We are tasked with evaluating the integral

$\int_0^4 y^3 \sqrt{9 - y^3} \, dy$

in terms of the Beta function. The general strategy here will be to use a substitution that transforms the integral into a form involving the Beta function.

Step 1: Substitution

Let’s make the substitution: $y^3 = 9 \sin^2(\theta)$ which implies that $y = (9 \sin^2(\theta))^{1/3}$ and [ dy = \frac{3}{9 \sin(\theta) \cos(\theta)} .