Region R bounded by y =
√
x, y = 0, x = 5, rotated about the
x-axis

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:

\[
V = \pi \left[ \frac{x^2}{2} \right]_{0}^{5}
= \pi \left( \frac{5^2}{2} - \frac{0^2}{2} \right)
= \pi \left( \frac{25}{2} \right)
\]

### 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.

Region R bounded by y = √x, y = 0, x = 5, rotated about the x-axis.

This detailed solution explains how to find the volume of a solid generated by rotating the region bounded by y = √x, y = 0, and x = 5 about the x-axis. Learn how to apply the disk method and definite integrals step-by-step, with formulas and solutions for advanced calculus students.

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