Circles and Ellipses!

We can find the area of a segment of an ellipse without integration. This is because we can consider an ellipse is a circle tilted at an angle. For this problem we can consider the circle of radius $a$ is tilted along the $x$ -axis. If the area of the segment of the circle is $A_\bigcirc$ , then the area of the segment of the ellipse is $A = \frac ba A_\bigcirc$ , where $a$ and $b$ are the major semiaxis and minor semiaxis respectively, or:

$A(x,y) = \frac ba \left(a^2 \tan^{-1} \left(\frac {bx}{ay} \right) - \frac ab xy\right) = ab \tan^{-1} \left(\frac {bx}{ay} \right) -xy$

Since $a=3$ and $b=1$ , the equation of the ellipse is $\dfrac {x^2}9 + y^2 = 1$ . For the pink area, we have $y = \dfrac 12$ , $\implies \dfrac {x^2}9 + \dfrac 14 = 1 \implies x = \dfrac {3\sqrt 3}2$ and:

$\begin{aligned} A_{\rm pink} & = 2 A \left(\frac {3\sqrt 3}2, \frac 12 \right) = 6 \tan^{-1} (\sqrt 3) - 2 \cdot \frac {3\sqrt 3}2 \cdot \frac 12 = 2 \pi - \frac {3\sqrt 3}2 \approx 3.685109096 \end{aligned}$

For the blue area, we get $x$ and $y$ from $(1): \ \dfrac {x^2}9 + y^2 = 1$ and $(2): \ x^2 + (y-1)^2 = 1$ . Then $9(1) - (2): \ 8y^2 + 2y - 9 = 0 \implies y = \dfrac {\sqrt{73}-1}8$ and $x = \dfrac {3\sqrt{2\sqrt{73}-10}}8$

$\begin{aligned} A_{\rm blue} & = 2(\sin^{-1}x - x(1-y) + A(x,y)) \\ & = 2 \left(\sin^{-1} x -x(1-y) + 3 \tan^{-1}\left(\frac x{3y} \right) - xy \right) \approx 3.06635633492 \end{aligned}$

Therefore the shaded area $A_{\rm shaded} = A_{\rm pink} + A_{\rm blue} \approx \boxed{6.75}$ .

For Region $2$ :

I chose the ellipses $(\dfrac{x^2}{9} + y^2 = 1$ above the line $y = 0)$ and $(\dfrac{x^2}{9} + (y - 1)^2 = 1$ below the line $y = 1)$ .

Solving for $y$ in both ellipses we want:

$f(x) = \dfrac{1}{3}\sqrt{9 - x^2}$ and $g(x) = 1 - \dfrac{1}{3}\sqrt{9 - x^2}$ .

$f(x) = g(x) \implies x = \pm\dfrac{3\sqrt{3}}{2} \implies$ the area $A_{2} = \displaystyle\int_{-\frac{3\sqrt{3}}{2}}^{\frac{3\sqrt{3}}{2}} f(x) - g(x) dx = \dfrac{2}{3}\displaystyle\int_{-\frac{3\sqrt{3}}{2}}^{\frac{3\sqrt{3}}{2}} \sqrt{9 - x^2} dx - 3\sqrt{3}$

For $I = \dfrac{2}{3}\displaystyle\int_{-\frac{3\sqrt{3}}{2}}^{\frac{3\sqrt{3}}{2}} \sqrt{9 - x^2} dx$

Let $x = 3\sin(\theta) \implies dx = 3\cos(\theta) \implies I = 6\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \cos^2(\theta) d\theta = 3\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 1 + \cos(2\theta) d\theta =$ $3(\theta + \dfrac{1}{2}\sin(2\theta))|_{-\frac{\pi}{3}}^{\frac{\pi}{3}} = 2\pi + \dfrac{3\sqrt{3}}{2} \implies A_{2} = 2\pi - \dfrac{3\sqrt{3}}{2} \approx \boxed{3.6851091}$

I chose the circles $x^2 + (y - 2)^2 = 1$ and $x^2 + (y + 1)^2 = 1$ .

Note by symmetry $A_{3} = A_{1}$ .

For Region $1$ :

Using the portion of the ellipse $(m(x) = 1 + \dfrac{1}{3}\sqrt{9 - x^2}$ above the line $y = 1)$ and the circle $(h(x) = 2 - \sqrt{1 - x^2}$ below the line $y = 2)$ and setting $h(x) = m(x) \implies$

$\sqrt{9 - x^2} + 3\sqrt{1 - x^2} = 3 \implies -10x^2 + 9 + 6\sqrt{9 - 10x^2 + x^4} = 0 \implies 324 - 360x^2 + 36x^4 = 100x^4 - 180x^2 + 81$

$\implies 64x^4 + 180x^2 - 243 = 0 \implies x^2 = \dfrac{9}{32}(-5 \pm \sqrt{73}) \implies x_{1},x_{2} = \pm\dfrac{3\sqrt{2}\sqrt{\sqrt{73} - 5}}{8}$ dropping the two imaginary roots.

$A_{1} = \displaystyle\int_{x_{1}}^{x_{2}} \dfrac{1}{3}\sqrt{9 - x^2} + \sqrt{1 - x^2} \:\ dx - \dfrac{3\sqrt{2}\sqrt{\sqrt{73} - 5}}{4}$ .

Let $x = \sin(\theta) \implies dx = \cos(\theta) d\theta \implies \displaystyle\int \sqrt{1- x^2} dx = \dfrac{1}{2}(\theta + \sin(\theta)\cos(\theta)) = \dfrac{1}{2}(\arcsin(x) + x\sqrt{1 - x^2})$

Let $x = \sin(\lambda) \implies dx = \cos(\lambda) d\lambda \implies \dfrac{1}{3}\displaystyle\int \sqrt{9- x^2} dx = \dfrac{3}{2}(\lambda + \sin(\lambda)\cos(\lambda)) = \dfrac{3}{2}(\arcsin(\dfrac{x}{3}) + \dfrac{x\sqrt{9 - x^2}}{3})$

$\implies I = \displaystyle\int_{x_{1}}^{x_{2}} \dfrac{1}{3}\sqrt{9 - x^2} + \sqrt{1 - x^2} \:\ dx = \dfrac{1}{2}(\arcsin(x) + 3\arcsin(\dfrac{x}{3}) + x(\sqrt{1 - x^2} + \dfrac{\sqrt{9 - x^2}}{3}))|_{x_{1}}^{x_{2}} =$

$\arcsin(x_{2}) + 3\arcsin(\dfrac{x_{2}}{3}) + x_{2}(\sqrt{1 - x_{2}^2} + \dfrac{\sqrt{9 - x_{2}^2}}{3})$

Using $x_{2} \approx 0.9983742 \implies I = 3.5296021 \implies A_{1} = 3.5296021 - 2 * x_{2} = 3.5296021 - 1.9967484 = \boxed{1.5328537}$

$\implies A_{Total} = A_{2} + 2A_{1} = 3.6851091 + 2(1.5328537) = \boxed{6.7508165}$ .

The answer that is counted as correct (6.7508165) and the answer of Chew-Seong (6.751712589) are both close, but differ from the 3rd decimal place. Below I derive yet a slighly different value analytically and confirm it numerically to the 10th decimal place.

Consider a unit circle centered at the origin, and a point $(x_0,y_0)=(\cos{φ_0}, \sin{φ_0})$ on it. The area within the circle above the line $y=y_0$ is found as $A =\int_{y_0}^1{2x}dy=\int_{φ_0}^{\frac{π}{2}}{2\cosφ} d\sin φ =\int_{φ_0}^π{2\cos^2φ}dφ =\int_{φ_0}^{π}{1+\cos 2φ}dφ =φ+\frac{1}{2}\sin 2φ \bigg \rvert_{φ_0}^\frac{π}{2}= \frac{π}{2}+\frac{\sin 2φ_0}{2} -φ_0$ By the double angle formula $\sin 2φ_0 =2\sin φ_0\cos φ_0=2y_0x_0$ this becomes $A=\frac{π}{2} -x_0y_0 -\arcsin y_0$

By scaling this to ellipses of any size, we get a more general function for the area above the line $y$ :

$f(a,b,x,y)=\frac{πab}{2}-ab\arcsin (\frac{y}{b}) - xy$

The pink area is found by twice such an area, setting $a=3,b=1, x=x_0=3\cdot\frac12\sqrt{3}, y=y_0=\frac{1}{2}$ $A_{pink}=2f(1,1,\frac{3}{2}\sqrt{3},\frac{1}{2})=3π-6\arcsin \frac{1}{2} -\frac{3}{2}\sqrt{3} = 2π-\frac{3}{2}\sqrt{3} \approx3.685...$

To find the blue area, consider a unit circle centered at $(0,1)$ and an ellipse centered at the origin. If we connect the two points of intersection by a line, we can calculate the parts bounded by the circle and the ellipse separately. To do this we need the coordinates of an intersection point $x_0$ and $y_0$ . Substitute the equation for the circle ( $x^2=1-(y-1)^2$ ) into the equation for the ellipse ( $x^2+9y^2=9$ ) to get $8y^2+2y-9=0$

We get $y_0=\frac{\sqrt{73}-1}{8}\approx 0.9430005, x_0=\sqrt{1-(y_0-1)^2}=\sqrt{\frac{9\sqrt{73}-45}{32}}\approx 0.9983742$

The $y_0$ we found is valid to get the area above $y=y_0$ in the ellipse, but since the circle is centered at (0,1), we need to plug $1-y_0$ into the formula to get the area below the line $y=y_0$ .

Total area is now given by $A=A_{pink}+A_{blue, circle}+A_{blue, ellipse}$ $=2f(1,1,\frac{3}{2}\sqrt{3}, \frac{1}{2})+2f(1,1,x_0,1-y_0)+2f(3,1,x_0,y_0)=\boxed{6.7514654307...}$

This is numerically confirmed by the code below, which outputs 6.7514654307...

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import math

def area():
    A=0
    x=-3
    dx=0.000001
    while x<3:
        h=0
        y_el=math.sqrt(1-x**2/9)
        h1=y_el-0.5
        if h1>0:
            h += h1
        if x>=-1 and x<=1:
            y_cr = math.sqrt(1-x**2)
            h2=y_cr + y_el -1
            if h2>0:
                h += h2
        A+=dx*h
        x+=dx
    return 2*A

print(area())