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.

 A: I would split the red area into two parts:
1) Draw a line from the right intersection point towards the origin. You now have two areas.
2) The triangle is easy: $A_1 = \frac{1}{2}xy = \frac{1}{2}\cdot4\cdot3 = 6$
3) The part of the circle is a bit more tricky. Generally, if you want to calculate the area of only a segment of a circle, you can use:
\begin{align}
A = \frac{1}{2}\Delta\phi r^2
\end{align}
where $\Delta\phi$ denotes the difference between the angles (in rad) where both segments start. In your case it would be 0 and the angle of your intersection. From Polar Coordinates, we know that the angle, given cartesian $x$ and $y$ in the first quadrant, is:
\begin{align}
\phi = \arctan \left(\frac{y}{x} \right) = \arctan \left(\frac{3}{4} \right) = 0.644
\end{align}
The radius $r$ can be computed via the Pythagorean Theorem:
\begin{align}
r^2 = x^2 + y^2 = 4^2 + 3^2 = 25
\end{align}
Putting it all together, we got
\begin{align}
A_2 = \frac{1}{2} \Delta\phi r^2 = 0.644 \cdot 25 \cdot \frac{1}{2}= 8.05 
\end{align}
After adding both areas, you got
\begin{align}
A_{\mathrm{tot}} = A_1 + A_2 = 6 + 8.05 = 14.05
\end{align}
I hope this is correct and it may help you.
A: The equation of the semicircle is found from the equation of the circle $x^2+y^2=25$.  Writing the equation in terms of $x^+$, we have $x=\sqrt{25-y^2}$.  We can simply integrate this with respect to $y$ and that is the area.
\begin{align*}
\int_0^3 \sqrt{25-y^2}\,dy
&=\int_0^{\arcsin(3/5)} 5\cos\theta\sqrt{25-25\sin^2(\theta)}\,d\theta\\
&=25\int_0^{\arcsin(3/5)} \cos^2(\theta)\,d\theta\\
&=25\int_0^{\arcsin(3/5)} \left[\frac12+\frac12\,\cos(2\theta)\right]d\theta\\
&=25\left(\frac12\theta+\frac14\sin(2\theta)\right)\bigg|_0^{\arcsin(3/5)}\\
&=25\left(\frac12\arcsin\left(\frac35\right)+\frac14\cdot\frac65\right)\\
&\approx 14.044.
\end{align*}
A: 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
$$6+\frac{25 \pi }{4}-\frac{25}{2} \arcsin\left(\frac{4}{5}\right). $$
A: You have already determined that for $0 \leq x \leq 4$, the pink region is upper bounded by $y = 3$ and lower bounded by $y = 0$.  This will give one integral for the area of the pink region.
For $4 \leq x \leq 5$, the pink region is upper bounded by $y = \sqrt{25-x^2}$ [footnote 1] and lower bounded by $y = 0$.  Therefore, the area of the pink region is
$$  \int_0^4 3 - 0 \,\mathrm{d}x + \int_4^5 \sqrt{25-x^2} - 0 \,\mathrm{d}x  \text{.}  $$
Note:  It is possible to write this area as a single integral if we integrate along the variable $y$.
[footnote 1]  Here we have solved the equation for the circle centered at the origin of radius $5$, $x^2 + y^2 = 5^2$, as \begin{align*}
y^2 &= 25 - x^2  \\
y &= \pm \sqrt{25 - x^2}  \text{.}
\end{align*}
The "$\pm$" gives the upper half of the circle when we choose "$+$" and the lower half when we choose "$-$".  Since the pink region is upper bounded by the upper half of the circle, we use $+\sqrt{25-x^2}$ in the integral.
