<s>[INST] How to find the overlapping area of a semicircle and a rectangle I am being asked to calculate the pink area that is overlapping between the semicircle and the rectangle. I was only given the radius of the circle (5), the equation of a circle $(x^2+y^2 = r^2).$ I have found the intersection points between the circle and the rectangle $(-4,3)$ and $(4,3)$ but I am lost afterwards, I no longer know what to do.

 [/INST] There are several ways you could approach this problem, but I think this would be the most straight-forward:


*

*Split up the pink region into two: $0\le x\le 4,$ and $4\le x\le 5.$ For the first region, you have a rectangle of area $3\cdot 4=12.$ 

*For the other region, you can perform an integral. You have that $5^2=x^2+y^2$ on the circle, so that $y=\sqrt{25-x^2}$ above the $x$ axis. That is, you calculate
$$\int_4^5\sqrt{25-x^2}\,dx. $$
This integral screams "trig substitution", and since the $25$ and $x^2$ have opposite signs, you would substitute as follows:
\begin{align*}
\frac{\sqrt{25-x^2}}{5}&=\sin(\theta)\\
\frac{x}{5}&=\cos(\theta)\\
dx&=-5\sin(\theta)\,d\theta.
\end{align*}
The integral becomes
$$\int_4^5\sqrt{25-x^2}\,dx=-25\int_{\arccos(4/5)}^{\arccos(1)}\sin^2(\theta)\,d\theta
=25\int_{0}^{\arccos(4/5)}\sin^2(\theta)\,d\theta. $$
You can integrate by parts to obtain
$$-6+\frac{25 \pi }{4}-\frac{25}{2} \arcsin\left(\frac{4}{5}\right), $$
for a final answer of
$$6+\frac{25 \pi }{4}-\frac{25}{2} \arcsin\left(\frac{4}{5}\right). $$
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