To find the volume of the given solid, we need to break it down step by step.

1. Base of the Region:

The base of the solid is the region bounded by the curves:

• $y = 6 - 38x^2$
• $y = 0$

This represents a parabola opening downward, and the base region extends from $y = 0$ to the curve.

2. Shape of the Cross Sections:

Each cross section of the solid parallel to the $x$-axis is a right triangle. The important points here are:

• The base of the triangle lies along the base of the region.
• The height of each triangle is equal to its base.

So, if the base of a triangle is $b(x)$, then the height is also $b(x)$.

3. Base of the Triangle:

The base of each triangle at a given value of $x$ is the vertical distance from the $x$-axis (where $y = 0$) to the curve $y = 6 - 38x^2$. This is simply $6 - 38x^2$.

So, the base and height of each triangle are both $6 - 38x^2$.

4. Area of the Cross Section:

For a right triangle, the area $A(x)$ is given by:

$A(x) = \frac{1}{2} \text{(base)} \times \text{(height)}$

Since the base and height are both $6 - 38x^2$, we have:

$A(x) = \frac{1}{2} (6 - 38x^2)^2$

5. Volume of the Solid:

To find the total volume, we integrate the area of the cross sections over the interval where the region exists. First, let's find the limits of integration.

Since $y = 6 - 38x^2$ is bounded by $y = 0$, we need to solve for the $x$-values where $y = 0$:

$6 - 38x^2 = 0$

$38x^2 = 6$

$x^2 = \frac{6}{38} = \frac{3}{19}$

$x = \pm \sqrt{\frac{3}{19}}$

Thus, the region extends from $x = -\sqrt{\frac{3}{19}}$ to $x = \sqrt{\frac{3}{19}}$.

Now, we compute the volume by integrating the area function:

$V = \int_{-\sqrt{\frac{3}{19}}}^{\sqrt{\frac{3}{19}}} \frac{1}{2} (6 - 38x^2)^2 \, dx$

Let me calculate this for you.The volume of the solid is approximately $7.63$ cubic units.

Would you like further clarification on the steps, or any more details?

Here are some related questions to deepen understanding:

1. How would the volume change if the height of each cross-section was double the base?
2. What would the integral look like if the cross-sections were equilateral triangles instead of right triangles?
3. Can you find the centroid of the base region described by $y = 6 - 38x^2$?
4. How does the shape of the parabola affect the overall volume?
5. How would the volume change if the bounds for $x$ were extended or narrowed?

Tip: Understanding how the cross-sectional shape impacts the volume is crucial for visualizing and solving more complex solids.