What is the Volume ?

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 $x$ axis is $V_{1} =\dfrac{\pi}{4}\displaystyle\int_{0}^{\dfrac{8}{5}} x^2 dx =$ $\dfrac{x^3}{3}|_ {0}^{\dfrac{8}{5}} = \boxed{\dfrac{128}{375}\pi}$ .

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

$V_{2} = \pi\displaystyle\int_{\dfrac{8}{5}}^{4} (4 - \sqrt{16 - (x - 4)^2})^2 dx =$

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

$\displaystyle\int_{\dfrac{8}{5}}^{4} (32 - (x - 4)^2) dx = 32x - \dfrac{(x - 4)^3}{3}|_{\dfrac{8}{5}}^{4}$ $= \dfrac{9024}{125}$

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

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

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

$\implies \displaystyle\int_{\dfrac{8}{5}}^{4} (\sqrt{16 - (x - 4)^2}) dx =$ $8\arcsin(\dfrac{3}{5}) + \dfrac{96}{25}$

$\implies V_{2} = \pi(\dfrac{9024}{125} - 64\arcsin(\dfrac{3}{5}) - \dfrac{768}{25}) =$ $\boxed{\pi(\dfrac{5184}{125} - 64\arcsin(\dfrac{3}{5}))}$

$\implies V = V_{1} + V_{2} = \pi(\dfrac{128}{375} + \dfrac{5184}{125} - 64\arcsin(\dfrac{3}{5})) =$ $64\pi(\dfrac{49}{75} - \arcsin(\dfrac{3}{5})) =$

$2^{2 * 3}(\dfrac{7^2}{3 * 5^2} - \arcsin(\dfrac{3}{5}))\pi =$ $a^{a * b}(\dfrac{c^a}{b * d^{a}} - \arcsin(\dfrac{b}{d}))\pi \implies a + b + c + d = \boxed{17}$