How do you find the limits of integration of polar curves?

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Find the area of the region in the plane enclosed by the cardioid r = 4+4\sin{\theta}

The book explains that "Because r sweeps out the region as {\theta} goes from 0 to 2{\pi}, these are our limits of integration."
 
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When working in Plane Polar coordinates, if you are talking about closed curves, then \theta always goes from 0 to 2\pi.
 
In another question, the limits change. How do I tell if the function is a closed curve?
The question is "Find the area inside the smaller loop of the limacon r = 1+2cos(\theta)."
The books gives the limits 2(\pi)/3 to 4(\pi)/3."
 
It's not always from 0 to 2pi. The best way to learn this is to really just familiarize yourself with polar curves. Do you know what a limacon looks like? If yes, you should be able to picture the small loop (there's only 1). Geometrically, this means that your polar curve crosses itself for somewhere. Algebraically, this means r = 0 for two values of theta between 0 and 2pi. So simply find these two values of theta. They will be your limits.
 
Is there a program that gives theta in terms of pi instead of decimals?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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