To evaluate the double integral $\iint_R f(x, y) \, dy \, dx$ over the region bounded by the x-axis, the ordinate $x = 2a$, and the curve $x^2 = 4ay$, we must first understand the region $R$ in the xy-plane.

Step 1: Analyze the boundaries

• The curve $x^2 = 4ay$ is a parabola that opens upwards with its vertex at the origin $(0, 0)$.
• The x-axis is $y = 0$.
• The vertical line is $x = 2a$.

Step 2: Express the region of integration

The region $R$ is bounded by:

• The x-axis $y = 0$
• The parabola $y = \frac{x^2}{4a}$
• The vertical line $x = 2a$

Step 3: Determine the limits of integration

For a fixed value of $x$ ranging from $0$ to $2a$, $y$ will range from the x-axis $y = 0$ to the parabola $y = \frac{x^2}{4a}$.

Thus, the integral is set up as follows: $\int_{x=0}^{2a} \int_{y=0}^{\frac{x^2}{4a}} f(x, y) \, dy \, dx$

Final Answer

The double integral over the given region is: $\iint_R f(x, y) \, dy \, dx = \int_{x=0}^{2a} \int_{y=0}^{\frac{x^2}{4a}} f(x, y) \, dy \, dx$

If you have a specific function $f(x, y)$ in mind, we can proceed to evaluate the integral by substituting it in.

Would you like to evaluate this integral for a specific function, or do you have any other questions?

Here are 5 related questions to expand on this:

1. How would the limits of integration change if the region were bounded by $y = 0$, $y = x^2/4a$, and $x = -2a$?
2. What if the function $f(x, y)$ were a simple polynomial like $f(x, y) = xy$? How would you solve the integral?
3. How would you approach the problem if the curve were $x^2 = 4ay$ but the region were bounded by $x = 3a$ instead of $x = 2a$?
4. Can this integral be evaluated using polar coordinates?
5. What is the geometric interpretation of this double integral?

Tip: Always sketch the region of integration first when solving double integrals to ensure the correct limits are chosen.