A volume is described as follows:
1.  the base is the region bounded by y=3-3x^2 and y=0
2.  every cross section parallel to the x-axis is a triangle whose height and base are equal.

Find the volume of this object.

To find the volume of the given object, we'll follow these steps:

### Step 1: Understand the base
The base of the solid is bounded by the curves:
- \( y = 3 - 3x^2 \) (a downward-opening parabola)
- \( y = 0 \) (the x-axis)

We first determine the limits of integration by finding where the curves intersect, which means solving for \( x \) when \( y = 0 \).

\[
3 - 3x^2 = 0
\]
\[
x^2 = 1
\]
\[
x = \pm 1
\]

So, the base is bounded by \( x = -1 \) and \( x = 1 \).

### Step 2: Describe the cross-sectional area
We are told that every cross-section parallel to the x-axis is a triangle whose height and base are equal. The length of the base of the triangle at any given \( x \) is the value of \( y \) at that point, which is \( 3 - 3x^2 \).

Since the height of the triangle equals its base, both the base and height of the triangle at each point \( x \) are \( 3 - 3x^2 \).

The area of a triangle is given by:

\[
A(x) = \frac{1}{2} \times \text{base} \times \text{height}
\]
\[
A(x) = \frac{1}{2} \times (3 - 3x^2) \times (3 - 3x^2)
\]
\[
A(x) = \frac{1}{2} \times (3 - 3x^2)^2
\]

### Step 3: Set up the integral
To find the volume of the solid, we integrate the cross-sectional area \( A(x) \) from \( x = -1 \) to \( x = 1 \):

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

### Step 4: Expand the integrand
First, expand \( (3 - 3x^2)^2 \):

\[
(3 - 3x^2)^2 = 9 - 18x^2 + 9x^4
\]

Now substitute this into the integral:

\[
V = \frac{1}{2} \int_{-1}^{1} (9 - 18x^2 + 9x^4) \, dx
\]

### Step 5: Evaluate the integral
Now, we can break this up into separate integrals:

\[
V = \frac{1}{2} \left[ \int_{-1}^{1} 9 \, dx - 18 \int_{-1}^{1} x^2 \, dx + 9 \int_{-1}^{1} x^4 \, dx \right]
\]

Evaluate each term:
- \( \int_{-1}^{1} 9 \, dx = 9 \times (1 - (-1)) = 18 \)
- \( \int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{2}{3} \)
- \( \int_{-1}^{1} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{-1}^{1} = \frac{2}{5} \)

Now, substitute these values back:

\[
V = \frac{1}{2} \left[ 18 - 18 \times \frac{2}{3} + 9 \times \frac{2}{5} \right]
\]
\[
V = \frac{1}{2} \left[ 18 - 12 + \frac{18}{5} \right]
\]
\[
V = \frac{1}{2} \left[ 6 + \frac{18}{5} \right]
\]
\[
V = \frac{1}{2} \times \frac{48}{5} = \frac{24}{5}
\]

### Final Answer:
The volume of the solid is \( \frac{24}{5} \) cubic units.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. What would change if the cross-sections were squares instead of triangles?
2. How do you find the volume if the height and base of the triangle had different ratios?
3. Can the method used here be applied to other solids with different base shapes?
4. How does changing the equation of the parabola (e.g., its coefficients) affect the volume?
5. What is the significance of symmetry in simplifying the integration process?

**Tip:** Symmetry in integrals can simplify calculations, as it often allows you to compute integrals over half the region and then double the result.

A volume is described as follows:
1. the base is the region bounded by y=3-3x^2 and y=0
2. every cross section parallel to the x-axis is a triangle whose height and base are equal.

Find the volume of this object.

This problem involves calculating the volume of a solid with a parabolic base bounded by y = 3 - 3x^2 and cross sections parallel to the x-axis that are triangles. Learn how to use integration and geometric principles to solve this step-by-step.

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