We need to find the area of the region bounded by the curves $y = 3x^2$ and $y = x^2 + 32$.

Step 1: Find the intersection points

We first set the two equations equal to each other to find the $x$-coordinates of the points where the curves intersect: $3x^2 = x^2 + 32$ $2x^2 = 32$ $x^2 = 16$ $x = \pm 4$ So, the curves intersect at $x = -4$ and $x = 4$.

Step 2: Set up the integral for the area

The area between the curves can be found by integrating the difference between the upper curve $y = x^2 + 32$ and the lower curve $y = 3x^2$ from $x = -4$ to $x = 4$: $\text{Area} = \int_{-4}^{4} \left[(x^2 + 32) - 3x^2\right] \, dx$ $\text{Area} = \int_{-4}^{4} \left(-2x^2 + 32\right) \, dx$

Step 3: Compute the integral

We can now compute the integral: $\int_{-4}^{4} (-2x^2 + 32) \, dx = \int_{-4}^{4} -2x^2 \, dx + \int_{-4}^{4} 32 \, dx$ The first part of the integral, $\int_{-4}^{4} -2x^2 \, dx$, is zero because $x^2$ is an even function and its integral from $-a$ to $a$ cancels out. So, we only need to compute the second part: $\int_{-4}^{4} 32 \, dx = 32 \cdot (4 - (-4)) = 32 \cdot 8 = 256$

Step 4: Conclusion

The area of the region is $\boxed{256}$.

Would you like more details on any step, or do you have further questions?

Here are five related questions:

1. How do we find intersection points of two curves?
2. What are even and odd functions, and how do they simplify integrals?
3. Why does the integral of an even function cancel out over symmetric limits?
4. How would this change if the curves were shifted vertically or horizontally?
5. Can we use other methods to find areas between curves?

Tip: Always check if any integrals involve odd functions over symmetric limits, as they might simplify your calculations!