To find the volume of the solid obtained by revolving the region bounded by the curve $y = \frac{8}{1 - \sin x}$ on the interval $[0, \frac{\pi}{4}]$ about the x-axis, we can use the disk method.

Step 1: Set up the integral for the volume

The formula for the volume $V$ when revolving around the x-axis using the disk method is:

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

Here:

• $f(x) = \frac{8}{1 - \sin x}$
• The bounds are $a = 0$ and $b = \frac{\pi}{4}$.

So, the volume is given by:

$V = \pi \int_{0}^{\frac{\pi}{4}} \left(\frac{8}{1 - \sin x}\right)^2 \, dx$

Step 2: Compute the integral

Let's compute this integral:

$V = \pi \int_{0}^{\frac{\pi}{4}} \frac{64}{(1 - \sin x)^2} \, dx$

This integral can be challenging to compute analytically, so it's typically evaluated using numerical methods. I'll compute the value numerically.

The volume of the solid obtained by revolving the region bounded by the curve $y = \frac{8}{1 - \sin x}$ on the interval $[0, \frac{\pi}{4}]$ about the x-axis is approximately $580.188$ cubic units, rounded to three decimal places.

Do you need further details or have any questions?

Here are some related questions you might consider:

1. What is the volume when the region is revolved around the y-axis instead?
2. How does the volume change if the function is $y = \frac{8}{1 + \sin x}$?
3. What if the interval is extended to $[0, \frac{\pi}{2}]$?
4. How would the method change if using the shell method instead of the disk method?
5. How can this integral be solved analytically if possible?

Tip: When dealing with volumes of revolution, ensure you correctly set up the bounds and the function being squared, as these are common sources of errors.