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

Show all work. (Also explain why you split up the regions)

User Guy Bowden
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1 Answer

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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=\frac12\implies\theta=\frac\pi6\text{ or }\theta=\frac{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}

User Lucazav
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