^{\pi}
\Theta\pars{-\cos\pars{\theta} - {r \over 6}}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
\\[3mm]&=\left. 2\int_{-\pi/2}^{\pi/2}
\Theta\pars{\sin\pars{\theta} - {r \over 6}}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
=2\left. \int_{\arcsin\pars{r/6}}^{\pi/2}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
\end{align}

$$
\left.\int_{0}^{2\pi}
\Theta\pars{\cos\pars{\theta} - {r \over 6}}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
=\left.\pi - 2\arcsin\pars{r \over 6}\right\vert_{0\ <\ r\ <\ 2}
\tag{2}
$$
In replacing $\pars{2}$ in $\pars{1}$, we find:

\begin{align}
{\cal A}&=\int_{0}^{2}\bracks{\pi - 2\arcsin\pars{r \over 6}}r\,\dd r
=2\pi - 2\int_{0}^{2}\arcsin\pars{r \over 6}r\,\dd r
\end{align}
  The last integral in the right member is an elementary one: It can be easily solved after integration by parts and the final result is
  $$
\color{#00f}{\large{\cal A} = 2\pi - 4\root{2} + 14\arcsin\pars{1 \over 3}}
\approx 5.3841
$$ 

