Find the red coloured area

Question

A circle is in a square of side 10 and a quadrant circle with radius 10 overlaps as shown in the figure. Find the red coloured area.

$\hskip2.4in$

I guess I could find the value by subtracting the area of circles from that of the square, but I can't because of the overlapping part.

Answer 1

@Leun Park

I can provide an approximation in case no one can get the exact answer and you need some answer.

$ \frac{1}{2} \left(100-\left(\frac{100 \pi }{4}+14.6381259\, +\frac{1}{4} (100-25 \pi )\right)\right)\approx .7285$

I think it is close.

Or notice the piece on the left is the same as the red piece. Now just get the intersection points of the arc and the circle and integrate between the two equations.

$\text{Area}=\int_0^{\frac{5}{4} \left(3-\sqrt{7}\right)} \left(5-\sqrt{10 x-x^2}\right) \, dx-\int_0^{\frac{5}{4} \left(3-\sqrt{7}\right)} \sqrt{20 x-x^2} \, dx $

$=\frac{1}{4} (-25) \left(-6+\sqrt{7}+10 \pi -4 \tan ^{-1}\left(2+\sqrt{7}\right)-16 \tan ^{-1}\left(4+\sqrt{7}\right)\right)$

$=.72850589607829715$

which agrees with my approximation.

Answer 2

The red area can be viewed as a trapezoid minus a triangle and two circular sectors.

Letting $O=(0,0)$ and $B=(1,1)$ we easily obtain $$P=\left({\sqrt{7}-1\over4},{\sqrt{7}+1\over4}\right)\ ,$$ so that the angles of the two sectors become $$\alpha=\arcsin{\sqrt{7}-1\over4},\quad \beta=\arcsin{5-\sqrt{7}\over8}\ .$$ The triangle $OAP$ has area $${1\over2}\>\vec{OA}\>\wedge\>\vec{OP}={1\over2} \ (1,-1)\wedge\left({\sqrt{7}-1\over4},{\sqrt{7}+1\over4}\right)={\sqrt{7}\over4}\ ,$$ and the encompassing trapezoid obviously has area ${3\over2}$.

At the end we have to multiply everything with $25$, so that we obtain $${\rm red\ area}=25\left({3\over2}-{\sqrt{7}\over4}-{1\over2}\arcsin{\sqrt{7}-1\over4}-2\arcsin{5-\sqrt{7}\over8}\right)\doteq0.728506\ .$$

Answer 3

An integral approach:

Call the red area $A$, then: $$ A = \{(x,y)^2 \mathbb{R}^2 : 0 \le x \le 10, 5 \le y \le 10, (x-10)^2+y^2 \ge 10^2, (x-5)^2 + (y-5)^2 \ge 5^2\}$$

(where you can change all $\le$ to $\lt$, if you don‘t want to "count the border", it won‘t change the result.) How did I get to this formula? Well, the first two conditions should be obvious. The formula for a circle with radius $r$ centered in $(x_0,y_0)$ is $((x-x_0)^2+(y-y_0)^2 = r^2$ and the red area is $above$ two circles.

The following requires some knowledge about the Lebesgue integral as Tonelli will be used. Though I hope most of it should be understandable with less knowledge.

The area of $A$ is $\int_A 1 \mathrm{d}(x,y)$.

$$ \int_A 1 \mathrm{d}(x,y) = \int_5^{10} \int_{\varphi(y)}^{10} 1 \mathrm{d}x \mathrm{d}y = \int_5^{10} 10 - \varphi(y) \mathrm{d}y $$

where $\varphi(y)$ gives the smallest $x$ for which $(x,y) \in A$. This is true because of the upper and lower bounds of the $x$ and $y$ values.

Two steps are remaining:

1. Find $\varphi(y)$ for all $y \in [5,10]$ To do so, write the conditions in the definition of $A$ as “$x \ge max(\dots,\dots)$”.
2. Solve the resulting one-dimensional integral. Honestly I do not know whether this is an easy task, as I have not done 1. (yet?). Hopefully you find an anti derivative here! (You should be able to assume the integral is a Riemann integral.)