We are given the region $R$ bounded by the following curves:

• $y = \frac{5}{\sqrt[4]{1 - x^2}}$,
• $y = 0$ (the x-axis),
• $x = 0$, and
• $x = \frac{1}{2}$.

We are asked to find the volume of the solid generated when this region is revolved around the x-axis. To do this, we can use the disk method.

Disk Method Formula:

The formula for the volume of a solid of revolution when revolving around the x-axis is:

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

Here:

• $f(x) = \frac{5}{\sqrt[4]{1 - x^2}}$,
• The bounds of integration are from $x = 0$ to $x = \frac{1}{2}$.

Thus, the volume is:

[ V = \pi \int_{0}^{\frac{1}{2}} \left[ \frac{5}{\sqrt[4]{1 - x^2}} \right]^2 , dx ]

Simplifying the integrand:

$\left( \frac{5}{\sqrt[4]{1 - x^2}} \right)^2 = \frac{25}{\sqrt{1 - x^2}}$

So the volume integral becomes:

$V = \pi \int_{0}^{\frac{1}{2}} \frac{25}{\sqrt{1 - x^2}} \, dx$

Now, we will solve this integral.

Step 1: Solve the integral

This is a standard integral:

$\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin(x)$

Therefore, the integral becomes:

$V = 25\pi \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^2}} \, dx = 25\pi \left[ \arcsin(x) \right]_0^{\frac{1}{2}}$

Step 2: Evaluate the integral at the bounds

$V = 25\pi \left[ \arcsin\left( \frac{1}{2} \right) - \arcsin(0) \right]$

We know that:

• $\arcsin\left( \frac{1}{2} \right) = \frac{\pi}{6}$,
• $\arcsin(0) = 0$.

Thus, the volume is:

$V = 25\pi \left( \frac{\pi}{6} - 0 \right) = \frac{25\pi^2}{6}$

Final Answer:

The volume of the solid is:

$V = \frac{25\pi^2}{6}$

Would you like further details, or do you have any questions?

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Here are five related questions to expand your understanding:

1. How does the disk method differ from the shell method in finding volumes of solids of revolution?
2. What is the geometric interpretation of the integrand in the disk method?
3. How would the volume change if the region were revolved around the y-axis instead?
4. Can you apply this technique to other regions bounded by different curves?
5. What would be the result if the upper limit of integration was extended to $x = 1$?

Tip: Always check the symmetry of the region when revolving around an axis to simplify computations!