To find the volume $V$ of the solid $S$, we will follow these steps:

Step 1: Identify the Base Region

The base of $S$ is the region enclosed by the parabola and the x-axis, which means we need to determine the limits for $x$.

The given parabola equation is: $y = 6 - 6x^2$ To find the points where it intersects the x-axis, set $y = 0$:

$6 - 6x^2 = 0$

$x^2 = 1$

$x = \pm1$

So, the base region is for $x$ in the interval $[-1,1]$.

Step 2: Cross-section Shape and Area

The cross-sections perpendicular to the x-axis are isosceles triangles, where the height equals the base.

• The base of each triangle at a given $x$ is given by $y = 6 - 6x^2$.
• Since the height equals the base, the height is also $y = 6 - 6x^2$.

The area $A(x)$ of an isosceles triangle with base $b$ and height $h$ is:

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

Substituting $b = h = 6 - 6x^2$:

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

Step 3: Compute the Volume Integral

The volume is found by integrating the cross-sectional area along the x-axis:

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

Substituting $A(x)$:

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

Expanding $(6 - 6x^2)^2$:

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

So,

$V = \int_{-1}^{1} \frac{1}{2} (36 - 72x^2 + 36x^4) \, dx$

$V = \frac{1}{2} \int_{-1}^{1} (36 - 72x^2 + 36x^4) \, dx$

Step 4: Evaluate the Integral

We integrate each term separately:

$\int_{-1}^{1} 36 \, dx = 36(x) \Big|_{-1}^{1} = 36(1 - (-1)) = 36(2) = 72$

$\int_{-1}^{1} 72x^2 \, dx = 72 \int_{-1}^{1} x^2 \, dx$

Since $x^2$ is an even function,

$\int_{-a}^{a} x^2 \, dx = \frac{2a^3}{3}$

So,

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

$\int_{-1}^{1} 72x^2 \, dx = 72 \times \frac{2}{3} = 48$

$\int_{-1}^{1} 36x^4 \, dx = 36 \int_{-1}^{1} x^4 \, dx$

Since $x^4$ is also even,

$\int_{-a}^{a} x^4 \, dx = \frac{2a^5}{5}$

So,

$\int_{-1}^{1} x^4 \, dx = \frac{2(1)^5}{5} = \frac{2}{5}$

$\int_{-1}^{1} 36x^4 \, dx = 36 \times \frac{2}{5} = \frac{72}{5}$

Step 5: Compute the Final Volume

$V = \frac{1}{2} (72 - 48 + \frac{72}{5})$

$= \frac{1}{2} \left( \frac{360}{5} - \frac{240}{5} + \frac{72}{5} \right)$

$= \frac{1}{2} \times \frac{192}{5}$

$= \frac{192}{10} = 19.2$

Thus, the volume of the solid is:

$\boxed{19.2}$

Would you like a further breakdown or visualization of this process?

Related Questions

1. How would the volume change if the cross-sections were equilateral triangles instead?
2. What if the base was a different parabola, such as $y = 4 - x^2$?
3. How can we generalize this approach for other types of cross-sections?
4. What happens if the solid extends from $x = -a$ to $x = a$ instead of $\pm1$?
5. How do you compute volume if the cross-sections were semicircles instead of triangles?

Tip: When dealing with cross-sections, always express the base in terms of $x$ before setting up the integral.