The problem asks to evaluate the double integral:

$\iint_R x y \, dx \, dy$

where $R$ is the region bounded by $x^2 = 4y$, $x = 2a$, and $y = 0$.

Step-by-step Solution:

1. Understand the region $R$:

• The parabola $x^2 = 4y$ opens upwards, with the vertex at the origin $(0, 0)$.
• $y = 0$ corresponds to the x-axis.
• The vertical line $x = 2a$ cuts through the parabola and defines the right boundary of the region.

The region $R$ is enclosed between $x^2 = 4y$, $x = 2a$, and $y = 0$.

2. Set up the bounds for the double integral:

• The parabola equation $x^2 = 4y$ can be rewritten as $y = \frac{x^2}{4}$.
• For a fixed $x$, $y$ varies from the x-axis ($y = 0$) to the parabola ($y = \frac{x^2}{4}$).
• $x$ varies from $0$ (the vertex) to $2a$ (right boundary).

Thus, the bounds are: $\int_0^{2a} \int_0^{\frac{x^2}{4}} x y \, dy \, dx$

3. Compute the inner integral:

For the inner integral: $\int_0^{\frac{x^2}{4}} x y \, dy$ Here, $x$ is treated as a constant. The integral becomes: $x \int_0^{\frac{x^2}{4}} y \, dy = x \left[\frac{y^2}{2}\right]_0^{\frac{x^2}{4}}$ Substitute the limits of $y$: $x \left[\frac{\left(\frac{x^2}{4}\right)^2}{2} - \frac{0^2}{2}\right] = x \cdot \frac{\left(\frac{x^4}{16}\right)}{2} = x \cdot \frac{x^4}{32} = \frac{x^5}{32}$

4. Compute the outer integral:

Now compute: $\int_0^{2a} \frac{x^5}{32} \, dx$ Factor out $\frac{1}{32}$: $\frac{1}{32} \int_0^{2a} x^5 \, dx$ The integral of $x^5$ is: $\frac{x^6}{6}$ Evaluate it from $0$ to $2a$: $\frac{1}{32} \left[\frac{(2a)^6}{6} - \frac{0^6}{6}\right] = \frac{1}{32} \cdot \frac{(2a)^6}{6}$ Simplify $(2a)^6 = 64a^6$: $\frac{1}{32} \cdot \frac{64a^6}{6} = \frac{64a^6}{192} = \frac{a^6}{3}$

Final Answer:

$\iint_R x y \, dx \, dy = \frac{a^6}{3}$

Let me know if you'd like further clarification or a detailed breakdown of any step!

Follow-up Questions:

1. How do you interpret the limits of integration in double integrals?
2. Why is $x$ treated as a constant during the inner integral?
3. What is the role of symmetry when solving integrals like this?
4. How would the result change if $x^2 = 4y$ was replaced by $y^2 = 4x$?
5. Can you generalize this approach for regions bounded by other parabolas?

Tip: Always sketch the region of integration for double integrals to ensure correct bounds.