A calculus problem by Rocco Dalto

$\dfrac{1}{2} * (\dfrac{dy}{dx})^2 - \dfrac{a}{y} = -\dfrac{1}{2} \implies$ $$

$\dfrac{dy}{dx} = \sqrt{\dfrac{2a - y}{y}} \implies x = \int \sqrt{\dfrac{y}{2a - y}} \: dy$ $$

Let $u^2 = y \implies 2u \: du = dy \implies$ $$

$x = 2 * \int \dfrac{u^2}{\sqrt{2a - u^2}} \: du$ $$

Let $u = \sqrt{2a} sin\theta \implies du = \sqrt{2a} cos\theta \: d\theta \implies$ $$

$x = 2a * \int (1 - cos(2\theta)) \: d\theta = 2a * (\theta - \dfrac{1}{2} sin(2\theta)) = a * (2\theta - sin(2\theta)) + C.$ $$

$x(\theta) = 0 \implies C = 0 \implies X = a * (2\theta - sin(2\theta))$

$x = u^2 = 2a * sin^2\theta = a * (1 - cos(2\theta)).$

Replacing $2\theta$ by $\theta$ we obtain: $$

$x = a * (\theta - sin\theta)$ , $y = a * (1 - cos\theta)$

Now we rotate the cycloid about the x - axis from $\theta = 0$ to $\theta = \pi$ . $$

$\dfrac{dx}{d\theta} = a * (1 - cos\theta)$ and $\dfrac{dy}{d\theta} = a * sin\theta \implies$ $$

Area $A = 2\pi * a^2 * \int_{0}^{\pi} (1 - cos\theta) \sqrt{2(1 - cos\theta)} \: d\theta$ = $$

$8\pi * a^2 * \int_{0}^{\pi} sin^2(\dfrac{\theta}{2}) * sin(\dfrac{\theta}{2}) \: d\theta$ $$

$8\pi * a^2 * \int_{0}^{\pi} (1 - cos^2(\dfrac{\theta}{2})) * sin(\dfrac{\theta}{2}) \: d\theta =$ $$

$8\pi * a^2 * \int_{0}^{\pi} sin(\dfrac{\theta}{2}) - cos^2(\dfrac{\theta}{2}) * sin(\dfrac{\theta}{2}) \: d\theta =$ $$

$8\pi * a^2 * (-2 cos(\dfrac{\theta}{2}) + \dfrac{2}{3} * cos^3(\dfrac{\theta}{2}))|_ {0}^{\pi} = \dfrac{32}{3}\pi * a^2 = \dfrac{m}{n}\pi * a^2 \implies$ $$

$m + n = 35.$ $$

Note: Replacing $2\theta$ by $\theta$ does not change the surface area.