To calculate the volume of the solid generated by revolving the region $R$ about the $x$-axis, we will use the disk method. Here's the step-by-step solution:

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Step 1: Understanding the boundaries

The region $R$ is bounded by:

1. $y = \sqrt{5x - 4}$
2. $y = 0$ (the $x$-axis)
3. $y = 4$ (horizontal line)
4. $x = 0$ (vertical line).

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Step 2: Find the bounds of $x$

To determine the bounds of $x$, note that: $y = \sqrt{5x - 4}.$

• At $y = 0$: $\sqrt{5x - 4} = 0 \implies 5x - 4 = 0 \implies x = \frac{4}{5}.$
• At $y = 4$: $\sqrt{5x - 4} = 4 \implies 5x - 4 = 16 \implies 5x = 20 \implies x = 4.$

Thus, $x$ ranges from $x = \frac{4}{5}$ to $x = 4$.

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Step 3: Set up the volume integral

The volume of the solid of revolution is given by: $V = \pi \int_{a}^{b} \big[f(x)\big]^2 \, dx,$ where $f(x) = \sqrt{5x - 4}$, $a = \frac{4}{5}$, and $b = 4$.

Substituting $f(x)$, the formula becomes: $V = \pi \int_{\frac{4}{5}}^{4} \big(\sqrt{5x - 4}\big)^2 \, dx.$

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Step 4: Simplify the integrand

$\big(\sqrt{5x - 4}\big)^2 = 5x - 4.$ Thus: $V = \pi \int_{\frac{4}{5}}^{4} (5x - 4) \, dx.$

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Step 5: Evaluate the integral

First, compute the integral of $5x - 4$: $\int (5x - 4) \, dx = \frac{5x^2}{2} - 4x + C.$

Now, evaluate the definite integral from $x = \frac{4}{5}$ to $x = 4$: $V = \pi \left[\frac{5x^2}{2} - 4x\right]_{\frac{4}{5}}^{4}.$

At $x = 4$: $\frac{5(4)^2}{2} - 4(4) = \frac{80}{2} - 16 = 40 - 16 = 24.$

At $x = \frac{4}{5}$: $\frac{5\left(\frac{4}{5}\right)^2}{2} - 4\left(\frac{4}{5}\right) = \frac{5 \cdot \frac{16}{25}}{2} - \frac{16}{5} = \frac{\frac{80}{25}}{2} - \frac{80}{25} = \frac{40}{25} - \frac{80}{25} = -\frac{40}{25} = -\frac{8}{5}.$

Now subtract: [ V = \pi \left[ 24 - \left(-\frac{8}{5}\right) \right] = \pi \left[ 24 + \frac{8}{5} \right]. ]

Simplify: $V = \pi \left[ \frac{120}{5} + \frac{8}{5} \right] = \pi \cdot \frac{128}{5}.$

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Final Answer:

$V = \frac{128\pi}{5} \, \text{cubic units}.$

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Do you want further clarification or a breakdown of any step? 😊

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Related Questions:

1. How is the disk method different from the shell method?
2. What is the geometric significance of the boundaries in this problem?
3. How would the volume change if the region was revolved about the $y$-axis?
4. Can this volume be computed using cylindrical shells? How?
5. What happens if the upper boundary is replaced with $y = x^2$?

Tip:

Always visualize the region and axis of revolution to avoid mistakes in setting up the integral!