The region $R$ is bounded by the curves $y = \sqrt{x}$, $y = 0$, and $x = 5$. When this region is rotated about the x-axis, we can use the method of disks to find the volume of the solid generated.

Step 1: Identify the boundaries

The curve is $y = \sqrt{x}$, and the boundaries are $y = 0$ (the x-axis) and $x = 5$. We will rotate the region from $x = 0$ to $x = 5$ about the x-axis.

Step 2: Set up the formula for the volume

The volume of the solid generated by rotating a region about the x-axis can be calculated using the disk method. The formula for the volume is:

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

In this case, $f(x) = \sqrt{x}$, $a = 0$, and $b = 5$.

Step 3: Set up the integral

Substitute $f(x) = \sqrt{x}$ into the formula:

$V = \pi \int_{0}^{5} (\sqrt{x})^2 \, dx$

Since $(\sqrt{x})^2 = x$, the integral becomes:

$V = \pi \int_{0}^{5} x \, dx$

Step 4: Solve the integral

Now, integrate $x$ with respect to $x$:

$\int x \, dx = \frac{x^2}{2}$

Evaluate the integral from 0 to 5:

= \pi \left( \frac{5^2}{2} - \frac{0^2}{2} \right) = \pi \left( \frac{25}{2} \right)$$ ### Step 5: Calculate the volume Now multiply by $$\pi$$: $$V = \frac{25\pi}{2}$$ Thus, the volume of the solid is: $$V = \frac{25\pi}{2} \, \text{cubic units}$$ --- Would you like any further clarification or details on this problem? ### Related Questions: 1. How would the result change if the region were rotated about the y-axis instead? 2. Can we use the shell method to solve this problem instead of the disk method? 3. How would the volume differ if the upper bound for $$x$$ were 10 instead of 5? 4. What if the curve was $$y = x^2$$ instead of $$y = \sqrt{x}$$? 5. How do we calculate the surface area of the solid generated by rotating this region? ### Tip: When using the disk or washer method, always check whether the function should be squared before integrating for volume, as this represents the area of the circular cross-sections.