The problem describes a 3D volume with a specific base and triangular cross sections. Let's break it down and solve for the volume step by step.

Step 1: Understanding the Base Region

The base of the object is bounded by the curve $y = 7 - \frac{7}{64}x^2$ and the line $y = 0$. This describes a parabolic region with a maximum height of 7 at $x = 0$ and symmetrical boundaries around the y-axis. The endpoints occur where $y = 0$, so we can find the x-boundaries by solving: $7 - \frac{7}{64}x^2 = 0 \quad \Rightarrow \quad \frac{7}{64}x^2 = 7 \quad \Rightarrow \quad x^2 = 64 \quad \Rightarrow \quad x = \pm 8$ Thus, the base is between $x = -8$ and $x = 8$.

Step 2: Understanding the Cross Sections

Each cross section parallel to the x-axis is a triangle whose base and height are equal. The base of the triangle at a specific x-coordinate is the vertical distance between $y = 0$ and the curve $y = 7 - \frac{7}{64}x^2$. Therefore, the length of the base at any point $x$ is: $\text{Base} = 7 - \frac{7}{64}x^2$ Since the height of the triangle is equal to the base, the area of each triangular cross section is given by: $A(x) = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times \left(7 - \frac{7}{64}x^2\right) \times \left(7 - \frac{7}{64}x^2\right) = \frac{1}{2} \left(7 - \frac{7}{64}x^2\right)^2$

Step 3: Setting up the Integral for Volume

To find the total volume, we integrate the area of these triangular cross sections along the x-axis from $x = -8$ to $x = 8$: $V = \int_{-8}^{8} A(x) \, dx = \int_{-8}^{8} \frac{1}{2} \left(7 - \frac{7}{64}x^2\right)^2 \, dx$

Step 4: Solving the Integral

We can now evaluate the integral to find the volume. I'll compute this next.

The volume of the object is approximately $209.07$ cubic units.

Would you like more details on any part of the solution, or do you have further questions?

Here are 5 related questions you might find interesting:

1. How would the volume change if the base of each triangular cross section was not equal to the height?
2. Can you compute the volume if the base region was bounded by a different parabola?
3. How would you find the surface area of this object instead of the volume?
4. What happens if we shift the parabola vertically or horizontally?
5. Can we generalize this method to other shapes of cross sections?

Tip: When solving for volumes using cross sections, always pay attention to the boundaries of the base and the shape of the cross sections to set up the integral correctly.