It's all Volumes

Using the diagram above the equation of line $AC$ is $y = \dfrac{1}{2}x$ or $x = 2y$ and for the circle centered at (4,4) with have $(x - 4)^2 + (y - 4)^2 = 16$

$x = 2y \implies (2y - 4)^2 + (y - 4)^2 = 16 \implies 5y^2 - 24y + 16 = 0 \implies$

$y = \dfrac{4}{5}, y = 4$ Since we already have $(8,4) \implies P(\dfrac{8}{5},\dfrac{4}{5})$

So the volume of the pink region when revolved about the $y$ axis is the volume of the cone
$V_{1} = \dfrac{1}{3}(\dfrac{8}{5})^2(\dfrac{4}{5})\pi = \boxed{\dfrac{256}{375}\pi}$ .

$(x - 4)^2 + (y - 4)^2 = 16 \implies x = 4 - \sqrt{16 - (y - 4)^2}$ which is the portion of the circle needed.

$V_{2} = \pi\displaystyle\int_{\dfrac{4}{5}}^{4} (4 - \sqrt{16 - (y - 4)^2})^2 dy =$

$\pi\displaystyle\int_{\dfrac{4}{5}}^{4} (32 - (y - 4)^2 - 8\sqrt{16 - (y - 4)^2}) dy$

$\displaystyle\int_{\dfrac{4}{5}}^{4} (32 - (y - 4)^2) dy = 32y - \dfrac{(y - 4)^3}{3}|_{\dfrac{4}{5}}^{4}$ $= \dfrac{34304}{375}$

For $I = \displaystyle\int \sqrt{16 - (y - 4)^2} dy$

Let $y - 4 = 4\sin(\theta) \implies dy = 4\cos(\theta) \implies I = 8\displaystyle\int (1 + \cos(2\theta)) d\theta$

$= 8(\theta + \sin(\theta)\cos(\theta)) = 8(\arcsin(\dfrac{y - 4}{4}) + \dfrac{(y - 4)\sqrt{16 - (y - 4)^2}}{16})$

$\implies \displaystyle\int_{\dfrac{4}{5}}^{4} (\sqrt{16 - (y - 4)^2}) dy =$ $8\arcsin(\dfrac{4}{5}) + \dfrac{96}{25}$

$\implies V_{2} = \pi(\dfrac{34304}{375} - 64\arcsin(\dfrac{4}{5}) - \dfrac{768}{25}) =$ $\boxed{\pi(\dfrac{22784}{375} - 64\arcsin(\dfrac{4}{5}))}$

$\implies V = V_{1} + V_{2} = \pi(\dfrac{256}{375} + \dfrac{22784}{375} - 64\arcsin(\dfrac{4}{5})) =$ $64\pi(\dfrac{24}{25} - \arcsin(\dfrac{4}{5})) =$

$2^{2 * 3}(\dfrac{2^3 * 3}{5^2} - \arcsin(\dfrac{2^2}{5}))\pi = a^{a * b}(\dfrac{a^{b} * b}{c^{a}} - \arcsin(\dfrac{a^a}{c}))\pi \implies a + b + c = \boxed{10}$ .