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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.

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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:

enter image description here

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.

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  • $\begingroup$ Excellent answer, and just as a riff off of this, you can also observe that the area of the quarter-circle is $\pi$. Then you can subtract $\int_{\arccos(1/3)}^{\pi/2}\int_{6\cos\theta}^2 r dr d\theta$ from $\pi$ to get half of the desired area. $\endgroup$ – John Moeller Feb 9 '14 at 22:37
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    $\begingroup$ Quite so (you get to dodge doing two integrals in that approach, since you can simply take one area measure from classical geometry). The important first step in these "area between two polar curves" problems is to have a good sketch of the region; as with so much other calculus problems, the picture is then useful in making choices about the calculation (and can suggest more than one method). $\endgroup$ – colormegone Feb 9 '14 at 22:41
  • $\begingroup$ Sorry, folks, I went back and amended my answer, since "between the curves" here didn't make it clear which region was to be covered (so now I have both...). $\endgroup$ – colormegone Feb 9 '14 at 22:57
  • $\begingroup$ I am very sorry for the late reply. I was asking for the area of intersection (between the two circles). I did not actually realize that we would account for a larger part of the area when integrating as an upper bound $r=2$, and for a very small part of the area as an upper bound $r=6cos{\theta}$. I guess in this case it is very helpful to draw a precise diagram (and not the deformed circles I tried to draw). $\endgroup$ – arcbloom Feb 11 '14 at 4:38
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    $\begingroup$ @AméricoTavares I use the Grapher utility on my Mac; the color "fills" are done with Photoshop. $\endgroup$ – colormegone Feb 15 '14 at 1:30
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 $$ \color{#00f}{\large{\cal A} = 2\pi - 4\root{2} + 14\arcsin\pars{1 \over 3}} \approx 5.3841 $$

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