To find the volume of the solid obtained by rotating the region bounded by the curves $y = 8 - 8x^2$ and $y = 0$ about the $x$-axis, we use the disk method.

The volume $V$ of the solid formed by revolving a curve $y = f(x)$ around the $x$-axis from $x = a$ to $x = b$ is given by the formula:

$V = \pi \int_{a}^{b} [f(x)]^2 \, dx$

Here, $f(x) = 8 - 8x^2$ and $y = 0$ is the boundary, so the region we're rotating lies between these two. Now let's go through the steps:

Step 1: Find the bounds $a$ and $b$

We need to find the $x$-values where the curve intersects the $x$-axis. This happens when $y = 0$, so:

$8 - 8x^2 = 0$ $8x^2 = 8 \implies x^2 = 1 \implies x = \pm 1$

Thus, the limits of integration are $x = -1$ and $x = 1$.

Step 2: Set up the integral

Using the formula for the disk method, we square $f(x)$ and set up the integral:

$V = \pi \int_{-1}^{1} (8 - 8x^2)^2 \, dx$

Step 3: Solve the integral

First, expand the square:

$(8 - 8x^2)^2 = 64 - 128x^2 + 64x^4$

Now, integrate term by term:

$V = \pi \int_{-1}^{1} \left( 64 - 128x^2 + 64x^4 \right) dx$

Step 4: Compute the integrals

Since the function is even (symmetric around the $y$-axis), we can double the integral from 0 to 1 to simplify the calculation:

$V = 2\pi \int_{0}^{1} \left( 64 - 128x^2 + 64x^4 \right) dx$

Now, calculate each integral:

$\int 64 \, dx = 64x$ $\int -128x^2 \, dx = -\frac{128x^3}{3}$ $\int 64x^4 \, dx = \frac{64x^5}{5}$

Evaluating these from 0 to 1:

$64(1) - \frac{128(1^3)}{3} + \frac{64(1^5)}{5} = 64 - \frac{128}{3} + \frac{64}{5}$

Now, simplify this expression:

$64 - \frac{128}{3} + \frac{64}{5} = \frac{960}{15} - \frac{640}{15} + \frac{192}{15} = \frac{512}{15}$

Finally, multiply by $2\pi$:

$V = 2\pi \times \frac{512}{15} = \frac{1024\pi}{15}$

Final Answer:

The volume of the solid is:

$V = \frac{1024\pi}{15} \, \text{cubic units}$

Would you like further details or clarifications?

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Here are some related questions you might explore:

1. How does the disk method differ from the washer method?
2. What happens if the region is rotated about a different axis, like the y-axis?
3. Can the volume be computed using cylindrical shells instead?
4. What is the effect of changing the equation to $y = 8 - 4x^2$?
5. How does the method work when revolving around a non-axis line?

Tip: Symmetry often simplifies integrals; use it when functions are even!