Viva La Revolucion!

The required integral is:

$I = \pi \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} {y^2dx} = \pi \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} {\left( \dfrac {1}{\sqrt{x}} - \dfrac{1}{\sqrt{x+1}} \right)^2 dx}$

$\quad = \pi \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} {\left( \dfrac {1}{x} - \dfrac{2}{\sqrt{x(x+1)}} + \dfrac {1}{x+1} \right) dx}$

$\quad = \pi \left( \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} { \dfrac {dx}{x}} -2 \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} { \dfrac {dx}{\sqrt{x(x+1)}}} + \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} { \dfrac {dx}{x+1}} \right)$

$\quad = \pi (I_1-2I_2 + I_3)$

$I_1 = \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} { \dfrac {dx}{x}} =\left[ \ln { x } \right] _{\frac{1}{8}} ^{\frac{9}{16}} = \ln {(\frac{9}{16})} - \ln {(\frac{1}{8})} = \ln {(\frac{9}{16}\times \frac{8}{1})} = \ln {(\frac {9}{2})}$

Similarly, $I_3 = \ln {(\frac {25}{18} )}$

$I_2 = \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} { \dfrac {dx}{\sqrt{x(x+1)}}} = \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} { \dfrac {dx}{\sqrt{(x+\frac {1}{2})^2-\frac {1}{4}}}} = \displaystyle \int _{\frac{1}{8}} ^{\frac{9}{16}} { \dfrac {2\space dx}{\sqrt{(2x+1)^2-1}}}$

Now substituting $2x+1 = \sec {\theta}$ , then:

$I_2 = \displaystyle \int _{\sec ^{-1} {\frac{5}{4}}} ^{\sec ^{-1} {\frac{17}{8}}} { \dfrac {2(\frac{1}{2}) \sec{\theta} \tan {\theta} \space d\theta}{\sqrt{\sec^2 {\theta}-1}}} = \displaystyle \int _{\sec ^{-1} {\frac{5}{4}}} ^{\sec ^{-1} {\frac{17}{8}}} {\sec {\theta} \space d\theta}$

$\quad = \ln {| \sec {\theta} + \tan {\theta} |} _{\sec ^{-1} {\frac{5}{4}}} ^{\sec ^{-1} {\frac{17}{8}}} = \ln {( \frac {17}{8} + \frac {15}{8})} - \ln {(\frac {5}{4} + \frac {3}{4} )}$

$\quad = \ln {4} - \ln {2} = \ln {2}$

$\Rightarrow I = \pi (I_1-2I_2+I_3) = \pi \left( \ln {(\frac {9}{2})} -2 \ln{2} + \ln {(\frac {25}{18} )} \right)$

$\quad \quad = \pi \ln {(\frac {9}{2} \times \frac {1}{2^2} \times \frac {25}{18})} = \pi \ln {( \frac {25}{16})} = 2\pi \ln {(\frac {5}{4})}$

$\Rightarrow A=2$ , $B=5$ and $C=4\quad \Rightarrow A\times B \times C = \boxed{40}$

@Steven Zheng This is not a solution, but I just thought I should point out a potential ambiguity. Note that $\pi*\ln(\frac{25}{16}) = 2\pi*\ln(\frac{5}{4})$ , thus $1*25*16 = 400$ and $2*4*5 = 40$ would both be valid answers. To avoid this ambiguity, you might want to specify that $A,B,C$ are integers with $A,B$ being square-free.

Edit: Now for an actual solution .....

Using the disk method, we have that the volume of revolution is

$\displaystyle\int_{a}^{b} \pi y^{2} dx = \pi \int_{a}^{b} (\dfrac{1}{x} + \dfrac{1}{x + 1} - \dfrac{2}{\sqrt{x(x + 1)}}) dx =\pi*\ln(x(x + 1)) - \pi*R$

evaluated from $a = \frac{1}{8}$ to $b = \frac{9}{16}$ , where

$R = 2 \displaystyle\int_{a}^{b} \dfrac{dx}{\sqrt{x^{2} + x + \frac{1}{4} - \frac{1}{4}}} = 2 \int \dfrac{du}{\sqrt{u^{2} - 1}}$ ,

where I have made the substitution $u = 2x + 1$ .

Now let $u = \sec(\theta)$ , making $du = \sec(\theta)\tan(\theta) d\theta$ and $\sqrt{u^{2} - 1} = \tan(\theta)$ . We then end up with

$R = 2 \displaystyle\int \sec(\theta) d\theta = 2*\ln(\sec(\theta) + \tan(\theta)) = 2*\ln(2x + 1 + 2\sqrt{x(x + 1)})$

evaluated from $a$ to $b$ . Plugging in the values for $a$ and $b$ yields a volume of

$\pi(\ln(\frac{225}{256}) - 2\ln(4) - (\ln(\frac{9}{64}) - 2\ln(2))) = 2\pi \ln(\frac{5}{4}).$

Thus $A*B*C = 2*5*4 = \boxed{40}$ .