b. $$ \int_{0}^{6} \int_{x / 3}^{x} x \sqrt{y^{3}+1} d y d $$

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Alright, I need to evaluate the double integral $$ \int_{0}^{6} \int_{x / 3}^{x} x \sqrt{y^{3}+1} \, dy \, dx $$. Let's break this down step by step to ensure I understand and solve it correctly.

First, I observe that the integral is with respect to $$ y $$ from $$ y = \frac{x}{3} $$ to $$ y = x $$, and then with respect to $$ x $$ from $$ x = 0 $$ to $$ x = 6 $$. The integrand is $$ x \sqrt{y^{3}+1} $$.

Maybe I should consider changing the order of integration to simplify the problem. Currently, the limits for $$ y $$ depend on $$ x $$, which might make the inner integral more complicated. Let's see if switching the order could help.

To change the order of integration, I need to express the region of integration in terms of $$ y $$ and $$ x $$. The original limits are:
- For $$ y $$: $$ \frac{x}{3} \leq y \leq x $$
- For $$ x $$: $$ 0 \leq x \leq 6 $$

From $$ \frac{x}{3} \leq y \leq x $$, I can solve for $$ x $$ in terms of $$ y $$:
- $$ y \leq x $$ implies $$ x \geq y $$
- $$ \frac{x}{3} \leq y $$ implies $$ x \leq 3y $$

So, the new limits for $$ x $$ in terms of $$ y $$ would be $$ y \leq x \leq 3y $$. The limits for $$ y $$ would then be from $$ 0 $$ to $$ 6 $$, since $$ x $$ ranges from $$ 0 $$ to $$ 6 $$, and $$ y $$ is bounded by $$ x $$.

Now, the integral becomes:
$$
\int_{0}^{6} \int_{y}^{3y} x \sqrt{y^{3}+1} \, dx \, dy
$$

This seems a bit cleaner. Now, I can integrate with respect to $$ x $$ first:
$$
\int_{y}^{3y} x \sqrt{y^{3}+1} \, dx = \sqrt{y^{3}+1} \int_{y}^{3y} x \, dx
$$
$$
= \sqrt{y^{3}+1} \left[ \frac{x^{2}}{2} \right]_{y}^{3y}
$$
$$
= \sqrt{y^{3}+1} \left( \frac{(3y)^{2}}{2} - \frac{y^{2}}{2} \right)
$$
$$
= \sqrt{y^{3}+1} \left( \frac{9y^{2}}{2} - \frac{y^{2}}{2} \right)
$$
$$
= \sqrt{y^{3}+1} \left( \frac{8y^{2}}{2} \right)
$$
$$
= 4y^{2} \sqrt{y^{3}+1}
$$

Now, the integral simplifies to:
$$
\int_{0}^{6} 4y^{2} \sqrt{y^{3}+1} \, dy
$$

To solve this, I can use substitution. Let $$ u = y^{3} + 1 $$, then $$ du = 3y^{2} dy $$, which implies $$ y^{2} dy = \frac{du}{3} $$.

Substituting into the integral:
$$
4 \int_{0}^{6} y^{2} \sqrt{y^{3}+1} \, dy = 4 \int_{u(0)}^{u(6)} \sqrt{u} \cdot \frac{du}{3}
$$
$$
= \frac{4}{3} \int_{1}^{217} u^{\frac{1}{2}} \, du
$$
$$
= \frac{4}{3} \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_{1}^{217}
$$
$$
= \frac{8}{9} \left( 217^{\frac{3}{2}} - 1^{\frac{3}{2}} \right)
$$
$$
= \frac{8}{9} \left( 217^{\frac{3}{2}} - 1 \right)
$$

Calculating $$ 217^{\frac{3}{2}} $$:
$$
217^{\frac{3}{2}} = (217)^{1.5} = 217 \times \sqrt{217}
$$
$$
\sqrt{217} \approx 14.73
$$
$$
217 \times 14.73 \approx 3190.41
$$

So,
$$
\frac{8}{9} \left( 3190.41 - 1 \right) = \frac{8}{9} \times 3189.41 \approx 2810.37
$$

Therefore, the value of the integral is approximately **2810.37**.

El valor de la integral doble es aproximadamente **2810.37**.

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