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I have difficulties calculating the area and setting the right boundaries of the following polar coördinates:

$$r=2(1+cos(\theta) ) $$

Thanks in advance

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  • $\begingroup$ Is $a$, the polar angle (generally referred as) $\theta$? $\endgroup$ – Anurag A Jun 3 at 17:38
  • $\begingroup$ yes it is, i did not know how to type it $\endgroup$ – Wouter Lommerse Jun 3 at 17:39
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    $\begingroup$ Since you haven't shown the detailed steps of your work we can only guess where you made your error. There are various ways you could make a mistake by a factor of $2$ in a problem like this. $\endgroup$ – David K Jun 3 at 17:51
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    $\begingroup$ Looks like you only squared part of your r @wouterlommerse $\endgroup$ – randomgirl Jun 3 at 19:51
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    $\begingroup$ It generally works out better (and I think is probably easier for you too) to edit the question in order to add information rather than putting a lot of formulas in comments. But it looks like the error has been identified now. $\endgroup$ – David K Jun 3 at 22:19
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The function $$\theta\mapsto r(\theta):=2(1+\cos\theta)$$ does not define an area per se. Now this function is $2\pi$-periodic, and graphing the curve $$\gamma:\quad\theta\mapsto\bigl(x(\theta),y(\theta)\bigr)=r(\theta)\,(\cos\theta,\sin\theta)\qquad(-\pi\leq\theta\leq\pi)$$ we obtain a "loop with an indent" enclosing a certain shape $A$, whereby $\gamma$ is astroidal with respect to the origin. The area of $A$ then can be calculated with the formula $${\rm area}(A)={1\over2}\int_{-\pi}^{\pi}r^2(\theta)\>d\theta=6\pi\ .$$

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