To compute the double integral $\iint_R f(x, y) \, dA$ using Fubini's theorem, we will evaluate the integral as an iterated integral. The function given is:

$f(x, y) = 12x \cdot 12y = 144xy,$ and the region $R = [1, 2] \times [1, 4]$.

Step 1: Setting up the iterated integral

Using Fubini's theorem, the double integral can be written as: $\iint_R f(x, y) \, dA = \int_1^2 \int_1^4 144xy \, dy \, dx.$

Step 2: Evaluating the inner integral

The inner integral is: $\int_1^4 144xy \, dy.$ Since $x$ is treated as a constant, this becomes: $144x \int_1^4 y \, dy.$ The integral of $y$ with respect to $y$ is: $\int y \, dy = \frac{y^2}{2}.$ So, the inner integral evaluates to: $144x \left[ \frac{y^2}{2} \right]_1^4 = 144x \left( \frac{4^2}{2} - \frac{1^2}{2} \right) = 144x \left( \frac{16}{2} - \frac{1}{2} \right) = 144x \cdot \frac{15}{2} = 1080x.$

Step 3: Evaluating the outer integral

The outer integral becomes: $\int_1^2 1080x \, dx.$ The integral of $x$ with respect to $x$ is: $\int x \, dx = \frac{x^2}{2}.$ So, the outer integral evaluates to: $1080 \int_1^2 x \, dx = 1080 \left[ \frac{x^2}{2} \right]_1^2 = 1080 \cdot \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = 1080 \cdot \left( \frac{4}{2} - \frac{1}{2} \right) = 1080 \cdot \frac{3}{2} = 1620.$

Final Answer:

The value of the double integral is: $\boxed{1620}.$

Would you like additional details or clarification?

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1. What is Fubini's theorem, and when can it be applied?
2. How do you determine the limits of integration for a given region $R$?
3. Why is it valid to separate the integral into iterated integrals?
4. What are other methods for evaluating double integrals besides Fubini's theorem?
5. Can the order of integration change the result? If so, how?

Tip: Always check the integrand and limits to ensure that the function is continuous over the given region before applying Fubini's theorem.