To find the volume of the solid generated when $R$ is revolved about the $x$-axis, we use the disk method, given by:

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

Step 1: Identify the Given Information

• The function describing the upper boundary is: $y = \frac{7}{\sqrt[4]{1 - x^2}}$
• The region is bounded by:
  • $y = 0$ (x-axis),
  • $x = 0$,
  • $x = \frac{1}{2}$.

Thus, the volume formula becomes:

$V = \pi \int_0^{1/2} \left( \frac{7}{\sqrt[4]{1 - x^2}} \right)^2 dx$

Step 2: Simplify the Integral

Squaring the function:

$\left( \frac{7}{\sqrt[4]{1 - x^2}} \right)^2 = \frac{49}{\sqrt{1 - x^2}}$

So the volume integral is:

$V = \pi \int_0^{1/2} \frac{49}{\sqrt{1 - x^2}} dx$

Step 3: Evaluate the Integral

The integral:

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

Applying this to our problem:

$V = 49\pi \int_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx$

$V = 49\pi \left[ \arcsin x \right]_0^{1/2}$

$V = 49\pi \left( \arcsin \frac{1}{2} - \arcsin 0 \right)$

Using known values:

$\arcsin \frac{1}{2} = \frac{\pi}{6}, \quad \arcsin 0 = 0$

$V = 49\pi \times \frac{\pi}{6}$

$V = \frac{49\pi^2}{6}$

Step 4: Conclusion

The exact volume of the solid is:

$\frac{49\pi^2}{6} \text{ cubic units}$

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Do you want further clarification or details?

Related Questions:

1. How would the volume change if the region were revolved about the $y$-axis instead?
2. What if the limits of integration changed to $x = 0$ to $x = 1$?
3. How do we compute the volume using the shell method instead?
4. What is the geometric interpretation of the disk method in this problem?
5. How does the function $y = \frac{7}{\sqrt[4]{1 - x^2}}$ behave as $x \to 1$?

Tip:

The disk method is useful when rotating around the x-axis, while the shell method is more useful when rotating around the y-axis. Always check which method simplifies the integral.