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 $u = \sqrt{2a} sinh(\theta) \implies du = \sqrt{2a} cosh(\theta) \: d\theta \implies$ $$

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

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

$x = u^2 = 2a * sinh^2(\theta) = a * (cosh(2\theta) - 1).$

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

$x = a * (sinh(\theta) - \theta)$ , $y = a * (cosh(\theta) - 1)$

Now we rotate the curve about the x - axis from $\theta = 0$ to $\theta = \ln(2)$ . $$

The Volume $V = \pi \int_{0}^{\ln(2)} (y(\theta))^2 \dfrac{dx}{d\theta} \: d\theta$ $$

$a^3\pi * \int_{0}^{\ln(2)} (cosh(\theta) - 1)^3 \:d\theta =$ $$

$a^3\pi * \int_{0}^{\ln(2)} (cosh^3(\theta) - 3cosh^2(\theta) + 3cosh(\theta) - 1) \: d\theta =$ $$

$a^3\pi * \int_{0}^{\ln(2)} ((1 + sinh^2(\theta)) cosh(\theta) - \dfrac{3}{2} (cosh(2\theta) + 1) + 3 cosh(\theta) - 1) \: d\theta =$ $$

$a^3\pi * \int_{0}^{\ln(2)} (4 cosh(\theta) + sinh^2(\theta) * cosh(\theta) -\dfrac{3}{2} cosh(\theta) - \dfrac{5}{2}) \: d\theta =$ $$

$a^3\pi * (4 sinh(\theta) + \dfrac{sinh^3(\theta)}{3} - \dfrac{3}{4} sinh(2\theta) - \dfrac{5}{2})|_{0}^{\ln(2)} =$ $$

$a^3\pi * (3 + \dfrac{9}{64} - \dfrac{45}{32} - \dfrac{5}{2} \ln(2)) =$ $$

$a^3\pi * (\dfrac{111}{64} - \dfrac{5}{2} \ln(2)) = a^3\pi * (\dfrac{m}{n} - \dfrac{p}{q} * \ln(q))$ $$

$\implies m + n + p + q = 182.$