Show that ∫ ∫ 𝑥2−𝑦2𝑥2+𝑦2𝑦𝑦24𝑎4𝑎0 𝑑𝑥𝑑𝑦 = 8𝑎2 (π2 − 53)

The integral you've given is a double integral in polar coordinates. The function to be integrated is a function of r (the radial coordinate) and θ (the angular coordinate). The limits of integration are also given in polar coordinates.

Here are the steps to solve the integral:

1. Convert the function to polar coordinates. The function is x² - y², which in polar coordinates becomes r²cos²θ - r²sin²θ = r²(cos²θ - sin²θ).

2. The double integral becomes ∫ (from 0 to 2π) ∫ (from 0 to 2a) r²(cos²θ - sin²θ) r dr dθ. The extra factor of r comes from the Jacobian determinant when changing to polar coordinates.

3. The inner integral can be solved by using the power rule for integration, ∫x^n dx = (1/n+1)x^(n+1). This gives ∫ (from 0 to 2a) r³ dr = [1/4]r⁴ (from 0 to 2a) = 4a⁴.

4. The outer integral can be solved by using the identities ∫ (from 0 to 2π) cos²θ dθ = π and ∫ (from 0 to 2π) sin²θ dθ = π. This gives ∫ (from 0 to 2π) (cos²θ - sin²θ) dθ = π - π = 0.

5. Therefore, the double integral is 0 * 4a⁴ = 0.

It seems there might be a mistake in the problem statement, as the result I obtained is 0, not 8a²(π² - 5/3). Please check the problem statement.