Finding the area of a region of circle

Question

Consider the following shape. It is requested to find the area of the shaded region. The radius of the circles and the overlapping half circles is 1 unit. The region is irregular and I couldn't find a way to find it.

Answer 1

Hint (from the picture):

By symmetry it is not difficult to see that: $ \angle(GAJ)=30°$ and $\angle(AIJ)=60°$.

I suppose that you can do from this.

Answer 2

I believe the area to be $$\frac{π+6\sqrt3}{12}.$$ One way is to realise that the smallest region of the interior of the largest square is the key. There are four such regions in your diagram (for clarity). Let this region have area $\epsilon$, then by focusing on one quadrant of the large square, you should be able to see that you can find this area by subtracting the areas of an equilateral triangle of unit side and two sectors of the unit circle each of arcs which are $1/12$ of the circumference of the unit circle from the area of one-fourth of the large square. If the sectors have area $S$ each and the triangle area $\triangle$, then $\epsilon$ is given by $$1-2S-\triangle=1-π/6-\sqrt3/4.$$ You should now to be able to see how I arrived at the quantity given above.

Of course, this is a very ugly way. The symmetry makes me think there should be a more elegant way, but that obviously eludes me.

Answer 3

Thanks to the hint by Emilio Novati, I solved the question as follows:

Solution: According to the shape above, I need to find $S$= The area of sector $KAJ$, and subtract two identical regions shown by $x$ from it.

$M = S - 2x$

The Area of $S$:

You can also see that $\Delta AIJ$ is an equilateral triangle because each side is equal to the radius, which is one unit, So, each angle is equal to $60^\circ$.

Based on this, $\angle KAJ = 270^\circ - (2*60^\circ) = 150^\circ \Rightarrow S = \frac{150}{360}\pi = \frac{5}{6}\pi$.

The areas of each x:

As mentioned, the area of triangle $\Delta AIJ$ is $\frac{\sqrt{3}}{4}$ since it is equilateral.

Since $\angle AIJ=60^\circ$, then the area of sector $AIJ$ is $\frac{\pi}{6}$.

$x$ = The area of sector $AIJ$ - The area of trainable $\Delta AIJ$ = $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$.

We have two of them so the answer is:

$M= S - 2x $ which leads to $$\frac{\pi + 6\sqrt{3}}{12}$$

Answer 4

Rotate the image, and place the origin of the Cartesian system at the center of the main circle. The other two circles will be centered at $(\pm 1/ \sqrt{2},-1/ \sqrt{2})$.

Obviously, it's enough to find half of the area for $x>0$.

For the green circle for $y>0$ we have:

$$y=\sqrt{1-x^2}$$

While for the right blue circle:

$$y =\sqrt{1-(x-1/ \sqrt{2})^2}-1/ \sqrt{2}$$

While polar coordinates are a good way to go, the problem can be solved as well in Cartesian.

We need to find the intersection point for $y>0$ of the two circles. Let's work with the equations:

$$y+1/ \sqrt{2} =\sqrt{1-x^2+\sqrt{2} x-1/2}=\sqrt{y^2+\sqrt{2} x-1/2}$$

Squaring:

$$y^2+\sqrt{2}y+1/2 =y^2+\sqrt{2} x-1/2$$

$$\sqrt{2}y =\sqrt{2} x-1$$

$$y =x-1/\sqrt{2}$$

From the second equation again:

$$y+1/ \sqrt{2} =\sqrt{1-y^2}$$

$$y^2+\sqrt{2}y+1/2 =1-y^2$$

$$2y^2+\sqrt{2}y-1/2 =0$$

$$y_0=\frac{\sqrt{3}-1}{2 \sqrt{2}}$$

$$x_0=\frac{\sqrt{3}+1}{2 \sqrt{2}}$$

Now all we need to do is integrate from $x=0$ to $x_0$ (subtracting the result for the blue circle from the green one):

$$S/2=\int_0^{x_0} \sqrt{1-x^2} dx-\int_0^{x_0}(\sqrt{1-(x-1/ \sqrt{2})^2}-1/ \sqrt{2}) dx=$$

$$=x_0/ \sqrt{2}+(x_0 \sqrt{1-x_0^2}+\arcsin x_0 )/2-(y_0 \sqrt{1-y_0^2}+\arcsin y_0 )/2+$$

$$+(-1/ \sqrt{2} \sqrt{1-1/2}+\arcsin (-1/ \sqrt{2}) )/2$$

$$S=\frac{\sqrt{3}+1}{2}+\frac{1}{4}+\frac{5 \pi}{12}-\frac{1}{4}-\frac{ \pi}{12}-\frac{1}{2}-\frac{ \pi}{4}$$

$$S=\frac{6 \sqrt{3}+\pi}{12}$$

Same result as Allawonder obtained.

This method is tedious (I have made liberal use of Wolfram Alpha), but we didn't need to think much.

Answer 5

Hint, by integral calculus.

Let the equation of the centered disk be $x^2+y^2\le1$, and exclude $(x-1)^2+y^2\le1$ and $x^2+(y-1)^2\le1.$

In polar coordinates (after simplification),

$$0,2\cos\theta,2\sin\theta\le\rho\le1.$$

$$A=\int_{\theta=0}^{2\pi}\int_{\rho=\min(1,\max(0,2\cos\theta,2\sin\theta))}^1\rho\,d\rho\,d\theta\\ =\frac12\int_{\theta=0}^{2\pi}\left[1-(\min(1,\max(0,2\cos\theta,2\sin\theta)))^2\right]d\theta.$$