To evaluate the given integral in terms of the Beta function, we start with the integral:

$\int_0^1 y^3 \sqrt{9 - y^3} \, dy$

To express this in terms of the Beta function, we'll use a substitution. Let $y^3 = 9 \sin^2(\theta)$, so $y = (9 \sin^2(\theta))^{1/3} = 3^{2/3} \sin^{2/3}(\theta)$.

Let's proceed with the substitution:

1. Substitute $y^3 = 9 \sin^2(\theta)$, hence $dy = \frac{2 \cdot 9 \sin(\theta) \cos(\theta)}{3y^2} d\theta = \frac{6 \sin(\theta) \cos(\theta)}{y^2} d\theta$.

2. The integral bounds change accordingly:

   • When $y = 0$, $\sin(\theta) = 0$ implying $\theta = 0$.
   • When $y = 1$, $\sin^2(\theta) = \frac{1}{9}$, so $\sin(\theta) = \frac{1}{3}$ implying $\theta = \arcsin\left(\frac{1}{3}\right)$.

Thus, the integral transforms into a form that can be related to the Beta function. However, the simplest approach to relate this to a Beta function is to set $y = 3u^{1/3}$, so that the integral becomes more straightforward:

Let $u = y^3 / 9$, hence $y = 3u^{1/3}$, and $dy = u^{-2/3} du / 3$.

Substituting into the integral:

$I = \int_0^1 (3u^{1/3})^3 \sqrt{9 - 9u} \cdot \frac{u^{-2/3}}{3} du$

Simplify:

$I = \int_0^1 9u \sqrt{9(1 - u)} du = 9 \int_0^1 u (3\sqrt{1-u}) du$

$I = 27 \int_0^1 u \sqrt{1 - u} du$

This integral is now in a form that matches the Beta function $B(x,y)$:

$I = 27 \cdot B\left(\frac{3}{2}, 2\right)$

Using the Beta function definition:

$B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt$

Thus:

$I = 27 \cdot \frac{\Gamma\left(\frac{3}{2}\right)\Gamma(2)}{\Gamma\left(\frac{7}{2}\right)}$

You can compute the Gamma functions to simplify further, but the integral is now expressed in terms of the Beta function.

Would you like more details on this process or explanation of the Beta and Gamma functions? Here are some related questions you might find interesting:

1. How is the Beta function related to the Gamma function?
2. What are common applications of the Beta function in mathematics?
3. How do we compute Beta functions with non-integer arguments?
4. What is the geometric interpretation of the Beta function?
5. How would this approach change if we had a different exponent in the integral?

Tip: The substitution method is powerful for transforming complex integrals into standard forms that match known functions like the Beta or Gamma functions.