What is the area of shaded region which is lies between outer and inner circle.

Question

There is a outer circle with radius 2r and another inner circle with radius r whose center is the middle of big circle.As depicted in the following figure.

Foo graph Image

There is a sector of 120 degree in inner circle which leads to shaded part between outer and inner circle. What is the area of shaded circle in terms of r ?

i.e. What is the value of K ?

Answer 1

WLOG let $r=1$.

Let us use a coordinate system centered on the large circle. Then the oblique line on the right has equation $x=\sqrt3(y-1)$ and meets the circle $x^2+y^2=4$.

Solving the quadratic equation, the intersection point is $(\dfrac{\sqrt{39}-\sqrt3}4,\dfrac{\sqrt{13}+3}4)$.

Hence, the aperture angle of the large sector

$$\theta=2\arctan\left(\frac xy\right)=1.19873\cdots \text{rad}=68.6821\cdots°.$$

The requested area is the difference between the large and the small sector and two triangles having base $1$ and height $x$.

$$K=\frac\theta2\,2^2-\frac\pi31^2-\frac22x=4\arctan\left(\frac{\sqrt{39}-\sqrt3}{\sqrt{13}+3}\right)-\frac\pi3-\dfrac{\sqrt{39}-\sqrt3}4=0.222026268\cdots.$$

Answer 2

something like this: $$ 2r^2\left(\int_0^{\frac{\sqrt{3}}2} \left(\sqrt{4-x^2} - 1 - \sqrt{1-x^2} \right)dx + \int_{\frac{\sqrt{3}}2}^{\frac{\sqrt{13}-1}4} \left(\sqrt{4-x^2} - 1-\frac{\sqrt{3}x}2 \right)dx \right) $$

Answer 3

Draw a line from the centre of the larger circle to the left-most corner of the shaded region. This has length $2r$. We then have an obtuse angled triangle with largest angle $\frac {2\pi}{3}$, due to symmetry. Let $\theta$ be the angle between the radii.

Using the sine rule, we have $$\sin\frac {2\pi}{3}=2\sin(\frac{\pi}{3}-\theta)$$ Expanding and rearranging, we have $$\sqrt3\cos\theta-\sin\theta=\frac {\sqrt3}{2}$$ Now a compound angle transformation gives $$2\cos(\theta+\frac {\pi}{3})=\frac {\sqrt3}{2}$$ Thus $$\theta=-\frac {\pi}{6}+\arccos(\frac {\sqrt3}{4})$$ Whereupon the area can also be written as: $$\frac 12(2r)^22\theta-\frac 13\pi r^2-2(\frac12)(2r)(r)\sin\theta$$ which simplifies to $$r^2\left(-\pi+4\arccos(\frac{\sqrt3}{4})+\frac{\sqrt3}{4}-\frac{\sqrt{39}}{4}\right)$$ This is the same as the answer already found, but in a slightly neater form.

Answer 4

You can also easily solve the problem by rotating the figure by $90^o$ clockwise so that the dotted line coincides with the x-axis & center of outer circle at the origin. Thus, equation of the outer circle: $x^2+y^2=4r^2$ & the equation radius above x-axis: $y-0=\tan60^o(x-r)$ or $y=\sqrt{3}(x-r)$ solving the equation for intersection point of line & outer circle as follows $$x^2+(\sqrt{3}(x-r))^2=4r^2 => 4x^2-6rx-r^2$$ now, solving for x-coordinate, we get $$x=\frac{6r\pm\sqrt{36r^2+16r^2}}{8}=\frac{r(3\pm \sqrt{13})}{4}$$ Consider a right triangle above x-axis obtained by joining the intersection point (in I-quadrant) to the center of outer circle, we get semi-aperture angle $\alpha$ at the center of outer circle as follows $$\alpha=\cos^{-1}\left(\frac{\frac{r(3+\sqrt{13})}{4}}{2r}\right)=\cos^{-1}\left(\frac{3+\sqrt{13}}{8}\right)=\sin^{-1}\left(\frac{\sqrt{42-6\sqrt{13}}}{8}\right)$$

Hence, the shaded area is $$=\frac{1}{2}(2\alpha)(2r)^2-2\left(\frac{1}{2}(r)(2r)(\sin\alpha)\right)-\frac{1}{2}\left(\frac{2\pi}{3}\right)(r^2)=r^2\left(4\cos^{-1}\left(\frac{3+\sqrt{13}}{8}\right)-\frac{\pi}{3}-\left(\frac{\sqrt{42-6\sqrt{13}}}{4}\right)\right)$$ Hence, we get $$K=4\cos^{-1}\left(\frac{3+\sqrt{13}}{8}\right)-\frac{\pi}{3}-\frac{\sqrt{42-6\sqrt{13}}}{4}\approx0.222026268$$