To find the volume $V$ of the solid $S$, we use the cross-sectional method.

Step 1: Identify the base region

The base of the solid is the region enclosed by the parabola: $y = 3 - 3x^2$ and the x-axis ($y = 0$). To find the limits of integration, solve for $x$ when $y = 0$:

$3 - 3x^2 = 0$

$x^2 = 1$

$x = \pm 1$

Thus, the base extends from $x = -1$ to $x = 1$.

Step 2: Identify the Cross-Section Shape

Each cross-section perpendicular to the $x$-axis is an isosceles triangle with height equal to its base.

From the given equation, the base of the triangle at a given $x$ is:

$\text{Base} = 2y = 2(3 - 3x^2) = 6 - 6x^2$

Since the height of the triangle is equal to the base:

$\text{Height} = 6 - 6x^2$

Step 3: Find the Area of a Cross-Section

The area $A(x)$ of an isosceles triangle with height equal to its base is:

$A = \frac{1}{2} (\text{Base}) (\text{Height})$

Substituting the values:

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

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

Step 4: Integrate to Find the Volume

The volume is given by:

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

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

Expanding the squared term:

$(6 - 6x^2)^2 = 36 - 72x^2 + 36x^4$

Thus,

$A(x) = \frac{1}{2} (36 - 72x^2 + 36x^4)$

$A(x) = 18 - 36x^2 + 18x^4$

Now, integrate:

$V = \int_{-1}^{1} (18 - 36x^2 + 18x^4) \,dx$

Since the integrand is symmetric about the $y$-axis, we can compute from 0 to 1 and double the result:

$V = 2 \int_0^1 (18 - 36x^2 + 18x^4) \,dx$

Computing each term separately:

$\int_0^1 18 \,dx = 18x \Big|_0^1 = 18(1 - 0) = 18$

$\int_0^1 36x^2 \,dx = 36 \cdot \frac{x^3}{3} \Big|_0^1 = 36 \cdot \frac{1}{3} = 12$

$\int_0^1 18x^4 \,dx = 18 \cdot \frac{x^5}{5} \Big|_0^1 = 18 \cdot \frac{1}{5} = \frac{18}{5}$

$V = 2 \left( 18 - 12 + \frac{18}{5} \right)$

$V = 2 \left( 6 + \frac{18}{5} \right)$

$V = 2 \left( \frac{30}{5} + \frac{18}{5} \right) = 2 \times \frac{48}{5} = \frac{96}{5}$

Final Answer:

$V = \frac{96}{5} \text{ cubic units}$

Would you like a more detailed explanation of any step?

Related Questions:

1. How would the volume change if the cross-sections were equilateral triangles instead?
2. How does the integral setup change if the cross-sections were semicircles instead of triangles?
3. What is the effect of modifying the parabola’s equation to $y = 4 - 4x^2$?
4. How do we find the centroid of this solid?
5. What happens if the cross-sections are rectangles with a fixed height instead?

Tip:

For solids with cross-sections, always determine the shape and dimensions of each cross-section before integrating!