Ellipses and Circles

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

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

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

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Note: I could not put any functions in the ellipse - circle diagram because the latex in Geogebra no longer works.