# Find the area of the region inside the limaçon

I'm struggling to figure out the answer to this:

Find the area of the region inside the limaçon, $r=3 + \sin(\theta)$

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It would help to plot it first.

Now use this formula $$A = \frac{1}{2}\int_{\alpha}^{\beta}f(\theta)^2\ d\theta$$

From the picture we see that the curve runs from $0$ to $2\pi$

$$A = \frac{1}{2}\int_{0}^{2\pi}(3 + \sin(\theta))^2\ d\theta = \frac{19\pi}{2}$$ The computation is trivial and it's something you can look up on WolframAlpha if you aren't sure

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Draw a picture. We use the standard formula for area in polar coordinates. In principle there could be a problem with the interval of integration, but here there is no problem, because $3+\sin\theta$ is always positive. So our area is $$\int_0^{2\pi}\frac{1}{2}(3+\sin\theta)^2\,d\theta.$$ For the details, expand. The integral of $\frac{9}{2}$ is easy. For the "middle" term, the integral is easily $0$. Finally, we need to deal with $\int_0^{2\pi}\frac{1}{2}\sin^2\theta\,d\theta$. There are many ways to do this. One of them is to use $\cos2\theta=1-2\sin^2\theta$. Another way is to use symmetry, and note that the integral over our interval of $\cos^2\theta$ is the same as the integral of $\sin^2\theta$. But the sum of these integrals is the integral of $1$.

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Recall the formula for finding an area integral in polar coordinates is given by $$\frac{1}{2}\int_{\alpha}^{\beta}r(\theta)^2\ d\theta$$ The curve is traced from $\alpha = 0$ to $\beta = 2\pi$. We then have $$A = \frac{1}{2}\int_{0}^{2\pi}\left(3 + \sin(\theta)\right)^2\ d\theta$$ This integral is quite standard so I leave you to finish it.

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If you've seen multiple integrals, calculating this area is relatively simple and doesn't require memorizing a single-integral formula. You just integrate $$\int_{\theta_{min}}^{\theta_{max}}\int_{r_{min}}^{r_{max}}rdrd\theta$$ with the limits of integration being the bounds of your function. Setting this up, we get $$\int_{0}^{2\pi}\int_{0}^{3+sin(\theta)}rdrd\theta$$ First we integrate with respect to $r$, and get $$\int_{0}^{2\pi}\frac{(3+sin(\theta))^2}{2}d\theta$$ Now we pull out the $\frac{1}{2}$ and expand to get $$\frac{1}{2}\int_0^{2\pi}9+6sin(\theta)+sin(\theta)^2d\theta$$ Using integration techniques, we simply this integral and find that $A=\frac{19\pi}{2}$

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I know this was posted almost four years ago, but--thank you! This is exactly what I needed. :) – user1739757 Apr 28 at 20:32