We are given a solid whose base is the region bounded by $y = 9 - x^2$ and $y = 0$, and every cross-section parallel to the $x$-axis is a right triangle with equal height and base, where the base of the right triangle lies along the base of the region.

Step 1: Determine the bounds of the base region.

The base of the region is bounded by the parabola $y = 9 - x^2$ and the $x$-axis ($y = 0$).

The region of interest spans from where the parabola intersects the $x$-axis. Setting $y = 0$ in the equation $9 - x^2 = 0$ gives:

$x^2 = 9 \implies x = \pm 3$

So, the region is bounded by $x = -3$ and $x = 3$.

Step 2: Cross-sectional area of the right triangle.

At each $x$, the cross-section is a right triangle. The base of this triangle is the distance between the $x$-axis and the curve $y = 9 - x^2$, which is $9 - x^2$. Since the height and base of the triangle are equal, the height is also $9 - x^2$.

The area $A(x)$ of a right triangle is given by:

$A(x) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (9 - x^2) \times (9 - x^2)$

$A(x) = \frac{1}{2} (9 - x^2)^2$

Step 3: Set up the integral for the volume.

The volume $V$ of the solid is the integral of the area of the cross-sections from $x = -3$ to $x = 3$:

$V = \int_{-3}^{3} A(x) \, dx = \int_{-3}^{3} \frac{1}{2} (9 - x^2)^2 \, dx$

Step 4: Simplify and evaluate the integral.

We can simplify the integral further by expanding $(9 - x^2)^2$:

$(9 - x^2)^2 = 81 - 18x^2 + x^4$

So the volume integral becomes:

$V = \frac{1}{2} \int_{-3}^{3} (81 - 18x^2 + x^4) \, dx$

Since the integrand contains only even powers of $x$, we can use symmetry to simplify the calculation. We can double the integral from $0$ to $3$:

$V = \frac{1}{2} \times 2 \int_{0}^{3} (81 - 18x^2 + x^4) \, dx$

$V = \int_{0}^{3} (81 - 18x^2 + x^4) \, dx$

Step 5: Compute the integral.

Now, we integrate each term:

$\int_{0}^{3} 81 \, dx = 81x \Big|_0^3 = 81(3) - 81(0) = 243$

$\int_{0}^{3} -18x^2 \, dx = -18 \left( \frac{x^3}{3} \right) \Big|_0^3 = -18 \left( \frac{27}{3} \right) = -18(9) = -162$

$\int_{0}^{3} x^4 \, dx = \frac{x^5}{5} \Big|_0^3 = \frac{243}{5}$

Step 6: Combine the results.

Summing up the integrals:

$V = 243 - 162 + \frac{243}{5}$

Now, simplify:

$V = 81 + \frac{243}{5} = \frac{405}{5} + \frac{243}{5} = \frac{648}{5}$

Thus, the volume of the solid is:

$\boxed{\frac{648}{5} \, \text{cubic units}}$

Would you like further details or clarification? Here are some related questions:

1. What if the cross-sections were equilateral triangles?
2. How would the volume change if the parabola were shifted vertically?
3. What if the cross-sections were semicircles instead of triangles?
4. How do we find the surface area of this object?
5. How would the bounds change if the curve were $y = x^2 - 9$?

Tip: When computing volumes using cross-sections, always verify the region and shape of the cross-section carefully.