A volume is described as follows:
The base is the region bounded by  y = 9 −1^x2 and  y= 0, and every cross section parallel to the  x -axis is a right triangle whose height and base are equal, and the base of the right triangle lies along the base of the region.

Find the volume of this object.

Volume =

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.

A volume is described as follows: The 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 whose height and base are equal, and the base of the right triangle lies along the base of the region. Find the volume of this object.

This problem involves finding the volume of a solid whose base is bounded by the parabola y = 9 - x^2 and the x-axis, with cross-sections that are right triangles. Learn how to set up and evaluate the integral to find the volume using calculus techniques, suitable for college-level students.

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