Problem Breakdown and Solution:

The problem involves calculating the volume of a solid under the surface $z = 4(x^2 + y^2)$ and above the rectangle $R$ in the $xy$-plane. The rectangle $R$ is defined by $1 \leq x \leq 3$ and $1 \leq y \leq 3$.

(a) Set up an iterated integral:

The volume of the solid is given by the double integral: $\int_R z \, dA = \int_R 4(x^2 + y^2) \, dx\,dy.$

The rectangle $R$ defines the bounds:

• For $x$: $1 \leq x \leq 3$,
• For $y$: $1 \leq y \leq 3$.

Thus, the iterated integral can be written as: $\int_{1}^{3} \int_{1}^{3} 4(x^2 + y^2) \, dx\,dy.$

(b) Evaluate the iterated integral:

The integral can be broken into two parts: $\int_{1}^{3} \int_{1}^{3} 4(x^2 + y^2) \, dx\,dy = \int_{1}^{3} \int_{1}^{3} 4x^2 \, dx\,dy + \int_{1}^{3} \int_{1}^{3} 4y^2 \, dx\,dy.$

Step 1: Evaluate $\int_{1}^{3} 4x^2 \, dx$: $\int_{1}^{3} 4x^2 \, dx = 4 \int_{1}^{3} x^2 \, dx = 4 \left[ \frac{x^3}{3} \right]_{1}^{3} = 4 \left( \frac{27}{3} - \frac{1}{3} \right) = 4 \cdot \frac{26}{3} = \frac{104}{3}.$

Step 2: Evaluate $\int_{1}^{3} 4y^2 \, dy$: This is symmetric with the $x^2$ case, so: $\int_{1}^{3} 4y^2 \, dy = \frac{104}{3}.$

Step 3: Combine results: Since the integrals over $x$ and $y$ are independent: $\int_{1}^{3} \int_{1}^{3} 4(x^2 + y^2) \, dx\,dy = \frac{104}{3} + \frac{104}{3} = \frac{208}{3}.$

Thus, the volume of the solid is: $\boxed{\frac{208}{3}}.$

Do you have any questions or want further details?

Related Questions:

1. How do you calculate the volume of a solid under a different surface, such as $z = x^2y$?
2. Can the volume integral be evaluated using polar coordinates?
3. What is the difference between iterated integrals and triple integrals in volume computation?
4. How would the integral change if the region $R$ were circular instead of rectangular?
5. What are the implications of symmetry in simplifying double integrals?

Tip:

When dealing with symmetric regions or functions, check if you can simplify calculations by splitting integrals or exploiting symmetry.