The problem asks to evaluate the definite integral $2\pi \int_{0}^{50} (12 + \frac{\sqrt{y^3(50-y)}}{40}) \sqrt{1 + (\frac{150y^2 - 4y}{80\sqrt{y^3(50-y)}})^2} dy$.

1. Problem Description

The problem asks to evaluate the definite integral

$2\pi \int_{0}^{50} (12 + \frac{\sqrt{y^3(50-y)}}{40}) \sqrt{1 + (\frac{150y^2 - 4y}{80\sqrt{y^3(50-y)}})^2} dy$.

2. Solution Steps

Let $f(y) = 12 + \frac{\sqrt{y^3(50-y)}}{40}$.

We need to find $f'(y)$ to simplify the expression under the square root in the integral.

$f(y) = 12 + \frac{(y^3(50-y))^{1/2}}{40} = 12 + \frac{(50y^3 - y^4)^{1/2}}{40}$

$f'(y) = \frac{1}{40} \cdot \frac{1}{2} (50y^3 - y^4)^{-1/2} (150y^2 - 4y^3) = \frac{150y^2 - 4y^3}{80\sqrt{50y^3 - y^4}} = \frac{150y^2 - 4y^3}{80\sqrt{y^3(50-y)}}$

Thus the integral becomes:

$2\pi \int_{0}^{50} f(y) \sqrt{1 + (f'(y))^2} dy = 2\pi \int_{0}^{50} (12 + \frac{\sqrt{y^3(50-y)}}{40}) \sqrt{1 + (\frac{150y^2 - 4y^3}{80\sqrt{y^3(50-y)}})^2} dy$

This looks like the surface area of revolution around the x-axis for the function $f(y)$ from $y=0$ to $y=50$.

The formula for the surface area of revolution around the x-axis is given by

$A = 2\pi \int_a^b f(y) \sqrt{1 + (f'(y))^2} dy$

where $f(y) = 12 + \frac{\sqrt{y^3(50-y)}}{40}$.

We can express the integral as

$2\pi \int_0^{50} (12 + \frac{\sqrt{y^3(50-y)}}{40}) \sqrt{1 + (\frac{150y^2 - 4y^3}{80\sqrt{y^3(50-y)}})^2} dy = 2\pi \int_0^{50} (12 + \frac{\sqrt{y^3(50-y)}}{40}) \sqrt{1 + \frac{(150y^2 - 4y^3)^2}{6400y^3(50-y)}} dy$

$= 2\pi \int_0^{50} (12 + \frac{\sqrt{y^3(50-y)}}{40}) \sqrt{\frac{6400y^3(50-y) + (150y^2 - 4y^3)^2}{6400y^3(50-y)}} dy$

$= 2\pi \int_0^{50} (12 + \frac{\sqrt{y^3(50-y)}}{40}) \frac{\sqrt{6400y^3(50-y) + (150y^2 - 4y^3)^2}}{80\sqrt{y^3(50-y)}} dy$

$= 2\pi \int_0^{50} \frac{1}{40}(480 + \sqrt{y^3(50-y)}) \frac{\sqrt{320000y^3 - 6400y^4 + 22500y^4 - 1200y^5 + 16y^6}}{80\sqrt{y^3(50-y)}} dy$

This integral is too complicated to solve analytically.

Based on the information available, I cannot simplify the integral to a solvable form. Therefore, I can't compute the final answer.

However, the expression resembles the formula for the surface area obtained by revolving a curve defined by $x = g(y)$ around the x-axis. In this case, the surface area formula is $2\pi \int_a^b f(y) \sqrt{1 + (f'(y))^2} dy$.

Therefore the integral can be seen as the surface area of the curve.

3. Final Answer

I am unable to provide a final numerical answer for this problem due to its complexity.