1. **State the problem:** We want to evaluate the integral $$\int_0^r \frac{1}{\pi r^2} \arccos\left(\frac{r x - x^2 - \sqrt{\frac{2 r^3}{x} - \frac{3}{2} x r^2 - x^2 - \frac{2 r^3}{x} + x^4}}{r^2}\right) dx.$$\n\n2. **Understand the integral:** The integrand is $$\frac{1}{\pi r^2} \arccos(\cdots)$$ where the argument of arccos is a complicated expression involving $x$ and $r$. The integral is from $0$ to $r$.\n\n3. **Simplify the expression inside the square root:** Inside the square root we have $$\frac{2 r^3}{x} - \frac{3}{2} x r^2 - x^2 - \frac{2 r^3}{x} + x^4.$$ Notice that $$\frac{2 r^3}{x} - \frac{2 r^3}{x} = 0,$$ so these terms cancel out. The expression simplifies to $$- \frac{3}{2} x r^2 - x^2 + x^4.$$\n\n4. **Rewrite the argument of arccos:** Now the argument is $$\frac{r x - x^2 - \sqrt{- \frac{3}{2} x r^2 - x^2 + x^4}}{r^2}.$$\n\n5. **Check the domain and behavior:** The square root must be real, so the expression inside must be non-negative. This is a complicated expression, but since the problem is given as is, we proceed formally.\n\n6. **Integral evaluation:** Given the complexity, this integral likely represents a geometric or probability problem involving $r$. Without further simplification or context, the integral cannot be expressed in elementary closed form easily.\n\n7. **Summary:** The integral is $$\int_0^r \frac{1}{\pi r^2} \arccos\left(\frac{r x - x^2 - \sqrt{- \frac{3}{2} x r^2 - x^2 + x^4}}{r^2}\right) dx.$$\n\nSince no further simplification is straightforward, the integral is expressed as above and can be evaluated numerically for given $r$.