Question

Calculate ∬y dA where R is the region between the disks x^2+y^2 <=1 & x^2+(y-1)^2 <=1

Answer 1

Let's first consider converting to polar coordinates.

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\begin{cases}x^2+y^2=1\iff r=1\\x^2+(y-1)^2=1\iff r=2\sin\theta\end{cases}$

We have

$1=2\sin\theta\implies\sin\theta=\dfrac12\implies\theta=\dfrac\pi6\text{ or }\theta=\dfrac{5\pi}6$

Then $\mathrm dA=r\,\mathrm dr\,\mathrm d\theta$ and the integral is

$\displaystyle\iint_Ry\,\mathrm dA=\int_{\pi/6}^{5\pi/6}\int_{2\sin\theta}^1r^2\sin\theta\,\mathrm dr\,\mathrm d\theta=\boxed{-\frac{\sqrt3}4-\frac{2\pi}3}$