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The question asks to find the area inside $r = 1 + \sin\theta$ and outside $r = 2 \sin\theta$ using double integrals.

In my attempt, I found the intersection to be $\theta = \frac{\pi}{2}$.

I bounded $r$ as $[2 \sin \theta, 1 + \sin \theta]$, since I couldn't find another intersection for the curves, I bounded $\theta$ as $[-\frac{\pi}{2}, \frac{\pi}{2}]$ to solve for half of the area required.

Then multiplied everything by $2$ to get the full area. I ended up with $\frac{-\pi}{2}$ while the correct answer is $\frac{\pi}{2}$.

What am I doing incorrectly?

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  • $\begingroup$ the curve r = 2 sin $\theta$ makes two cycles as $\theta$ goes from 0 to $2\pi$ while r = 1 + sin $\theta$ makes only one. $\endgroup$ – Doug M Mar 12 '16 at 1:32
  • $\begingroup$ Yes, I understand that. Is this consideration why I got the wrong answer? Can you explain in a bit more detail? $\endgroup$ – Sec Mar 14 '16 at 5:11
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The polar equation $$r(\theta):=2\sin\theta\qquad(0\leq\theta\leq 2\pi)$$ describes a circle of radius $1$ centered at $(0,1)$, covered twice. On the other hand, the area inside the curve $$r(\theta):=1+\sin\theta\geq0\qquad(0\leq\theta\leq 2\pi)$$ is given by $${1\over2}\int_0^{2\pi}(1+2\sin\theta+\sin^2\theta)\>d\theta={1\over2}(2\pi+0+\pi)={3\pi\over2}\ .$$ The area $A$ in question therefore has the value ${\displaystyle{\ 3\pi\over2}-\pi={\pi\over2}}$.

If you want to compute $A$ in terms of a double integral you should stay with polar coordinates. You then obtain $$A=2\left(\int_{-\pi/2}^0\int_0^{1+\sin\theta} r\>dr\ d\theta+\int_0^{\pi/2}\int_{2\sin\theta}^{1+\sin\theta} r\>dr\ d\theta\right)\ .$$

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  • $\begingroup$ Is it possible to solve with double integral instead of single? Or can you give me hints on why my solution is incorrect? $\endgroup$ – Sec Mar 14 '16 at 5:11
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$$\frac12\int_\frac{-\pi}{2}^\frac{\pi}{2}((1+sin\theta)^2-(2\sin\theta)^2)\,d\theta$$

$$= \frac{\pi}{4}$$

$$=\frac{\pi}{4} \cdot 2$$

$$=\frac{\pi}{2}$$

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  • $\begingroup$ The answer is correct, but the set-up is not, and the middle is missing. $\endgroup$ – Doug M Mar 12 '16 at 1:42

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