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I have to find the area of the region that lies between the curves $r=1+\sin\theta; r=1+\cos\theta$ . The answer the book gave was $\frac {3\pi}{2}-2\sqrt{2}$ . I tried generating the curve for $r=1+\cos\theta$ from $\frac {\pi}{4}$ to $\pi$, which generated the proper part of the curve, and then generating the curve for $r=1+\sin\theta$ from $\frac {-3\pi}{4}$ to $\frac {\pi}{4}$ which generated the correct part of that curve. I then attempted to put that into the formula given for area between polar curves, $A=2(\int_\alpha^\beta \frac {1}{2}f(\theta)^2 d\theta + \int_\gamma^\alpha \frac {1}{2}g(\theta)^2 d\theta)$ and used the bounds I previously used to generate my curve as the bounds on the integrals, respectively. However, I still am unable to get the answer the book gave. I have thought of trying to divide the section into two and find the area of one half and then doubling my result to find the area of the whole, but I am unsure of how to go about that. Does anyone have any suggestions? The point of intersection I found for the curves is $(1+ \frac {\sqrt{2}}{2} , \frac {\pi}{4})$.

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  • $\begingroup$ There are 2 distinct regions where these two curves have areas that lie between them. The range you have allowed for the curve $r=1+\cos\theta$ function only includes one of these regions. Have you tried $\frac{\pi}{4}$ to $\frac{5\pi}{4}$ $\endgroup$ – Peter Jun 29 '15 at 15:11
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There are two distinct regions where your curves overlap. They do overlap for the intervals of $\theta$ that you give, but that gives only the large overlap at the upper right of the origin. You left out the smaller overlap at the lower left of the origin.

enter image description here

That overlap is covered by $\frac{5\pi}4\le \theta\le \frac{3\pi}2$ for the $r=1+\cos\theta$ curve and by $\pi\le \theta\le \frac{5\pi}4$ for the $r=1+\sin\theta$ curve. To get the correct answer you will need to include that little piece.

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  • $\begingroup$ Thank you for pointing that out. I did not see that on the graph. I believe that you meant $\pi \le \theta \le \frac {5\pi}{4}$ for the $r=1+\cos\ \theta$ curve. Was the formula I listed above the correct one to use for computing this particular area? A friend suggested modifications may be made but I am not sure which. $\endgroup$ – Carly Sawatzki Jun 29 '15 at 14:40
  • $\begingroup$ @CarlySawatzki: Thanks for the correction. I need to run now and can't do more checking: perhaps later. $\endgroup$ – Rory Daulton Jun 29 '15 at 14:45
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I found the answer by separating the figure into four symmetrical parts, two corresponding to the curve $r=1+\sin\ \theta$ and one corresponding to the curve $r=1+\cos\ \theta$. This way the intersecting areas in both quadrants one and four were accounted for. Then, I found the points at which the radius (r) of the curve would be zero, focusing only on the curve $r=1+\cos\ \theta$ as well as using the points I found for the points of intersection, which were at $(1+ \frac {1}{\sqrt{2}}, \frac{\pi}{4})$ and $(1- \frac {1}{\sqrt{2}}, \frac{5\pi}{4})$. I used the formula listed above, which was $A=2(\int_\alpha^\beta \frac {1}{2}f(\theta)^2 d\theta + \int_\gamma^\alpha \frac {1}{2}g(\theta)^2 d\theta)$. Thus, the setup for the formula is as follows for the curve $r=1+ \cos\ \theta$: $A=2(\int_\pi^\frac {5\pi}{4} \frac {1}{2}(1+ \cos\ \theta)^2 d\theta + \int_\frac {\pi}{4}^\pi \frac {1}{2}(1+\cos\ \theta)^2 d\theta)$ I then solved each integral separately and added them together, as follows: $(-\sqrt{2}+\frac{9\pi}{8}-\frac{7}{4})+(-\sqrt{2}+\frac{3\pi}{8}+\frac{7}{4})=(\frac{3\pi}{2}-2\sqrt{2})$, which was the answer given by the book. Thank you all for your help!

Note: This solution was not tested if you use the halves of the curves generated by the $r=1+\sin\ \theta$ curve. I assume it would work similarly if you solved for the values of $\theta$ that would make $\sin\ \theta=-1$ true and used this in your bound for the integral.

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Why don't you simply take an integral for the upper half of the region which is bounded by $r_2=1+\cos\theta$ with $\dfrac{\pi}{4}<\theta<5\dfrac{\pi}{4}$.

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Due to symmetry about $ \theta = \pi/4$ line,

$$ 2\cdot \int_{\pi/4}^{\pi/4} \frac12 (1+\cos(\theta))^2 d \theta $$

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