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I am trying to solve this problem. I know that the solution is really simple, but I need to solve it using integral calculus.

I have two circles which define the following area between them: $$x^2+(y-1/2)^2>1/4$$ $$x^2+(y-1)^2<1$$

I've tried solving it using transformation to polar coordinates, but i don't know how to do it correctly, when there are multiple intersecting objects.

EDIT: Thanks to the comment I used $$x=r*\cos{\varphi}\\y=r*\sin{\varphi}$$ and I managed to get $$ |r| > \sin{\varphi}\\|r|<2\sin{\varphi}$$ So the integral I created is $$2\int_0^\pi\int_{\sin{\varphi}}^{2\sin{\varphi}}rdrd\varphi$$

It gives the correct solution. However, can't it be computed differently? (Using different bounds). I needed to know how to integrate $\sin^2{x}$ in this case, which is complicated a bit.

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    Just plot the graphs. – 2017-01-03
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    Put $x=rcost $ and $y=rsint$ – 2017-01-03
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    If it matters, you needn't antidifferentiate $\sin^{2}\phi\, d\phi$, you only need the identity $\int_{0}^{\pi} \sin^{2}\phi\, d\phi = \pi/2$, which is fairly clear from symmetries of the trig functions. – 2017-01-03

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As you have already mentioned in your question, you want to perform the following double integral:

$$ 2 \int_{0}^{\pi} \int_{sin\phi}^{2sin\phi} rdrd\phi = 2 \int_{0}^{\pi} \frac{r^2}{2} \bigg|_{sin\phi}^{2sin\phi} d\phi = \int_{0}^{\pi} 4 sin^2\phi - sin^2\phi d\phi = 3 \int_{0}^{\pi} sin^2\phi d\phi $$

You can use the fact that $ cos(2\phi) = cos^2\phi - sin^2\phi = 1 - 2sin^2\phi \rightarrow sin^2\phi = \frac{1}{2} \big( 1 - cos2\phi \big) $ to simplify this integral, i.e.

$ 3 \int_{0}^{\pi} sin^2\phi d\phi = 3 \int_{0}^{\pi} \frac{1}{2} \big( 1 - cos2\phi \big) d\phi = \frac{3}{2} \int_{0}^{\pi} (1 - cos2\phi) d\phi = \frac{3}{2} \Big( \phi - \frac{1}{2}sin2\phi \Big) \Big|_{0}^{\pi} = \frac{3}{2} \Big( \pi - 0 - (0 - 0) \Big) = \frac{3}{2}\pi $