A calculus problem by Rocco Dalto

$\dfrac{1}{2} \left(\dfrac{dy}{dx}\right)^2 - \dfrac{a}{y} = 0 \implies$ $$

$\dfrac{dy}{dx} = \sqrt{\dfrac{2a}{y}} \implies$ $x = \dfrac{1}{\sqrt{2a}} * \int \sqrt{y} \: dy =$ $\dfrac{1}{3} \sqrt{\dfrac{2}{a}} * y^{\dfrac{3}{2}} + C$ $$

$x(0) = 0 \implies x = \dfrac{1}{3} \sqrt{\dfrac{2}{a}} * y^{\dfrac{3}{2}}$ $$

and $$

$\dfrac{dx}{dy} = \dfrac{1}{\sqrt{2a}} * \sqrt{y}.$

Rotating the curve $x = \dfrac{1}{3} \sqrt{\dfrac{2}{a}} * y^{\dfrac{3}{2}}$ about the x axis from $y = 0$ to $y = 2a$ , the surface area $A = \sqrt{\dfrac{2}{a}}\pi * \int_{0}^{2a} y \sqrt{2a + y} \: dy$ $$

Let $u = y$ and $dv = \sqrt{2a + y} \: dy \implies du = dy$ and $v = \dfrac{2}{3} * (2a + y)^{\dfrac{3}{2}} \implies$ $$

$A = \dfrac{2}{3}\sqrt{\dfrac{2}{a}}\pi * ((2a + y)^{\dfrac{3}{2}} * y - \dfrac{2}{5} * (2a + y)^{\dfrac{5}{2}})|_{0}^{2a} =$ $$

$= \dfrac{2\sqrt{2}}{3}\pi * a^2 * (\dfrac{16 + 8\sqrt{2}}{5}) =$ $$

$\dfrac{32}{15} \sqrt{2}\pi * a^2 + \dfrac{32}{15}\pi * a^2 =$ $\dfrac{32}{15} * (\sqrt{2} + 1)\pi * a^2 = \dfrac{m}{n} * (\sqrt{q} + r)\pi * a^2$ $\implies m + n + p + r = 50.$