MHB What is the area of region $S$?

[Ackbach]

Here is this week's POTW:

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Sorry for the shamefully late POTW this week. To compensate, I'll make it a bit easier.

Find the area of the region $S=\{(x,y):x\ge 0, y\le 1, x^2+y^2\le 4y\}.$

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[Ackbach]

Congratulations to kiwi for his correct solution, which follows:

Spoiler

If $x^2+y^2\le 4y$, then $\displaystyle A=\int_0^1 x \, dy=\int_0^1\sqrt{4-(y-2)^2} \, dy$. Let $y-2=2\sin(\alpha)$, which implies that $dy=2\cos(\alpha) \, d\alpha$. So
\begin{align*}
A&=\int_{-\pi/2}^{-\pi/6}2\cos(\alpha) 2\cos(\alpha) \, d\alpha \\
&=2\int_{-\pi/2}^{-\pi/6}[1+\cos(2\alpha)] \, d\alpha \\
&=2\left[\alpha+\frac12 \sin(2\alpha)\right]_{-\pi/2}^{-\pi/6} \\
&=\left[-\frac{\pi}{3}+\sin\left(-\frac{\pi}{3}\right)\right]-\left[-\pi+\sin(-\pi)\right] \\
&=\frac{2\pi}{3}-\sin\left(\frac{\pi}{3}\right) \\
&=\frac{2\pi}{3}-\frac{\sqrt{3}}{2}.
\end{align*}