https://file.aiquickdraw.com/mathbot/file/1726393472658dzcz699g.jpg

The image shows a double integral expression:

\[
2a \sqrt{2a x - x} \int_0^{2a} \int_0^{\sqrt{2ax - x}} xy \, dy \, dx
\]

This is a double integral over a region bounded by the limits in both \(x\) and \(y\) with the integrand being \(xy\). 

### Steps to solve:
1. **Outer Integral**: \( \int_0^{2a} (\text{inner integral}) dx \)
2. **Inner Integral**: \( \int_0^{\sqrt{2ax - x}} xy \, dy \)

Let me calculate this for you step by step.The solution to the double integral is:

\[
\frac{4a^3(2a - 1)}{3}
\]

This represents the result of evaluating the given expression over the defined limits.

Would you like more details on the integration process or have any other questions? Here are five related questions:

1. What is the geometric interpretation of a double integral?
2. How do you apply the method of changing variables in double integrals?
3. What are the typical applications of double integrals in physics?
4. How can polar coordinates simplify certain double integrals?
5. How would this integral change if the limits of \(x\) and \(y\) were different?

**Tip:** When solving double integrals, always start by evaluating the inner integral first, then proceed to the outer integral, and ensure to pay attention to the order of integration.

Evaluate the double integral expression: 2a * sqrt(2a x - x) * integral from 0 to 2a and integral from 0 to sqrt(2a x - x) of xy dy dx.

This problem focuses on evaluating a double integral of the form 2a * sqrt(2a x - x) * integral from 0 to 2a and integral from 0 to sqrt(2a x - x) of xy dy dx. Learn how to solve step-by-step using methods of double integration and fundamental calculus.

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