We are tasked with calculating the following iterated integral:

$\int_0^1 \int_0^1 6xy \sqrt{x^2 + y^2} \, dy \, dx$

This represents a double integral with the inner integral over $y$ (from 0 to 1) and the outer integral over $x$ (also from 0 to 1).

Step 1: Compute the inner integral (over $y$)

The inner integral is:

$I_1(x) = \int_0^1 6xy \sqrt{x^2 + y^2} \, dy$

To solve this, we need to perform integration with respect to $y$, while treating $x$ as a constant. The presence of the square root suggests a substitution method.

Let’s use the substitution $u = x^2 + y^2$, so that:

$du = 2y \, dy \quad \text{or} \quad y \, dy = \frac{1}{2} du$

When $y = 0$, $u = x^2$, and when $y = 1$, $u = x^2 + 1$. Substituting these into the integral:

$I_1(x) = 6x \int_0^1 y \sqrt{x^2 + y^2} \, dy = 3x \int_{x^2}^{x^2+1} \sqrt{u} \, du$

Now, integrate $\sqrt{u}$:

$\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}$

Evaluating from $u = x^2$ to $u = x^2 + 1$:

$I_1(x) = 3x \left[ \frac{2}{3} (x^2 + 1)^{3/2} - \frac{2}{3} x^{3} \right]$

Simplifying:

$I_1(x) = 2x \left[ (x^2 + 1)^{3/2} - x^3 \right]$

Step 2: Compute the outer integral (over $x$)

Now, we integrate this expression with respect to $x$ from 0 to 1:

$I_2 = \int_0^1 2x \left[ (x^2 + 1)^{3/2} - x^3 \right] \, dx$

We split the integral into two parts:

$I_2 = 2 \int_0^1 x (x^2 + 1)^{3/2} \, dx - 2 \int_0^1 x^4 \, dx$

Step 3: Solve the first integral

For the first part, let’s use the substitution $v = x^2 + 1$, so $dv = 2x \, dx$, and when $x = 0$, $v = 1$, and when $x = 1$, $v = 2$.

Thus:

$\int_0^1 x (x^2 + 1)^{3/2} \, dx = \frac{1}{2} \int_1^2 v^{3/2} \, dv$

The integral of $v^{3/2}$ is:

$\int v^{3/2} \, dv = \frac{2}{5} v^{5/2}$

Evaluating from 1 to 2:

$\frac{1}{2} \left[ \frac{2}{5} (2^{5/2} - 1^{5/2}) \right] = \frac{1}{5} \left( 2^{5/2} - 1 \right)$

Step 4: Solve the second integral

The second part is simpler:

$\int_0^1 x^4 \, dx = \frac{1}{5}$

Thus:

$-2 \int_0^1 x^4 \, dx = -\frac{2}{5}$

Step 5: Combine the results

Putting everything together, we have:

$I_2 = 2 \times \frac{1}{5} \left( 2^{5/2} - 1 \right) - \frac{2}{5}$

Simplifying:

$I_2 = \frac{2}{5} \left( 2^{5/2} - 1 - 1 \right) = \frac{2}{5} \left( 2^{5/2} - 2 \right)$

This is the value of the iterated integral.

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Would you like any further details or explanations on any of the steps? Here are some follow-up questions:

1. Can you provide a general strategy for solving iterated integrals?
2. How do you recognize when a substitution is appropriate in integrals?
3. What are the common techniques for solving integrals involving square roots?
4. How does the order of integration affect the result of a double integral?
5. What are the advantages of using numerical methods for complex integrals?

Tip: Always double-check the limits of integration and make sure any substitutions match the correct bounds.