The problem asks to compute the volume of the solid bounded above by the elliptic paraboloid $z = 64 - x^2 - 3y^2$ and the square region $R = [0, 2] \times [0, 2]$ in the $xy$-plane.

Solution Approach:

To find the volume of the solid, we will set up a double integral. The volume of the solid is given by:

$V = \int\int_R z(x,y) \, dA$ where $z(x,y) = 64 - x^2 - 3y^2$ and $R = [0, 2] \times [0, 2]$.

So, the volume integral becomes: $V = \int_0^2 \int_0^2 (64 - x^2 - 3y^2) \, dx \, dy$

Step-by-step Calculation:

1. Integrate with respect to $x$:

$\int_0^2 (64 - x^2 - 3y^2) \, dx = \int_0^2 (64 - 3y^2) \, dx - \int_0^2 x^2 \, dx$

The first part: $\int_0^2 (64 - 3y^2) \, dx = (64 - 3y^2) \cdot x \Big|_0^2 = (64 - 3y^2) \cdot 2 = 128 - 6y^2$

The second part: $\int_0^2 x^2 \, dx = \frac{x^3}{3} \Big|_0^2 = \frac{8}{3}$

So, the integral with respect to $x$ is: $128 - 6y^2 - \frac{8}{3}$ Simplifying: $\frac{384}{3} - \frac{8}{3} - 6y^2 = \frac{376}{3} - 6y^2$

2. Integrate with respect to $y$:

Now, we need to integrate with respect to $y$:

$\int_0^2 \left( \frac{376}{3} - 6y^2 \right) \, dy$

First part: $\int_0^2 \frac{376}{3} \, dy = \frac{376}{3} \cdot y \Big|_0^2 = \frac{376}{3} \cdot 2 = \frac{752}{3}$

Second part: $\int_0^2 6y^2 \, dy = 6 \cdot \frac{y^3}{3} \Big|_0^2 = 6 \cdot \frac{8}{3} = 16$

So, the total volume is: $\frac{752}{3} - 16 = \frac{752}{3} - \frac{48}{3} = \frac{704}{3}$

Thus, the exact volume is: $\boxed{\frac{704}{3}}$

Would you like more details on any of the steps? Here are five related questions that might help expand your understanding:

1. How would the volume change if the region $R$ were different?
2. What happens to the volume if the equation of the paraboloid changes?
3. Can you set up the integral in polar coordinates instead of Cartesian?
4. How would the limits of integration change if the region $R$ were a different shape?
5. What if the paraboloid intersected the $xy$-plane at different points?

Tip: When solving double integrals, breaking them into smaller integrals often simplifies the problem.